Class 6 Maths - Fractions

Represent $\dfrac{-5}{6}$ on the number line.

The number can be represented on the number line by moving in the negative direction on the number line and dividing each point in

$\dfrac{1}{6}$ and stop at

$5^{th}$ point

Convert the fraction in its simplest form.
$108 : 100$ is $\displaystyle \frac{27}{25}$
if true then enter $1$ and if false then enter $0$ .

Given ratio is $108 : 100$

Simplifying it by dividing by $4$ , we get the ratio $= 27 : 25$

In fraction form it can be written as, $\dfrac {27}{25}$

Hence given statement is true

The simplest form of the ratio $25:100$ is ____

Given ratio is $25 : 100$

Now

$25 : 100$

$=\dfrac{25\div25}{100\div25}$

$= 1 : 4$

Hence, the simplest form is $1:4$

Reduce to standard form $\displaystyle \frac{36}{-24}$

The HCF of 36 and 24 is 12
Thus its standard form would be obtained by dividing by 12

$\displaystyle \frac{36}{-24}$ =

$\displaystyle \frac{36\div (12)}{-24\div (12)}$
=

$\displaystyle \frac{-3}{2}$

Satypal walks $\displaystyle \frac{2}{3}$ km from a place P towards east and then there $\displaystyle 1\frac{5}{7}$ km towards west Where will he be now from P ?

Let us denote the distance travelled towards east by positive sign so thus distance of Satypal from the point P would be

$\displaystyle \frac{2}{3}$ +

$\displaystyle (-1\frac{5}{7})$

$= \displaystyle \frac{2}{3}$ +

$\displaystyle \frac{-12}{7}$

$= \displaystyle \frac{2\times 7}{3\times 7}$ +

$\displaystyle \frac{(-12)\times 3}{7\times 3}$

$= \displaystyle \frac{14-36}{21}$

$= \displaystyle \frac{-22}{ 21 }$

$= \displaystyle -1\frac{1}{21}$

Since it is negative it means Satypal is at a distance

$\displaystyle1 \frac{1}{21}$ km towards west of P

Verify the following
$\left( \dfrac{5}{4} + \dfrac{-1}{2} \right) + \dfrac{-3}{2} = \dfrac{5}{4} + \left( \dfrac{-1}{2} + \dfrac{-3}{2} \right)$

Given,

we have to verify

$\dfrac { 5 }{ 4 } +\dfrac { -1 }{ 2 } +\dfrac { -3 }{ 2 } =\dfrac { 5 }{ 4 } +\dfrac { -1 }{ 2 } +\dfrac { -3 }{ 2 }$

LHS

$\dfrac { 5 }{ 4 } +\dfrac { -1 }{ 2 } +\dfrac { -3 }{ 2 }$

$=\dfrac { 5+\left( -2 \right) }{ 4 } +\dfrac { -3 }{ 2 }$

$=\dfrac { 3 }{ 4 } +\dfrac { -3 }{ 2 }$

$=\dfrac { 3-6 }{ 4 }$

$=\dfrac { -3 }{ 4 }$

RHS

$\dfrac { 5 }{ 4 } +\left( \dfrac { -1 }{ 2 } +\dfrac { -3 }{ 2 } \right)$

$=\dfrac { 5 }{ 4 } +\dfrac { -1-3 }{ 2 }$

$=\dfrac { 5 }{ 4 } +\dfrac { -4 }{ 2 }$

$=\dfrac { 5-8 }{ 4 }$

$=\dfrac { -3 }{ 4 }$

hence proved LHS=RHS

$\dfrac { 5 }{ 4 } +\dfrac { -1 }{ 2 } +\dfrac { -3 }{ 2 } =\dfrac { 5 }{ 4 } +\dfrac { -1 }{ 2 } +\dfrac { -3 }{ 2 }$

Represent $\dfrac {5}{3}$ and $-\dfrac {5}{3}$ on the number line

Draw an integer line representing

$-2, -1, 0, 1, 2$ .
Divide each unit into three equal parts to the right and left sides of zero respectively. Take five parts out of these. The fifth point on the right of zero represents

$\dfrac {5}{3}$ and the fifth one to the left of zero represents

$\dfrac {-5}{3}$ .
1058525_569239_ans_9c496cdfb14a403ab62b79185b62d394.jpg

Simplify $\dfrac{2}{5} + \dfrac{3}{7} + \left(\dfrac{-6}{5} \right ) + \left( \dfrac{-13}{7}\right )$

Rearrange the given fractions keeping similar fractions together.

$\dfrac{2}{5} + \dfrac{3}{7} + \left( \dfrac{-6}{5} \right) + \left( \dfrac{-13}{7} \right ) = \dfrac{2}{5} + \dfrac{3}{7} - \dfrac{6}{5} - \dfrac{13}{7}$

$= \left( \dfrac{2}{5} - \dfrac{6}{5} \right ) + \left( \dfrac{3}{7} - \dfrac{13}{7} \right)$ (by commutative law of addition)

$= \dfrac{2 - 6}{5} + \dfrac{3 - 13}{7}$

$= \dfrac{-4}{5} + \dfrac{-10}{7} = \dfrac{-4}{5} - \dfrac{10}{7}$

$= \dfrac{-4 \times 7 - 10 \times 5}{35} = \dfrac{-28 - 50}{35} = \dfrac{-78}{35}$

$\dfrac{9}{7}$

1205275_564896_ans_96df93cc78d44d2aaf4633e0339e601e.png

If $\displaystyle\frac{x}{b+c-a}=\displaystyle\frac{y}{c+a-b}=\frac{x}{a+b-c}$ , then rove that $\displaystyle\frac{x+y+z}{a+b+c}=\displaystyle\frac{x(y+z)+y(z+x)+z(x+y)}{2(ax+by+cz)}$ .

$\dfrac{x}{b+c-a}=\dfrac{y}{c+a-b}=\dfrac{z}{b+a-c}$ .......

$(i)$

$\dfrac{\text{sum of numerators}}{\text{sum of denominator}}=\dfrac{x+y+z}{b+c-a+c+a-b+b+a-c}$

$=\dfrac{x+y+z}{a+b+c}$ ........

$(ii)$

Multiply both numerator and denominator of

$(i)$ by

$y+z,z+x$ and

$x+y$ respectively, we get

$\dfrac{x(y+z)}{(y+z)(b+c-a)}=\dfrac{y(z+x)}{(z+x)(c+a-b)}=\dfrac{z(x+y)}{(x+y)(a+b-c)}$

$\dfrac{\text{sum of numerators}}{\text{sum of denominator}}$

$=\dfrac{x(y+z)+y(z+x)+z(x+y)}{(y+z)(b+c-a)+(z+x)(c+a-b)+(x+y)(a+b-c)}$

$=\dfrac{x(y+z)+y(z+x)+z(x+y)}{2ax+2by+2cz}$ .....................(2)

As per (1) and (2) we get

$\dfrac{x+y+z}{a+b+c}=\dfrac{x(y+z)+y(z+x)+z(x+y)}{2(ax+by+cz)}$

Find the factor which will rationalise:
$\sqrt [ 3 ]{ 3 } -\sqrt { 2 }$

We need to find the factor which when multiplied by

$(\sqrt[3]{3} - \sqrt{2})$ will lead to a rational number

Let

$x = 3^{\frac{1}{3}}$ and

$y = 2^{\frac{1}{2}}$ , then

$x^6$ and

$y^6$ are both rational

Also,

$x^6 - y^6 = (x-y)(x^5+x^4y+x^3y^2+x^2y^3+xy^4+y^5)$

Hence,

$x^5+x^4y+x^3y^2+x^2y^3+xy^4+y^5$ is the required factor.

Substituting the value for

$x$ and

$y$ , then the required factor is

$3^{\frac{5}{3}}+3^{\frac{4}{3}}\cdot 2^{\frac{1}{2}}+3^{\frac{3}{3}}\cdot 2^{\frac{2}{2}}+3^{\frac{2}{3}}\cdot 2^{\frac{3}{2}}+3^{\frac{1}{3}}\cdot 2^{\frac{4}{2}}+2^{\frac{5}{2}}$

$\Rightarrow 3^{\frac{5}{3}}+3^{\frac{4}{3}}\cdot 2^{\frac{1}{2}}+6+3^{\frac{2}{3}}\cdot 2^{\frac{3}{2}}+4\cdot 3^{\frac{1}{3}}+2^{\frac{5}{2}}$

Find the factor which will rationalise:
$\sqrt [ 3 ]{ 3 } -1$

We need to find the factor which when multiplied by

$(\sqrt[3]{3} - 1)$ will lead to a rational number

Let

$x = 3^{\frac{1}{3}}$ and

$y = 1$ , then

$x^3$ and

$y^3$ are both rational

Also,

$x^3 - y^3 = (x-y)(x^2+xy+y^2)$

Hence,

$x^2+xy+y^2$ is the required factor.

Substituting the value for

$x$ and

$y$ , then the required factor is

$3^{\frac{2}{3}}+3^{\frac{1}{3}}+1$

Find two proper fractions in their lowest terms having $12$ and $8$ for their denominators and such that their difference is $\dfrac{1}{24}$ .

Let the two fractions be

$\dfrac{x}{12}$ and

$\dfrac{y}{8}$ .

Then,

$\dfrac{x}{8} - \dfrac{y}{12} = \dfrac{1}{24} \ \text{or} \ \dfrac{x}{12} - \dfrac{y}{8} = \dfrac{1}{24}$

$\Rightarrow 3x - 2y = 1\ \text{or} \ 2y-3x = 1$

Since we want the lowest fractions

$x$ and

$y$ are odd integers and

$x \lt 12 , \, y \lt 8$

The only ordered pair

$(x,y)$ that satisfies the above equation is

$(1,1); (5,7) \ \text{or} \ (3,5); (7,11)$

Hence, the two fractions are

$\dfrac{1}{8}, \dfrac{1}{12}; \dfrac{5}{8}, \dfrac{7}{12}$ and

$\dfrac{3}{8}, \dfrac{5}{12}; \dfrac{7}{8}, \dfrac{11}{12}$

Find the factor which will rationalise:
$\sqrt [ 6 ]{ 5 } +\sqrt [ 3 ]{ 2 }$

We need to find the factor which when multiplied by

$(\sqrt[6]{5} + \sqrt[3]{2})$ will lead to a rational number

Let

$x = 5^{\frac{1}{6}}$ and

$y = 2^{\frac{1}{3}}$ , then

$x^6$ and

$y^6$ are both rational

Also,

$x^6 - y^6 = (x+y)(x^5-x^4y+x^3y^2-x^2y^3+xy^4-y^5)$

Hence,

$x^5-x^4y+x^3y^2-x^2y^3+xy^4-y^5$ is the required factor.

Substituting the value for

$x$ and

$y$ , then the required factor is

$5^{\frac{5}{6}}-5^{\frac{4}{6}}\cdot 2^{\frac{1}{3}}+5^{\frac{3}{6}}\cdot 2^{\frac{2}{3}}-5^{\frac{2}{6}}\cdot 2^{\frac{3}{3}}+5^{\frac{1}{6}}\cdot 2^{\frac{4}{3}}-2^{\frac{5}{3}}$

$\Rightarrow 5^{\frac{5}{6}}-5^{\frac{4}{6}}\cdot 2^{\frac{1}{3}}+5^{\frac{3}{6}}\cdot 2^{\frac{2}{3}}-5^{\frac{2}{6}}\cdot 2+5^{\frac{1}{6}}\cdot 2^{\frac{4}{3}}-2^{\frac{5}{3}}$

Find two fractions having $7$ and $9$ for their denominators, and such that their sum is $1\dfrac{10}{63}$ .

$1\dfrac{10}{63}=\dfrac{73}{63}=\dfrac{28+45}{63}=\dfrac49+\dfrac57$

Convert the following mixed fraction into improper fraction.
$\displaystyle 10\frac{2}{9}$ .

Mixed fraction

$10\dfrac{2}{9}$

$\dfrac{10\times 9+2}{9}=\dfrac{92}{9}$ .

1162637_700445_ans_b885c061fdc2485fbda3ed8375cc01d2.jpg

Convert the following mixed fraction into improper fraction.
$\displaystyle 1\frac{2}{7}$ .

Mixed fraction

$1\dfrac{2}{7}$ can be converted to improper fraction

$1\dfrac{2}{7}=\dfrac{7\times 1+2}{7}\\\ \ \ \ \ \ =\dfrac{9}{7}$ .

Convert the following mixed fraction into improper fraction.
$\displaystyle 8\frac{7}{9}$ .

Mixed fraction

$8\dfrac{7}{9}$

$\dfrac{8\times 9+7}{9}=\dfrac{77}{9}$ .

1162639_700449_ans_2ee4790dc6784e15b38a281dbce877ce.jpg

Convert the following improper fraction into mixed fraction.
$\displaystyle\frac{9}{4}$ .

Improper fraction:

$\dfrac{9}{4}$

$4)9(2$

$8$

________

$1$

________

$\Rightarrow 2\dfrac{1}{4}$ .

Convert the following improper fraction into mixed fraction.
$\displaystyle\frac{27}{4}$ .

Using division theorem:

$27=4(6)+3$

Dividing both sides by

$4$ :

$\dfrac{27}{4}=6+\dfrac{3}{4}$

$\implies \dfrac{27}{4}=6\dfrac{3}{4}$

Solve the following:
$\cfrac { 5 }{ 8 } +\cfrac { 1 }{ 6 }$ = $\dfrac ab$ . Find a + b

1694596_703826_ans_6257cfe6a9624e05acde55e351b3eb69.jpg

Fill in the blank, with appropriate answer by doing addition and subtraction.

From the figure,

$\dfrac{2}{8}+\dfrac{4}{8}=\dfrac{6}{8}$ and

$\dfrac{2}{8}-\dfrac{1}{8}=\dfrac{1}{8}$

Similarly, we get the remaining as shown in the figure above.

Which rational number should be subtracted from $\dfrac{-5}{6}$ to get $\dfrac{4}{9}$ ?

Let the rational number be

$x$

$\dfrac{-5}{6}-x=\dfrac{4}{9}\implies -x=\dfrac{4}{9}+\dfrac{5}{6}\implies x=-\dfrac{23}{18}$

Using suitable rearrangement and find the sum,
${4 \over 7} + \left( {{{ - 4} \over 9}} \right) + {3 \over 7} + \left( {{{ - 13} \over 9}} \right)$

$\dfrac{4}{7}+\bigg(\dfrac{-4}{9}\bigg)+\dfrac{3}{7}+\bigg(\dfrac{-13}{9}\bigg)=\dfrac{4+3}{7}-\dfrac{13+4}{9}=1-\dfrac{17}{9}=\dfrac{9-17}{9}=\dfrac{-8}{9}$

Solve
${\left( {{1 \over 2}} \right)^3} + {\left( {{1 \over 3}} \right)^3} - {\left( {{5 \over 6}} \right)^3}$

$\bigg(\dfrac{1}{2}\bigg)^{3}+\bigg(\dfrac{1}{3}\bigg)^{3}-\bigg(\dfrac{5}{6}\bigg)^{3}=\dfrac{3^{3}+2^{3}-5^{3}}{6^{3}}=\dfrac{27+8-125}{216}=-\dfrac{90}{216}=-\dfrac{5}{12}$

If $\dfrac{9}{5}$ + $\dfrac{11}{6}$ is $\dfrac{m}{30}$ then m equals to:

Given:

$\dfrac{9}{5}+\dfrac{11}{6}$

$\dfrac{54+55}{5\times6}$

$\dfrac{99}{30}$

$\Rightarrow m= 99$

Add the following rational numbers:
$\dfrac{1}{2}+\dfrac{4}{5}$

Given:

$\dfrac{1}{2}+\dfrac{4}{5}$

LCM of

$2,5=2\times 5 = 10$

Thus,

$\dfrac{1}{2}+\dfrac{4}{5}=\dfrac{1\times 5+4\times 2}{10}$

$=\dfrac{13}{10}$

Add
${3 \over 5}{\rm{ and }}{2 \over 3}$

$\dfrac{3}{5}+\dfrac{2}{3}=\dfrac{9+10}{15}=\dfrac{19}{15}$

What should be subtracted from $-2$ to get $\frac{3}{8}$ ?

Let the rational number be

$x$

$-2-x=\dfrac{3}{8}\implies -x=\dfrac{3}{8}+2\implies x=-\dfrac{19}{8}$

Add
${\rm{2 and - 1}}{1 \over 2}$

$2-1\dfrac{1}{2}=2-\dfrac{3}{2}=\dfrac{1}{2}$

If $\left[ \frac{-2}{5} +\frac{1}{5} \right] - \frac{2}{3}=\dfrac{-m}{15}$ .Find $m$

$\\\left((\frac{-2}{5})+(\frac{1}{5})\right)-(\frac{2}{3})=(\frac{-m}{15})\\(\frac{-2+1}{5})-(\frac{2}{3})=(\frac{-m}{5})\\(\frac{-1}{5})-(\frac{2}{3})=(\frac{-m}{5})\\(\frac{-3-10}{15})=(\frac{-m}{15})\\\therefore\>m=+13$

Find
${{53} \over {14}} - {{201} \over {196}}$

$\dfrac{53}{14}-\dfrac{201}{196}=\dfrac{53\times 14-201}{196}=\dfrac{742-201}{196}=\dfrac{201}{196}$

Rationalise the denominator of $\dfrac{5}{\sqrt{3}-\sqrt{5}}$

$\dfrac { 5 }{ \sqrt { 3 } -\sqrt { 5 } }$

$=\dfrac { 5 }{ \sqrt { 3 } -\sqrt { 5 } } \times \dfrac { \sqrt { 3 } +\sqrt { 5 } }{ \sqrt { 3 } +\sqrt { 5 } }$

$=\dfrac { 5(\sqrt { 3 } +\sqrt { 5 } ) }{ 3-5 }$

$=-\dfrac { 5(\sqrt { 3 } +\sqrt { 5 } ) }{ 2 }$

What should be added to $-2$ to get $\frac{3}{8}$ ?

Let the rational number be

$x$

$-2-x=\dfrac{3}{8}\implies -x=\dfrac{3}{8}+2\implies x=-\dfrac{19}{8}$

Evaluate:
$(i) \dfrac{3}{7} + \dfrac{-4}{9} - \dfrac{-11}{7} - \dfrac{7}{9}$

$(ii) \dfrac{2}{3} + \dfrac{-4}{5} - \dfrac{1}{3} - \dfrac{2}{5}$

$(iii) \dfrac{4}{7} -\dfrac{-8}{9} - \dfrac{-13}{7} + \dfrac{17}{9}$

Evaluate:

$(i)\ \dfrac{3}{7} + \dfrac{-4}{9} - \dfrac{-11}{7} - \dfrac{7}{9}$

$=\dfrac{3}{7}-\left(\dfrac{-11}{7}\right)+\left(\dfrac{-4}{9}\right)-\dfrac{7}{9}=\dfrac{3}{7}+\dfrac{11}{7}+\left(\dfrac{-4}{9}-\dfrac{7}{9}\right)$

$=\dfrac{14}{7}-\dfrac{11}{9}=\dfrac{126-77}{63}=\dfrac{49}{63}=\dfrac{7}{9}$

$(ii)\ \left[\dfrac{2}{3} + \left(\dfrac{-4}{5}\right) - \dfrac{1}{3} - \dfrac{2}{5}\right]=\left(\dfrac{2}{3}-\dfrac{1}{3}\right)+\left(\dfrac{-4}{5}\right)+\left(-\dfrac{2}{5}\right)$

$=\dfrac{1}{3}-\dfrac{6}{5}=\dfrac{5-18}{15}=\dfrac{-13}{15}$

$(iii)\ \dfrac{4}{7} -\left(\dfrac{-8}{9}\right) - \left(\dfrac{-13}{7}\right) + \dfrac{17}{9}=\dfrac{4}{7}-\left(\dfrac{-13}{7}\right)-\left(\dfrac{-8}{9}\right)+\dfrac{17}{9}$

$=\dfrac{(4+13)}{7}+\dfrac{17}{9}+\dfrac{8}{9}$

$=\dfrac{17}{7}+\dfrac{17+8}{9}=\dfrac{17}{7}+\dfrac{25}{9}$

$=\dfrac{153+175}{63}=\dfrac{328}{63}$

Convert it to improper fraction $3\frac{3}{4}$

$3\frac{3}{4}$

$=\dfrac{(3\times4)+3}{4}$

$=\dfrac{12+3}{4}$

$=\dfrac{15}{4}$

solve $\left(2 - \dfrac{3}{ 5}\right)$

$2-\dfrac{3}{5}=\dfrac{10-3}{5}=\dfrac{7}{5}$

Show the fraction $\dfrac{2}{5},\dfrac{3}{5},\dfrac{4}{5}$ and $\dfrac{5}{5}$ on a number line.

Fractions

$\dfrac{2}{5}, \dfrac{3}{5}, \dfrac{4}{5}, \dfrac{5}{5}$

Since the denominators of each fraction is

$5$ . Divide the space of two consecutive integers (on the number line) in

$5$ equal parts.

Add: $\dfrac{5}{12} +\dfrac{3}{8}$

$\dfrac{5}{12}+\dfrac{3}{8}$

Addition of unlike factors.

Take LCM of denominators. LCM of

$12,8$ is

$24$

$\therefore \dfrac{5}{12}+\dfrac{3}{8}=\dfrac{10+9}{24}=\dfrac{19}{24}$

Write the additive inverse of each of $\dfrac{2}{8}$ and $\dfrac{-5}{9}$

Additive inverse of

$A$ is

$-A$

$A+(-A)=0$

Hence additive inverse of

$\dfrac{2}{8}$ is

$-\dfrac{2}{8}=-\dfrac{1}{4}$

Additive inverse of

$\dfrac{-5}{9}$ is

$-\dfrac{-5}{9}=\dfrac{5}{9}$

Express $\dfrac {17}{7}$ as mixed fraction.

$\displaystyle \frac{7\left|\begin{matrix} 1 & 7 \\ 1 & 4 \end{matrix} \right| 2}{3} \Rightarrow 2 \frac{3}{7} =$ Mixed fraction.

Find (i) $5-3\dfrac {1}{4}=$ ...............

$5-3\dfrac{1}{4}=\left [ 5-\left(3+\dfrac{1}{4}\right) \right ]=5-\dfrac{13}{4}=\dfrac{7}{4}$

Convert it to mixed fraction $\frac{34}{5}$

$\dfrac{34}{5}$

$=\dfrac{30+4}{5}$ ---------(split the numerator as the maximum number which is - (i)exactly divisible by denominator (ii) less than numerator)

$=\dfrac{(6\times5)+4}{5}$

$=6\frac{4}{5}$

$21+\dfrac {7}{8}$

$=\dfrac{21\times 8+7}{8}=\dfrac{175}{8}$

What is : $\dfrac{3}{20}-\dfrac{5}{8}$ ?

$\dfrac{3}{20}-\dfrac{5}{8}$

LCM of 8 and 20=40

$\dfrac{3\times 2}{20\times 2}-\dfrac{5\times 5}{8\times 5}=\dfrac{6-25}{40}=\dfrac{-19}{40}$

Hence the correct answer is

$\dfrac{-19}{40}$

$7-3\dfrac{1}{2}=$ _______

$\displaystyle 7-3\frac{1}{2}=\left [ 7-\left ( 3+\frac{1}{2} \right ) \right ]=7-\frac{7}{2}=\frac{7}{2}$

$36+\dfrac {12}{5}$

$=\dfrac{36\times 5+12}{5}=\dfrac{192}{5}$

Solve : $9+\dfrac {3}{2}$

$=\dfrac{18+3}{2}=\dfrac{21}{2}$

Solve: $\dfrac{3}{5} + \dfrac{2}{7}$

$\begin{array}{l} \dfrac { 3 }{ 5 } +\dfrac { 2 }{ 7 } \\ =\dfrac { { 7\times 3+5\times 2 } }{ { 35 } } \\ =\dfrac { { 21+10 } }{ { 35 } } \\ =\dfrac { { 31 } }{ { 35 } } \end{array}$

Add $\dfrac{-2}{5}$ and its multiplicative inverse.

$\cfrac{{ - 2}}{5} + \cfrac{-5}{2}$

$= \cfrac{{ - 2}}{5} - \cfrac{5}{2}$

$= \cfrac{{ - 4 - 25}}{{10}}$

$= \cfrac{{ - 29}}{{10}}$

Solve $8\dfrac{x}{6} + 3\dfrac{3}{8} = 12\dfrac{5}{4}$

$\dfrac{48+x}{6}+\dfrac{27}{8}=\dfrac{53}{4}$

$\dfrac{x}{6}=\dfrac{79}{8}-8$

$x=\dfrac{45}{4}$

$15+\dfrac {5}{7}$

$=\dfrac{15\times 7+5}{7}=\dfrac{110}{7}$

If $\dfrac {16}{17}+\dfrac {2}{17}=\dfrac{m}{17}$ .Find $m$

Solution:-

$\cfrac{16}{17} + \cfrac{2}{17} = \cfrac{m}{17}$

$\Rightarrow \; \cfrac{16 + 2}{17} = \cfrac{m}{17}$

$\Rightarrow \; \cfrac{18}{17} = \cfrac{m}{17}$

$\Rightarrow \; m = 18$

Hence, the correct answer is 18.

Add the following
$\dfrac{4}{7}$ and $\dfrac{5}{7}$

$\cfrac{4}{7} + \cfrac{5}{7}$

$\cfrac{{4 + 5}}{7}$

$= \cfrac{9}{7}$

Solve
$2-\dfrac{3}{5}$

$2-\cfrac { 3 }{ 5 } =\cfrac { 2\left( 5 \right) -3 }{ 5 } =\cfrac { 10-3 }{ 5 } =\cfrac { 7 }{ 5 }$

Simplify: $1\dfrac{7}{8}\times 3\dfrac{1}{5}$

$1\dfrac {7}{8} \times 3\dfrac {1}{5}$

$= \dfrac {{1 \times + 7}}{8} \times \dfrac {{3 \times 5 + 1}}{5}$

$= \dfrac {{8 + 7}}{8} \times \dfrac {{15 + 1}}{5}$

$= \dfrac {{15}}{8} \times \dfrac {{16}}{5}$

$= 3 \times 2$

$= 6$

Write the following rational number in decimal form
$\dfrac {23}{2^{3}\cdot 5^{2}}$

Given rational number:

$\dfrac{23}{2^3\times 5^2}$

$\Rightarrow\dfrac{23}{2\times 4\times 25}$

$\Rightarrow \dfrac{23}{2\times 100}$

$\Rightarrow \dfrac{11.5}{ 100}$

$\Rightarrow 0.115$

This is the required decimal form.

The sum of two rational numbers is $\dfrac { 4 }{ 21}$ . If one of them is $\dfrac { 5 }{ 7 }$ , find the other.

For given the sum of two rational number is

$\cfrac{4}{21}$

Where one of the rational number is

$\cfrac{5}{7}$

Hence the other rational number is

$=\cfrac{4}{21}-\cfrac{5}{7}=\cfrac{4-15}{21}=-\cfrac{11}{21}$

Express each of the following as a fraction in simplest form:
$66\dfrac { 2 }{ 3 }\%$

$66\dfrac{2}{3}\%$

$=\dfrac{200}{3}\%$

$=\dfrac{200}{300}=\dfrac{2}{3}$

Simplify $\dfrac {23}{16}$

$\dfrac{23}{16}$

$=1\dfrac{7}{16}$

$=1.43$

Simplify $\dfrac{1}{2} + \dfrac{5}{2} + \dfrac{3}{{10}}$

Consider the given expression.

$=\dfrac{1}{2}+\dfrac{5}{2}+\dfrac{3}{10}$

$=\dfrac{5+25+3}{10}$

$=\dfrac{33}{10}$

Hence, the value of this expression is $\dfrac{33}{10}$ .

Solve $2\dfrac{1}{2} - 1\dfrac{4}{5}$

$2\dfrac{1}{2}-1\dfrac{4}{5}$

$\Rightarrow$

$\dfrac{(2\times 2)+1}{2}-\dfrac{(1\times 5)+4}{5}$

$\Rightarrow$

$\dfrac{5}{2}-\dfrac{9}{5}$

$\Rightarrow$

$\dfrac{5\times 5}{2\times 5}-\dfrac{9\times 2}{5\times 2}$

$\Rightarrow$

$\dfrac{25}{10}-\dfrac{18}{10}$

$\Rightarrow$

$\dfrac{25-18}{10}$

$\Rightarrow$

$\dfrac{7}{10}$

Write the additive inverse of $\dfrac{{43}}{9}\,$

$\Rightarrow$ The additive inverse is what we add to a number to get zero.

$\Rightarrow$ So, additive inverse of

$\dfrac{43}{9}$ is

$\dfrac{-43}{9}$ , because

$\dfrac{43}{9}+\left(\dfrac{-43}{9}\right)=0$

$\therefore$ The required additive inverse is

$\dfrac{-43}{9}$

Soham poured some water in two fish bowls. He poured $4\dfrac{5}{8}$ liters of water in the first dish bowl and $2\dfrac{5}{6}$ liters in the second bowl. How much water Soham pour in the first bowls altogether?

In first bowl he poured = $4{\dfrac5 8}l$

$={\dfrac{37} 8}l.$

In second bowl $= 2{\dfrac 56}l$

$={\dfrac{17}6}l$

altogether he bowled $= {\dfrac{37} 8} + {\dfrac 76}$

$={\dfrac{111 + 68} {24}}$

$={\dfrac{179}{24}} = 7{\dfrac{11} {24}}l$

Work out : $4\dfrac{9}{8} - 2\dfrac{4}{5}$

$4{\dfrac 98} - 2{\dfrac 4 5}$

$={\dfrac{41}8} - {\dfrac{14} 5} = {\dfrac{205 - 112} {40}} = {\dfrac{93} {40}} = 2{\dfrac{13}{40}}$

Express each of the following as a fraction in simplest form:
$33\dfrac { 1 }{ 2 }\%$

$33\dfrac{1}{2}\%$

$=\dfrac{67}{2}\%$

$=\dfrac{67}{200}$

Subtract: $\dfrac { 31 }{ 2 } -\dfrac { 2 }{ 3 }$

$\dfrac{{31}}{2} - \dfrac{2}{3}$

$= \dfrac{{93 - 4}}{6}$

$= \dfrac{{89}}{6}$

How do represent fraction on number line
$\dfrac{3}{4}$

Consider the given number, $\dfrac{3}{4}$ .

We know that it is proper dfraction which lie between $0$ and $1$

It is represent in number line as shown in fig.

1049146_1056947_ans_d92aba44d741498796657ef3b9127440.png

Add the following fractions.
(a) $\dfrac{7}{12}$ and $\dfrac{3}{8}$ (b) $\dfrac{9}{16}$ and $\dfrac{7}{12}$ .

(a)

$\dfrac{7}{12}$ and

$\dfrac{3}{8}$

$=\dfrac{14+9}{24}=\dfrac{23}{24}$

(b)

$\dfrac{9}{16}$ and

$\dfrac{7}{12}$

$=\dfrac{27+28}{48}=\dfrac{55}{48}$

Add.
$6\dfrac{1}{3}+2\dfrac{1}{3}$

$6\dfrac{1}{3}$ is a mixed fraction and can be written as

$6\dfrac{1}{3} = \dfrac{6\times3 + 1}{3} = \dfrac{19}{3}$

$2\dfrac{1}{3}$ is a mixed fraction and can be written as

$2\dfrac{1}{3} = \dfrac{2\times3 + 1}{3} = \dfrac{7}{3}$

$\therefore 6\dfrac{1}{3} + 2\dfrac{1}{3} = \dfrac{19}{3} + \dfrac{7}{3} = \dfrac{26}{3}$

If $\dfrac{5}{7}+\dfrac{9}{11}=\dfrac{118}{11a}$ . Find $a$

$(\dfrac{5}{7})+(\dfrac{9}{11})\\=(\dfrac{55+63}{77})\\=(\dfrac{118}{77})\\by\>comparison,\>a\>=\>7$

What number should be added in $\dfrac{-7}{8}$ to get $\dfrac{4}{9}$

Let the number to be added be

$x$

According to question,

$\dfrac{-7}{8} + x = \dfrac{4}{9}$

$\Rightarrow x = \dfrac{4}{9} + \dfrac{7}{8}$

$\Rightarrow x = \dfrac{32+63}{8\times9}$

$\therefore x = \dfrac{95}{72}$

Evaluate $\dfrac{6}{{96}} + \dfrac{4}{{192}}$

$\dfrac{6}{96}+\dfrac{4}{192}$

$=\dfrac{1}{16}+\dfrac{1}{48}$

$=\dfrac{3}{48}+\dfrac{1}{48}$

$=\dfrac{4}{48}$

$\dfrac{6}{96}+\dfrac{4}{192}$

$=\dfrac{1}{12}$

In a camp, there are $420$ boys and $180$ girls. Find the ratio of boys to girls:

In the camp there are

$420$ boys and

$180$ girls.

Then the ratio of boys to girls is

$\dfrac{420}{180}=\dfrac{7}{3}$ .

Add $\dfrac{6}{{96}} + \dfrac{4}{{92}}$

$\dfrac{6}{96}+\dfrac{4}{92}$

$\dfrac{1}{16}+\dfrac{1}{23}=\dfrac{23+16}{368}$

$=\dfrac{39}{368}$

What fraction is $270g$ of $3 kg$ .

We have

$270$ gm

$=\dfrac{270}{3000}=\dfrac{9}{100}$ part of

$3$ kg. [ Since

$3$ kg

$=3,000$ gm]

Subtract $\dfrac {2}{5}-5$

Solution:-

$\cfrac{2}{5} - 5 = \cfrac{2 - \left( 5 \times 5 \right)}{5} = \cfrac{2 - 25}{5} = \cfrac{-23}{5}$

Hence, the correct answer is

$\cfrac{-23}{5}$ .

Simplify $:\,8\dfrac{5}{6} - 3\dfrac{3}{8} + 1\dfrac{7}{2}$ .

$\begin{array}{l}8\dfrac{5}{6} - 3\dfrac{3}{8} + 1\dfrac{7}{2}\\ = \dfrac{{8 \times 6 + 5}}{6} - \dfrac{{3 \times 8 + 3}}{8} + \dfrac{{1 \times 2 + 7}}{2}\\ = \dfrac{{53}}{6} - \dfrac{{27}}{8} + \dfrac{9}{2}\\ = \dfrac{{4 \times 53 - 3 \times 27 + 12 \times 9}}{{24}}\\ = \dfrac{{212 - 81 + 108}}{{24}}\\ = \dfrac{{239}}{{24}} = 9\dfrac{{23}}{{24}} \approx 10\end{array}$

Simplify: $2\dfrac{2}{3} +3\dfrac{1}{2}$

Given:

$2\dfrac{2}{3} +3\dfrac{1}{2}$

$\Rightarrow \dfrac{8}{3}+\dfrac{7}{2}=\dfrac{2(8)+3(7)}{6}=\dfrac{37}{6}$

Mother bought two chickens. One weight $1\dfrac{3}{4}\ kg$ and the other $1\dfrac{1}{2}\ kg$ . How much did they both weigh together?

One weight of chicken

$=1\dfrac{3}{4}$ kg

$=\dfrac{7}{4}$ kg

Other

$=1\dfrac{1}{2}$ kg

$=\dfrac{3}{2}$ kg.

So together

$=\dfrac{7}{4}+\dfrac{3}{2}$

$=\dfrac{7}{4}+\dfrac{6}{4}=\dfrac{13}{4}$ kg.

How much less is $28$ km than $42.6$ km?

We have

$28$ km is

$(42.6-28)=14.6$ km less than

$42.6$ km.

Solve : $4\dfrac{1}{2} - \dfrac{1}{4} + \dfrac{1}{5} + \dfrac{3}{5}$

$4\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{3}{5}$

$=\dfrac{9}{2}-\dfrac{1}{4}+\dfrac{1+3}{5}$

$=\dfrac{9}{2}-\dfrac{1}{4}+\dfrac{4}{5}$

$=\dfrac{90-5+16}{20}$

$=\dfrac{101}{20}$

$- \frac{2}{{21}} + \frac{3}{7}$

$-\dfrac{2}{21}+\dfrac{3}{7}$

$=-\dfrac{2}{21}+\dfrac{3\times 3}{7\times 3}$

$=\dfrac{-2+9}{21}=\dfrac{7}{21}=\dfrac{1}{3}$

Solve $4\dfrac{2}{3} - 1\dfrac{{11}}{{12}}$

Given that,

$4\dfrac{2}{3}-1\dfrac{11}{12}$

$=\dfrac{14}{3}-\dfrac{23}{12}$

$=\dfrac{14\times 4-23\times 1}{12}$

$=\dfrac{56-23}{12}$

$=\dfrac{33}{12}$

$=\dfrac{11}{4}$

$=2\dfrac{3}{4}$

Hence, the value is

$2\dfrac{3}{4}$

Add $1\dfrac{3}{4}, 2\dfrac{1}{10}, 3\dfrac{1}{3}$ .

$\Rightarrow$ Here Summation will be S,

$S = 1\dfrac{3}{4} + 2\dfrac{1}{10} + 3\dfrac{1}{3}$

now, we can write mixed function as below,

$a\dfrac{b}{c} = \dfrac{ac+b}{c} = a+\dfrac{b}{c}$

So,

$S = 1+\dfrac{3}{4}+ 2+\dfrac{1}{10}+3+\dfrac{1}{3}$

$= 6+ \dfrac{3}{4} + \dfrac{1}{10} +\dfrac{1}{3}$

(take LCM of 4, 10, 3, we get 60).

$= 6+\dfrac{3(15)+1(6)+1(20)}{60}$

$= 6+\dfrac{45+6+20}{60}$

$= 6 +\dfrac{71}{60}$

$= 6+ \dfrac{60+11}{60}$

$= 6+1 +\dfrac{11}{60}$

$= 7+\dfrac{11}{60}$

$= 7\dfrac{11}{60}$

So,

$\boxed{1\dfrac{3}{4}+2\dfrac{1}{10}+3\dfrac{1}{3} = 7\dfrac{11}{60}}$

1209403_1069770_ans_69e6e5f9afb94fa29f7a34eaeffde3df.jpg

Prove that : $\dfrac{a^{-1}}{a^{-1} + b^{-1}} + \dfrac{a^{-1}}{a^{-1} - b^{-1}} = \dfrac{-2b^2}{b^2 - a^2}$

L.H.S.

$=\dfrac{a^{-1}}{a^{-1}+b^{-1}}+\dfrac{a^{-1}}{a^{-1}-b^{-1}}$

$=\dfrac{\dfrac{1}{a}}{\dfrac{1}{a}+\dfrac{1}{b}}+\dfrac{\dfrac{1}{a}}{\dfrac{1}{a}-\dfrac{1}{b}}$

$=\dfrac{\dfrac{1}{a}}{\dfrac{b+a}{ab}}+\dfrac{\dfrac{1}{a}}{\dfrac{b-a}{ab}}$

$=\dfrac{1}{a}\times \dfrac{ab}{b+a}+\dfrac{1}{a}\times \dfrac{ab}{b-a}$

$=\dfrac{b}{b+a}+\dfrac{b}{b-a}$

$=\dfrac{b(b-a)+b(b+a)}{b^{2}-a^{2}}$

$=\dfrac{b^{2}-ab+b^{2}+ab}{b^{2}-a^{2}}$

$=\dfrac{2b^{2}}{b^{2}-a^{2}}$

$=$ R.H.S.

A family eats $1\dfrac{2}{3}$ breads at a meal. Will $3$ breads be enough for $2$ meals?

$1\dfrac{2}{3}$ for one meal

$\Rightarrow 1$ meal

$=\dfrac{5}{3}$ breads

For

$2$ meals

$=\dfrac{5}{3}\times 2$ breads

$=\dfrac{10}{3}\approx 3.33$

$\therefore$

$3$ breads are not enough for

$2$ meals.

List five rational numbers between
1) -1 and 0
2) -2 and -1
3) $\dfrac{-4}{5}$ and $\dfrac{-2}{3}$
4) $\dfrac{1}{2}$ and $\dfrac{2}{3}$

1) 1 =

$\dfrac{6}{6} so , \dfrac{1}{6} \dfrac{2}{6} \dfrac{3}{6} \dfrac{4}{6} \dfrac{5}{6}$

2) -2 =

$\dfrac{-12}{6} and -1 = \dfrac{-6}{6} so, \dfrac{-7}{6} \dfrac{-8}{6} \dfrac{-9}{6} \dfrac{-10}{6} \dfrac{-11}{6}$

3) Similarly

$\dfrac{-119}{150}$ $$\dfrac{-118}{150}

\dfrac{-117}{150}

\dfrac{-116}{150}

\dfrac{-115}{150}$$

4) similarly

$\dfrac{31}{60} \dfrac{32}{60} \dfrac{33}{60} \dfrac{34}{60} \dfrac{35}{60}$

Convert the following into proper fraction: $4\dfrac{3}{5}$

$4\dfrac{3}{5}= \dfrac{4\times5+3}{5}$

$=\dfrac{20+3}{5}$

$=\dfrac{23}{5}$

$4\dfrac{3}{5}=\dfrac{23}{5}$

Add $-2 + \dfrac{15}{2} + \dfrac{11}{4} =$

$-2 + \dfrac{15}{2} + \dfrac{11}{4}$

$= \dfrac{15-4}{2} + \dfrac{11}{4}$

$= \dfrac{11}{2} + \dfrac{11}{4}$

$=\dfrac{33}{4}$ .

Draw number lines and locate the points on them:
$\frac{2}{5}, \frac{3}{5}, 1, \frac{4}{5}$

divide space between

$0$ to

$1$ in

$5$ equal parts

1420593_1076003_ans_70c81c83bf3b44bfa129174c04bdefef.png

Solve:
$\dfrac{16}{9}-{{x}^{2}}=0$

Given that

$\dfrac{16}{9}-{{x}^{2}}=0$

${{x}^{2}}=\dfrac{16}{9}$

$x=\sqrt{\dfrac{16}{9}}=\pm \dfrac{4}{3}$

$x=\pm \dfrac{4}{3}$

This is the answer

Simplify: $\dfrac{3}{7} + \left( {\dfrac{{ - 6}}{{11}}} \right) + \left( {\dfrac{{ - 8}}{{21}}} \right) + \left( {\dfrac{5}{{22}}} \right)$

We have,

$\dfrac{3}{7}+\left( \dfrac{-6}{11} \right)+\left( \dfrac{-8}{21} \right)+\dfrac{5}{22}$

$=\dfrac{3}{7}-\dfrac{8}{21}+\dfrac{5}{22}-\dfrac{6}{11}$

$=\dfrac{9-8}{21}+\dfrac{5-12}{22}$

$=\dfrac{1}{21}-\dfrac{7}{22}$

$=\dfrac{22-147}{21\times 22}$

$=\dfrac{-125}{462}$

Hence, the value is $\dfrac{-125}{462}$ .

Solve: $\dfrac{3}{8}-\dfrac{1}{12}$

Given that:

$\dfrac{3}{8}-\dfrac{1}{12}$

$=\dfrac{3\times 3-1\times 2}{24}$

$=\dfrac{9-2}{24}$

$=\dfrac{7}{24}$

Hence, the value is

$\dfrac{7}{24}$

Draw number lines and locate the points on them:
$\dfrac{1}{4}, \dfrac{2}{4}, \dfrac{3}{4}, \dfrac{4}{4}$

Distance between two consecutive adjacent so in equal

1420605_1075996_ans_f8615829fa9e4ac5b125e314cde60016.png

$\left(2\frac{1}{2}+3\frac{1}{3}\right)+\frac{2}{9}\left(3\frac{1}{8}-2\frac{1}{12}\right)$

$(2(\frac{1}{2})+3(\frac{1}{3}))+(\frac{2}{9})(3(\frac{1}{8})-2(\frac{1}{12}))\\= ((\frac{5}{2})+(\frac{10}{3}))+(\frac{2}{9})((\frac{25}{8})-(\frac{25}{12}))\\=(\frac{15+30}{6})+(\frac{2}{9})((\frac{75-50}{24}))\\(\frac{45}{6})+(\frac{2}{9})\times (\frac{25}{24})\\=(\frac{15}{2})+(\frac{25}{108})\\=(\frac{(15\times 54)+25}{108})=(\frac{835}{108})$

Evaluate: $\frac{1}{-15}+\frac{5}{-12}$

$(\frac{1}{-15})+(\frac{5}{-12})\\=(\frac{-1}{15})-(\frac{5}{12})\\=(\frac{-4-25}{60})\\=(\frac{-29}{60})$

Is $\dfrac{3}{4}$ is a proper fraction?

We can define three types of dfraction.

Proper fraction –the number is less than the denominator

Example: $\dfrac{1}{2},\dfrac{5}{8},\dfrac{3}{7}$ .

Improper fraction –the numerator is greater than(0r equal to) the denominator .

Example: $\dfrac{3}{2},\dfrac{7}{4},\dfrac{5}{5}$ .

Mixed fraction –A whole number and proper fraction together .

Example : $2\dfrac{1}{2},7\dfrac{3}{4},1\dfrac{4}{5}$

So, $\dfrac{3}{4}$ is proper fraction.

Evaluate: $\frac{-8}{9}+\frac{-5}{12}$

$-(\frac{8}{9})+(\frac{-5}{12})\\=(\frac{-8}{9})-(\frac{5}{12})\\=(\frac{-8\times4-5\times3}{36})\\=(\frac{-32-15}{36})\\=(\frac{-47}{36})$

Evaluate: $4+\frac{3}{-5}$

$4+(\frac{-3}{5})\\=(\frac{4\times5-3}{5})\\=(\frac{20-3}{5})\\=(\frac{17}{5})$

Evaluate: $\frac{5}{9}+\frac{3}{-4}$

$(\frac{5}{9})+(\frac{3}{-4})\\=(\frac{5}{9})-(\frac{3}{4})\\=(\frac{20-27}{36})\\=(\frac{-7}{36})$

What numbers should be added to $\cfrac { -5 }{ 8 }$ so as to get $\cfrac { -3 }{ 2 } ?$

According to the problem

we assume x be added so,

$-\dfrac {5}{8} + x = -\dfrac {3}{2}$

$x = -\dfrac {3}{2}+\dfrac {5}{8}$

$x = \dfrac {{ -12 + 5}}{8}$

$x = - \dfrac {7}{8}$

Evaluate: $0+\frac{-2}{7}$

$0+(\frac{-2}{7})=(\frac{-2}{7})\\as\>adding\>zero\>to\>a\>number\>doesn't\>change\>its\>value$

Evaluate: $\frac{5}{-11}+0$

$(\frac{5}{-11})+0\\=(\frac{-5}{11})+0\\=(\frac{-5}{11})\\as\>adding\>zero\>to\>a\>number\>doesn't\>change\>its\>value$

Evaluate: $2+\frac{-3}{5}$

$2+(\frac{-3}{5})\\=2-(\frac{3}{5})\\=(\frac{2\times5-3}{5})\\=(\frac{7}{5})$

Evaluate: $\frac{4}{-9}+1$

$(\frac{4}{-9})+1\\=(\frac{-4}{9})+1\\=(\frac{-4+9}{9})\\=(\frac{5}{9})$

Find:
$\dfrac{3}{5} + \dfrac{2}{7}$

$\dfrac{3}{5} + \dfrac{2}{7}$

$=\dfrac{{7 \times 3 + 5 \times 2}}{{35}}$

$=\dfrac{{21 + 10}}{{35}}$

$=\dfrac{{31}}{{35}}$

Represent the following number on the number line
$\dfrac {7}{4}$

Consider the given improper dfraction,

$\dfrac{7}{4}=1\dfrac{3}{4}=1+\dfrac{3}{4}$

Hence, $\dfrac{7}{4}$ lie between 1 and 2 .

1055660_1085806_ans_770eca1b4c634674badb6a4963cae449.png

Convert into improper fractions.
$8\frac{1}{2}$

$8\frac{1}{2}=\dfrac{2\times8+1}{2}=\dfrac{17}{2}$

Convert into improper fractions.
$4\dfrac{5}{6}$

$4\dfrac{5}{6}=\dfrac{6\times4+5}{6}=\dfrac{29}{6}$

Represents these number on number line
$\dfrac {3}{5}$

Solution

Consider the following question

$\dfrac{3}{5}$

To mark $\dfrac{3}{5}$ ; move three parts on the right-side of zero.

Hence, this is the correct answer.

1053336_1085552_ans_30fdd36a97444b18a8487769cf66516e.png

Represent the following number on the number line
$\dfrac {3}{5}$

Consider the given proper fraction $\dfrac{3}{5}$ .which lie between 0 and 1.

Because it is a proper fraction. proper fraction $<$ $1$

1055781_1085810_ans_f4064a2d1e3a4e54be41453a5360e44e.png

Evaluates : $\left(\dfrac{1}{3}\right)^{-1} + \left(\dfrac{1}{5}\right)^{-1} - \left(\dfrac{1}{6}\right)^{-1}$

$\left ( \cfrac{1}{3} \right )^{-1}+\left ( \cfrac{1}{5} \right )^{-1}-\left ( \cfrac{1}{6} \right )^{-1}$

$=\cfrac{3}{1}+\cfrac{5}{1}-\cfrac{6}{1}$

$\left ( x^{-1} =\cfrac{1}{x}\right )$

$=8-6=2$

Solve: $9\dfrac{1}{9}-3\dfrac{4}{7}$

we have,

$\Rightarrow 9\dfrac{1}{9}-3\dfrac{4}{7}$

$\Rightarrow \dfrac{82}{9}-\dfrac{25}{7}$

$\Rightarrow \dfrac{82\times 7-25\times 9}{63}$

$\Rightarrow \dfrac{574-225}{63}$

$\Rightarrow \dfrac{349}{63}$

Hence, this is the answer.

Add $1\dfrac{2}{3} + 2\dfrac{4}{5}$

we can also write

$\dfrac{5}{3}+\dfrac{14}{5}$

now,

$=>\dfrac{25+42}{15}$

$=>\dfrac{67}{15}$

Add $\cfrac{8}{6} + \cfrac{5}{3}$

$=>\dfrac{8+10}{6}$

$=>3$

Find $\dfrac{{\sqrt 3 }}{2} \times \dfrac{{\sqrt 3 }}{2} + \dfrac{1}{2} \times \dfrac{1}{2}$

We have,

$\dfrac{\sqrt{3}}{2}\times \dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\times \dfrac{1}{2}$

$=\dfrac{3}{4}+\dfrac{1}{4}$

$=\dfrac{3+1}{4}$

$=\dfrac{4}{4}=1$

Hence, the value is $1$ .

$a=\dfrac{1}{6}$ and $\dfrac{6}{7}=\dfrac{5}{b}$ , what is the value of $a+b$ ?

$a=\dfrac{1}{6}$

$b=\dfrac{5}{6} \times 7 \Rightarrow b = \dfrac{30}{7}$

$a+b = \dfrac{1}{6}+ \dfrac{30}{7}$

$=\dfrac{187}{42}$

$\Rightarrow a+b=\dfrac{187}{42}$

What should be added to $\left[ {\dfrac{{ - 3}}{5} + \dfrac{{ - 5}}{3}} \right]$ to get $1$ ?

We have,

$\left[ \dfrac{-3}{5}+\dfrac{-5}{3} \right]$

Let we add $x$ in above expression, we have

$-\dfrac{3}{5}-\dfrac{5}{3}+x=1$

$x=1+\dfrac{3}{5}+\dfrac{5}{3}$

$x=\dfrac{15+9+25}{15}$

$x=\dfrac{49}{15}$

Hence, the value of $x$ is $\dfrac{49}{15}$ .

Add $\cfrac{5}{4} + \left( {\cfrac{{ - 11}}{4}} \right)$

We can also write

$=>\dfrac{5}{4}-\dfrac{11}{4}$

$=>\dfrac{5-11}{4}$

$=>\dfrac{-3}{2}$

Which of the following are proper fractions?
$\dfrac{1}{8},\dfrac{2}{5},\dfrac{{16}}{{21}},\dfrac{{23}}{{10}},3,\dfrac{8}{{17}}$

We know that,

A proper fraction has numerator less then its denominator.

Then,

$\dfrac{1}{8},\dfrac{2}{5},\dfrac{16}{21}$ and $\dfrac{8}{17}$ are proper fractions.

$\dfrac{23}{10}$ and

$3$ are not proper fractions.

Solve : $2+\dfrac {3}{4}$

We have,

$2+\dfrac{3}{4}$

$=\dfrac{11}{4}$

Hence, this is the answer.

What number should be added to $\dfrac {-6}{5}$ as so to get $\dfrac {1}{3}$ ?

Let the number be $x$ .

According to the question,

$-\dfrac{6}{5}+x=\dfrac{1}{3}$

$\dfrac{5x-6}{5}=\dfrac{1}{3}$

$3\left( 5x-6 \right)=5$

$15x-18=5$

$15x=23$

$x=\dfrac{23}{15}$

Hence, this is the answer.

Solve
$\dfrac {1}{18}+\dfrac {1}{18}$

Consider the following Equation.

$\dfrac{1}{18}+\dfrac{1}{18}=\dfrac{2}{18}$

$=\dfrac{1}{9}$

Hence, this is the correct answer.

Solve
$\dfrac {5}{8}+\dfrac {3}{8}$

Consider the following Equation.

$\dfrac{5}{8}+\dfrac{3}{8}=\dfrac{8}{8}$

$=1$

Hence, this is the correct answer.

In a bag, there are $20 kg$ of fruits. If $7\dfrac {1}{6} kg$ of these fruits be oranges and $8\dfrac {2}{3} kg$ of these are apples and rest are grapes. Find the mass of the grapes in the bag.

Total mass of fruits in the bag

$=20$ kg

Mass of oranges

$=7\dfrac{1}{6}=\dfrac{43}{6}$ kg

Mass of the apples

$8\dfrac{2}{3}=\dfrac{26}{3}$ kg

Therefore, Mass of the grapes

$=20-\dfrac{43}{6}-\dfrac{26}{3}$

$=\dfrac{120-43-52}{6}$

$=\dfrac{25}{6}$

$=4\dfrac{1}{6}$ kg

Show $\dfrac{1}{2}$ on a number line.

We know,

$\dfrac{1}{2}=0.5$
1337170_1112178_ans_daf69552e29e4d128c901cd446854570.png

Solve:
$\frac{2}{{21}} + \frac{3}{7}$
$\frac{2}{5} + \frac{1}{3}$

We have,

(1): $\dfrac{2}{21}+\dfrac{3}{7}$

$=\dfrac{2+9}{21}$

$=\dfrac{11}{21}$

(2): $\dfrac{2}{5}+\dfrac{1}{3}$

$=\dfrac{6+5}{15}$

$=\dfrac{11}{15}$

Hence, this is the answer.

Express as a mixed fraction

(i) $\dfrac{37}{8}$

(ii) $\dfrac{28}{5}$

$\dfrac{37}{8}=\dfrac{32+5}{8}=4+\dfrac{5}{8}=4\dfrac{5}{8}$

ii)

$\dfrac{28}{5}=\dfrac{25+3}{5}=5+\dfrac{3}{5}=5\dfrac{3}{5}$

Find: $- \dfrac{6}{{11}} - \left( { - \dfrac{7}{{15}}} \right)$

Solve

$-\dfrac{6}{11}-\left( -\dfrac{7}{15} \right)$

$=-\dfrac{6}{11}+\dfrac{7}{15}$

$=\dfrac{-90+77}{165}$

$=\dfrac{-13}{165}$

Hence, this is the answer.

Find the sum $\dfrac{5}{3}+\dfrac{3}{5}$

$\dfrac{5}{3}+\frac{3}{5}=\dfrac{5\times 5+3\times 3}{3\times 5}=\dfrac{34}{15}$

Find the sum $\dfrac{-2}{3}+0$

$\dfrac{-2}{3}+0=\dfrac{-2}{3}$

Add and express sum of the following rational numbers: $\dfrac{-11}{4}$ and $\dfrac{24}{7}$

The given numbers are

$\dfrac{-11}{4}$ and

$\dfrac{24}{7}$

$=\dfrac{24\times4-11\times7}{28}$

$=\dfrac{96-77}{28}$

$=\dfrac{19}{28}$

Solve:-
$3\frac{3}{4} + 5\frac{1}{6} + 4\frac{1}{3}$

Consider the given expression.

$3\dfrac{3}{4}+5\dfrac{1}{6}+4\dfrac{1}{3}$

$=\dfrac{15}{4}+\dfrac{31}{6}+\dfrac{13}{3}$

$=\dfrac{45+62+52}{12}$

$=\dfrac{159}{12}$

$=\dfrac{53}{4}$

Hence, this is the answer.

The sum of a fraction and its reciprocal is $1{\frac{1}{2}}$ more than the fraction. Find the fraction.

Let the fraction be $\dfrac{x}{y}$ .

According to the question,

$\dfrac{x}{y}+\dfrac{y}{x}=\dfrac{x}{y}+1\dfrac{1}{2}$

$\dfrac{x}{y}+\dfrac{y}{x}=\dfrac{x}{y}+\dfrac{3}{2}$

$\dfrac{y}{x}=\dfrac{3}{2}$

$\dfrac{x}{y}=\dfrac{2}{3}$

Hence, the fraction is $\dfrac{2}{3}$ .

What should be added to $5\dfrac{1}{5}$ to get $7\dfrac{3}{7}$ ?

Let the number = x

According to problem

$\displaystyle x+5\frac{1}{5}=7\frac{3}{7}$

$\displaystyle x+\frac{26}{5}=\frac{52}{7}$

$\displaystyle x=\frac{52}{7}-\frac{26}{5}$

$\displaystyle x=\frac{260-182}{35}$

$\displaystyle x=\frac{78}{35}$

$\therefore$ The number is

$\dfrac{78}{35}$

Solve:
$\dfrac{5}{12}+\dfrac{7}{16}$

$\begin{array}{l} \frac{5}{{12}} + \frac{7}{{16}} = \frac{{5*4 + 7*3}}{{48}} = \frac{{20 + 21}}{{48}}\\ = \frac{{41}}{{48}} \end{array}$

Solve the
$\dfrac{7}{9}-\dfrac{2}{5}$

$\dfrac{7}{9}-\dfrac{2}{5}=\dfrac{7\times 5-2\times 9}{45}=\dfrac{17}{45}$

Solve: $\dfrac{56}{8}-2\dfrac{3}{7}$ .

The mixed fraction,

$2\dfrac{3}{7}=\dfrac{2\times 7+3}{7}$

$=\dfrac{17}{7}$

So,

$\dfrac{56}{8}-\dfrac{17}{7}=\dfrac{56\times 7-17\times 8}{7\times 8}$

$=\dfrac{392-136}{56}$

$= \dfrac{256}{56}$

$= \dfrac{32}{7}$

$\therefore \dfrac{56}{8}-2\dfrac{3}{7}=\dfrac{32}{7}$

Solve:
$\dfrac{19}{5}+\dfrac{-7}{5}$

$=\dfrac{19}{5}+\dfrac{-7}{5}$

$=\dfrac{19-7}{5}$

$=\dfrac{12}{5}$

$=2.4$ .

Write the fraction whose numerator is one and equivalent to $\dfrac{9}{36}$ ?

$\begin{array}{l} \dfrac{9}{{36}}\\ = \dfrac{{9 \div 9}}{{36 \div 9}}\\ = \dfrac{1}{4} \end{array}$

Solve the
$\dfrac{-3}{7}+\dfrac{2}{3}$

$\dfrac{-3}{7}+\dfrac{2}{3}= \dfrac{-3\times 3+2\times7}{21}=\dfrac{5}{21}$

Solve $3\dfrac{2}{5}-2\dfrac{1}{4}$ .

$3\dfrac{2}{5} - 2\dfrac{1}{4} = \dfrac{{17}}{5} - \dfrac{9}{4} = \dfrac{{17 \times 4}}{{5 \times 4}} + \dfrac{{9 \times 5}}{{4 \times 5}} = \dfrac{{68}}{{20}} + \dfrac{{45}}{{20}} = \dfrac{{23}}{{20}}$

Solve
$\dfrac{-5}{6}+\dfrac{3}{11}$

$\dfrac{-5}{6}+\dfrac{3}{11}=\dfrac{-5\times 11 + 3\times 6}{66}=\dfrac{-55+18}{66}=\dfrac{-37}{66}$

Find $17-12\dfrac{2}{8}$ .

According to the problem,

$17 - 12 \times \dfrac{2}{8}$

$= 17 - 3$

$= 14$

Solve the
$\dfrac{-13}{7}+\dfrac{6}{7}$

$\dfrac{-13}{7} +\dfrac{6}{7}= \dfrac{-13+6}{7}=\dfrac{-7}{7} =-1$

Solve :
$\dfrac {5}{3} + \dfrac {3}{5}$ =?

$\Rightarrow \dfrac{5\times 5}{3 \times 5}+\dfrac{3}{5}\times \dfrac{3}{3}$

$\Rightarrow \dfrac{25}{15}+\dfrac{9}{15}$

$\Rightarrow \dfrac{34}{15}$

$\therefore \boxed {\dfrac{5}{3}+\dfrac{3}{5}=\dfrac{34}{15}}$

Covert $\dfrac{2}{3}, \dfrac{5}{7}$ pair of fraction into like fractions?

To convert

$\dfrac{2}{3},\dfrac{5}{7}$ into Like fractions, we need to multiply & divide both fractions with a number with which we get the same denominator.

$\dfrac{2}{3}\times \dfrac{7}{7}=\dfrac{14}{21}$

$\dfrac{5}{7}\times \dfrac{3}{3}=\dfrac{15}{21}$

$\therefore$ Like fractions are

$\dfrac{14}{21},\dfrac{15}{21}$

Simplify: $\dfrac {7}{24} - \dfrac {17}{36}$

Given:

$\dfrac{7}{24}-\dfrac{17}{36}$

$LCM (24,36)=2 \times 2 \times 2 \times 3 \times 1 \times 3$

$=72$

Now,

$\dfrac{7}{24}-\dfrac{17}{36}$

$=\dfrac{7 \times 3}{24 \times 3}-\dfrac{17 \times 3}{36 \times 2}$

$=\dfrac{21}{72}-\dfrac{34}{72}=\dfrac{21-34}{72}=\dfrac{-13}{72}$

Find equivalent fraction $\dfrac{3}{14}$ having

a) numerator $42$ b) denominator $132$

(a) We know that

$3\times 14=42$

$\therefore$ Multiplying numerator and denominator by 14, we get

$\dfrac{3\times 14}{12\times 14}=\dfrac{42}{168}$

$\therefore$ Equivalent fraction of

$\dfrac{3}{12}$ having numerator 42 is =

$\dfrac{42}{168}$

(b) We know that

$12\times 11=132$

$\therefore$ Multiplying numerator & denominator by 11, we get

$\dfrac{3\times 11}{12\times 11}=\dfrac{33}{132}$

$\therefore$ Equivalent fraction of

$\dfrac{3}{12}$ having denominator 132 is

$=\dfrac{33}{132}$

Convert $1\dfrac{4}{11}$ into improper fraction?

$1\dfrac{4}{11}$

$=\dfrac{(11\times 1)+4}{11}$

$=\dfrac{11+4}{11}$

$=\dfrac{15}{11}$

$\therefore 1\dfrac{4}{11}=\dfrac{15}{11}$

Substact the sum of $\dfrac{1}{2}$ and $-\dfrac{1}{4}$ from the sum of $\dfrac{3}{8}$ and $\dfrac{1}{4}$

Required value =

$(\dfrac{3}{8}+\dfrac{1}{4})-(\dfrac{1}{2}-\dfrac{1}{4})$

$=\dfrac{5}{8}-\dfrac{1}{4}$

$=\dfrac{3}{8}$

Solve :
$\dfrac {-2}{3} + 0$

$\dfrac{-2}{3}+0$

$\Rightarrow \dfrac{-2}{3}+\underline{0}\left ( \dfrac{3}{3} \right )$

$\Rightarrow \dfrac{-2+0(3)}{3}$

$\Rightarrow \dfrac{-2+0}{3}\Rightarrow \boxed{\dfrac{-2}{3}}$

Find $\dfrac{3}{7} + \left(\dfrac{-6}{11} \right) + \left(\dfrac{-8}{21} \right) + \dfrac{5}{22}$

$\cfrac{3}{7}+\left ( -\cfrac{6}{11} \right )+\left ( -\cfrac{8}{21} \right )+\cfrac{5}{22}$

$=\cfrac{3}{7} -\cfrac{6}{11} -\cfrac{8}{21}+\cfrac{5}{22}$

$=\left (\cfrac{3}{7} -\cfrac{8}{21} \right )+\left ( \cfrac{5}{22}-\cfrac{6}{11}\right )$

$=\cfrac{9-8}{21}+\cfrac{5-12}{22}$

$=\cfrac{1}{21}+\cfrac{-7}{22}$

$=\cfrac{1}{21}-\cfrac{7}{22}$

$=\cfrac{22-147}{21\times 22}$

$=-\cfrac{125}{462}$

Find :
$\dfrac {-3}{8} - \dfrac {7}{11}$

$\dfrac{-3}{8}-\dfrac{7}{11}$

$\Rightarrow\dfrac{-3\times 11}{8\times 11}-\dfrac{7\times 8}{11\times 8}$

$\Rightarrow\dfrac{-33}{88}-\dfrac{56}{88}$

$\Rightarrow\dfrac{-89}{88}$

$\therefore \boxed{\dfrac{-3}{8}-\dfrac{7}{11}=\frac{-89}{88}}$

Find :
$-2\dfrac {1}{3} + 4 \dfrac {3}{5}$

$\Rightarrow-2\dfrac{1}{3}+4\dfrac{3}{5}$

$\Rightarrow-\dfrac{(3\times 2)+1}{3}+\dfrac{(4\times 5)+3}{5}$

$\Rightarrow-\dfrac{7}{3}+\dfrac{23}{5}$

$\Rightarrow\dfrac{-7\times 5}{3\times 5}+\dfrac{23\times 3}{5\times 3}$

$\Rightarrow\dfrac{35}{15}+\dfrac{69}{15}$

$\Rightarrow\dfrac{34}{15}\Rightarrow\dfrac{(15\times 2)+4}{15}\Rightarrow2\dfrac{4}{15}$

$\therefore \boxed{-2\dfrac{1}{3}+4\dfrac{3}{5}=2\dfrac{4}{15}}$

Add
$\dfrac{\left ( -447 \right )}{495}+\dfrac{745}{\left ( 990 \right )}$

$\dfrac{{\left( { - 447} \right)}}{{495}} + \dfrac{{745}}{{990}}$

implies that,

$= \dfrac {{ - 864 + 745}}{{990}}$

$= \dfrac {{ - 149}}{{990}}$

Solve : $\dfrac{(1.3)^2}{6 \times 0.315} + \dfrac{0.315}{2}$

$\dfrac{(1.3)^2}{6\times 0.315}+\dfrac{0.315}{2}$

$\dfrac{1.69}{1.89}+0.1575$

$0.8941+0.1575$

$=1.0516$

Solve:
(i) $1 \dfrac{2}{3} - 3\dfrac{5}{6}$ (ii) $4 \dfrac{1}{2} - 3\dfrac{1}{3}$

(i)

$1\dfrac{2}{3}-3\dfrac{5}{6}$

$=\dfrac{(3\times 1)+2}{3}-\dfrac{(6\times 3)+5}{6}$

$=\dfrac{3+2}{3}-\dfrac{18+5}{6}$

$=\dfrac{5}{3}-\dfrac{23}{6}$

$=\dfrac{10-23}{6}$

$=\dfrac{-13}{6}$

$=-2\dfrac{1}{6}$

$\therefore 1\dfrac{2}{3}-3\dfrac{5}{6}=-2\dfrac{1}{6}$ .

(ii)

$4\dfrac{1}{2}-3\dfrac{1}{3}$

$=\dfrac{(2\times 4)+1}{2}-\dfrac{(3\times 3)+1}{3}$

$=\dfrac{8+1}{2}-\dfrac{9+1}{3}$

$=\dfrac{9}{2}-\dfrac{10}{3}$

$=\dfrac{27-20}{6}$

$=\dfrac{7}{6}$

$=1\dfrac{1}{6}$

$\therefore 4\dfrac{1}{2}-3\dfrac{1}{3}=1\dfrac{1}{6}$ .

Express the following as improper fractions:
$7\dfrac{3}{4}$
$5\dfrac{6}{7}$
$2\dfrac{5}{6}$
$10\dfrac{3}{5}$

$\Rightarrow 7\dfrac{3}{4}=\dfrac{31}{4}$

$\Rightarrow 5\dfrac{6}{7}=\dfrac{41}{7}$

$\Rightarrow 2\dfrac{5}{6}=\dfrac{17}{6}$

$\Rightarrow 10\dfrac{3}{5}=\dfrac{53}{3}$ .

1198026_1158276_ans_8a63f33284e24ebdb529c5586c53d678.jpg

$75 \ paise$ is ________ part of a rupee

$\rightarrow$ 1 Rs = 100 P

$\therefore 75 P = x$ part of 100 P

$\therefore x = \dfrac{75 P }{100 P}$

$\therefore x = \dfrac{3}{4}$

Ans :-

$(\dfrac{3}{4})^{th}$

1115291_1165621_ans_91f67af5f0144a19af7488ff875a6708.jpeg

Solve: $1 \dfrac{2}{3} + 2\dfrac{4}{5}$

$1\dfrac{2}{3}+2\dfrac{4}{5}$

$=\dfrac{(3\times 1)+2}{3}+\dfrac{(5\times 2)+4}{5}$

$=\dfrac{5}{3}+\dfrac{14}{5}$

$=\dfrac{25+42}{15}$

$=\dfrac{67}{15}$

$=4\dfrac{7}{15}$

$\therefore 1\dfrac{2}{3}+2\dfrac{4}{5}=4\dfrac{3}{15}$ .

Solve:
$\dfrac {5}{4} + \left( \dfrac {-11}{4} \right)$

$\dfrac{5}{4} + \left(\dfrac{-11}{4}\right) = \dfrac{5 - 11}{4} = \dfrac{- 6}{4} = \dfrac{-3}{2}$ (Ans)

Verify the following equality
$\left(\dfrac{-3}{5}+\dfrac{1}{6}\right)+\dfrac{9}{10}=\dfrac{-3}{5}+\left(\dfrac{1}{6}+\dfrac{9}{10}\right)$

Using BODMAS rule,

Solving LHS, we get

$\dfrac{-13}{30}+\dfrac{9}{10}$

$\dfrac{14}{30}=\dfrac{7}{15}$

Now solving RHS, we get

$\dfrac{-3}{5}+\dfrac{16}{15}$

$\dfrac{7}{15}$ = LHS

Hence proved.

Solve $\dfrac{7}{18}+\dfrac{11}{18}-1$

$\dfrac{7}{{18}} + \dfrac{{11}}{{18}} - 1$

$= \dfrac{{7 + 11 - 18}}{{18}}$

$= \dfrac{0}{{18}}$

$= 0$

Find the sum:
(i) $\frac { 5 }{ 4 } +\left( \frac { -11 }{ 4 } \right)$

By taking LCM, we get

$\implies\dfrac{5-11}{4}$

$\implies\dfrac{-6}{4}$

$\implies\dfrac{-3}{2}$

$1\dfrac{1}{2}$ .convert to mix fraction.

Solution:-

Any fraction is given in the form

$x \dfrac{y}{z}$ can be converted into

mixed fraction by following the way

$\Rightarrow x\dfrac{y}{z} = \dfrac{(x\times z+y)}{z}$

Now given

$\Rightarrow 1\dfrac{1}{2} = \dfrac{(2\times1)+1}{2}=\dfrac{3}{2}$ Ans

Solve : $\left[\dfrac{4}{5} +\left\{\dfrac{3}{5} + \left(\dfrac{2}{5} - \dfrac{1}{6} \right) \right\} + \dfrac{1}{2}\right] - \dfrac{2}{3}$

$\left [\dfrac{4}{5}+\left \{ \dfrac{3}{5}+\left ( \dfrac{2}{5}-\dfrac{1}{6} \right ) \right \} +\dfrac{1}{2} \right ]-\dfrac{2}{3}$

$\left [\dfrac{4}{5}+\left \{ \dfrac{3}{5}+\dfrac{7}{30} \right \} +\dfrac{1}{2} \right ]-\dfrac{2}{3}$

$\left [\dfrac{4}{5}+ \dfrac{5}{6}+\dfrac{1}{2} \right ]-\dfrac{2}{3}$

$\dfrac{32}{15}-\dfrac{2}{3}$

$\dfrac{22}{15}$

Find the value.
$\dfrac{-3}{7}=\dfrac{...}{14}=\dfrac{9}{...}=\dfrac{-6}{...}$

$\dfrac{{ - 3}}{7} = \dfrac{{ - 3 \times 2}}{{7 \times 2}} = \dfrac{{ - 6}}{{14}} = \dfrac{{ - 6 \times \dfrac{9}{{ - 6}}}}{{14 \times \dfrac{9}{{ - 6}}}} = \dfrac{9}{{ - 21}} = \dfrac{{ - 6}}{{14}}$

Fractions:-
(ii) $\dfrac{9}{5}-\dfrac{6}{7}+\dfrac{3}{2}-\dfrac{5}{4}.$

$\dfrac{9}{5}-\dfrac{6}{7}+\dfrac{3}{2}-\dfrac{5}{4}$

$=\dfrac{9\times 7}{5\times 7}-\dfrac{6\times 5}{7\times 5}+\dfrac{3\times 2}{2\times 2}-\dfrac{5}{4}$

$=\dfrac{63-30}{35}+\dfrac{6-5}{4}$

$\dfrac{33}{35}+\dfrac{1}{4}=\dfrac{33\times 4}{35\times 4}+\dfrac{1\times 35}{4\times 35}$

$=\dfrac{132+35}{140}=\dfrac{167}{140}$

Solve $\dfrac{2}{3} - \dfrac{1}{9}$

$\frac{2}{3}-\frac{1}{9}$

let us multiply 3 in both numerator and denominator of

$\frac{2}{3}$

$\frac{2\times 3}{3\times 3}-\frac{1}{9}$

$\frac{6}{9}-\frac{1}{9}$

$\frac{6-1}{9}$

$\frac{5}{9}$

Fractions:-
(i) $\dfrac{3}{5}+\dfrac{6}{4}.$

$\dfrac{3}{5} + \dfrac{6}{4} = \dfrac{3\times4}{5\times4} + \dfrac{6\times5}{4\times5} = \dfrac{12 + 30}{20} = \dfrac{42}{20} = \dfrac{21}{10}.$

Work out the following. Simplify your answers where possible.

a) $7 + \frac{7}{8} = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_$
b) $14 + \frac{7}{9} = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_$

$a)\ 7+\dfrac{7}{8}=\dfrac{7\times8}{8}+\dfrac{7}{8}=\dfrac{56+7}{8}=\dfrac{63}{8}$

$b)\ 14+\dfrac{7}{9}=\dfrac{14\times9}{9}+\dfrac{7}{9}=\dfrac{126+7}{9}=\dfrac{133}{9}$

What is the fraction of Rs. $1$ to $50$ paise

In Rs

$1$ there is

$100$ paise

So, the fraction will be

$= \dfrac {100}{50}$

$= {2}$

If $\dfrac{1}{5}$ of a number is subtracted from $\dfrac{1}{3}$ of that number, the result is $6$ . Find the number.

Let the required number be

$x$ .

Now according to the question

$\cfrac{x}{3} - \cfrac{x}{5} = 6$

$\cfrac{5x - 3x}{15} = 6$

$2x = 15 \times 6$

$x = \cfrac{90}{2} = 45$

Hence the required number is

$45$ .

Fill $>, <$ or $=$ sign
$\cfrac{15}{42}$ $\square$ $\cfrac{15}{19}$

$\displaystyle \frac{15}{42} \square \frac{15}{19}$

$(15\times 19) < \dfrac{15}{19}$

$(15\times 19) \boxed{<} (15\times 42)$

$\therefore \dfrac{15}{42} < \dfrac{15}{19}$

Find the value:-
$\frac{3}{{13}} + \frac{2}{7}$

$\dfrac{3}{13}+\dfrac{2}{7}$

$=\dfrac{3}{13}\times \dfrac{7}{7}+\dfrac{2}{7}\times\dfrac{13}{13}$

$=\dfrac{21}{91}+\dfrac{26}{91}$

$=\dfrac{21+26}{91}=\dfrac{47}{91}$

Find the sum :
$\dfrac{-3}{-11}+\dfrac{5}{9}$

$\dfrac{-3}{-11} + \dfrac{5}{9} = \dfrac{3}{11} + \dfrac{5}{9} = \dfrac{(3\times 9)+(5\times 11)}{99} = \dfrac{27+55}{99} = \dfrac{82}{99}$

Solve:-
$\left ( \dfrac{1}{2} \right )^{3}+\left ( \dfrac{1}{3} \right )^{3}-\left ( \dfrac{5}{6} \right )^{3}$

According to the problem,

${\left( {\dfrac{1}{2}} \right)^3} + {\left( {\dfrac{1}{3}} \right)^3} - {\left( {\dfrac{5}{6}} \right)^3}$

$= \dfrac{1}{8} + \dfrac{1}{{27}} + \dfrac{{125}}{{216}}$

$= \dfrac{{27 + 8 + 125}}{{216}}$

$= \dfrac{{160}}{{216}}$

$= 0.740$

Add: $\dfrac{5}{3} + \dfrac{3}{5}$

Given that:

$\dfrac{5}{3} + \dfrac{3}{5}$

taking LCM

$=\dfrac{5\times5+3\times3}{15}$

$=\dfrac{25+9}{15}$

$=\dfrac{34}{15}$

Solve:
$\frac{{ - 2}}{3} + 0$

$\dfrac{-2}{3} + 0$

$\dfrac{-2}{3} + \frac{0}{1}$

$\dfrac{-2 + 0}{3}$

$\dfrac{-2}{3}$

Solve :
$\dfrac{-9}{10}+\dfrac{22}{15}$

$\dfrac{-9}{10} + \dfrac{22}{15} = \dfrac{(-9\times 3) + (22 \times 2)}{30} = \dfrac{-27 + 44}{30} = \dfrac{17}{30}$ (Ans)

Fill in the boxes with the symbols '<' or '>' to make the given statements true:
$\dfrac{5}{11} \Box \dfrac{3}{7}$
$\dfrac{8}{15} \Box \dfrac{3}{5}$
$\dfrac{11}{14} \Box \dfrac{29}{35}$
$\dfrac{13}{27} \Box \dfrac{15}{48}$

$(a)\ \dfrac{5}{11} \boxed{>} \dfrac{3}{7}$

By taking LCM, we get

$\dfrac{35>33}{77}$

$(b)\ \dfrac{8}{15} \boxed{<} \dfrac{3}{5}$

By taking LCM, we get

$\dfrac{8<9}{15}$

$(c)\ \dfrac{11}{14} \boxed{<} \dfrac{29}{35}$

By taking LCM, we get

$\dfrac{55<58}{70}$

$(d)\ \dfrac{13}{27} \boxed{>} \dfrac{15}{48}$

$\dfrac{208>135}{432}$

Arrange the following rational number in descending order
$\frac{17}{-30},\frac{11}{-15},\frac{-7}{10},\frac{3}{5}.$

Rational numbers can be arranged in descending

all ascending order by making their denominates equal.

Therefore, LCM of 30, 15, 10 and 5 is 30.

So,

$\frac{-17}{30} = \frac{-17}{30}; \frac{-7}{10}= \frac{21}{30}$

$\frac{-11}{15} = \frac{-22}{30} ; \frac{3}{5} = \frac{18}{30}$

so, in descending order

$\rightarrow$

$\frac{18}{30}> \frac{-17}{30}> \frac{-21}{30}> \frac{-22}{30}$

$= \frac{3}{5}> \frac{-17}{30} > \frac{-7}{10}> \frac{-11}{15}$

1114578_1201727_ans_5e26e98e080f4231918544fac097a9f7.jpg

Solve:
$\frac{{ - 9}}{{10}} + \frac{{32}}{{15}}$

Given that:

$\frac{-9}{10}+\frac{32}{15}$

Taking LCM=30

$=\frac{(-9\times3)+32\times2}{30}$

$=\frac{-27+64}{30}$

$=\frac{37}{30}$

Answer

$7\frac{1}{2} - \left[ {2\frac{1}{4} + \left\{ {1\frac{1}{4} - \frac{1}{2}\left( {\frac{3}{2} - \frac{1}{3} - \frac{1}{6}} \right)} \right\}} \right]$

$=7 \frac {1}{2}-[2 \frac {1}{4}+$ {

$1 \frac {1}{4}- \frac {1}{2}(\frac {3}{2}-\frac {1}{3}-\frac {1}{6})$ }

$]$

By simplifying mixed fractions

$= \frac {15}{2}-[ \frac {9}{4}+$ {

$\frac {5}{4}- \frac {1}{2}(\frac {3}{2}-\frac {1}{3}-\frac {1}{6})$ }

$]$

According to BODMAS

$= \frac {15}{2}-[ \frac {9}{4}+$ {

$\frac {5}{4}- \frac {1}{2}(\frac {3(3)-1(2)-1}{6})$ }

$]$

$= \frac {15}{2}-[ \frac {9}{4}+$ {

$\frac {5}{4}- \frac {1}{2}(\frac {9-2-1}{6})$ }

$]$

$= \frac {15}{2}-[ \frac {9}{4}+$ {

$\frac {5}{4}- \frac {1}{2}(\frac {6}{6})$ }

$]$

$= \frac {15}{2}-[ \frac {9}{4}+$ {

$\frac {5}{4}- \frac {1}{2}(1)$ }

$]$

$= \frac {15}{2}-[ \frac {9}{4}+$ {

$\frac {5}{4}- \frac {1}{2}$ }

$]$

$= \frac {15}{2}-[ \frac {9}{4}+$ {

$\frac {5(1)-1(2)}{4}$ }

$]$

$= \frac {15}{2}-[ \frac {9}{4}+$ {

$\frac {3}{4}$ }

$]$

$= \frac {15}{2}-[ \frac {9}{4}+$

$\frac {3}{4}$

$]$

$= \frac {15}{2}-[ \frac {12}{4}]$

$= \frac {15}{2}-[ \frac {6}{2}]$

$= \frac {15}{2}- \frac {6}{2}$

$= \frac {15-6}{2}$

$= \frac {9}{2}$

$= 4\frac {1}{2}$

Evaluate : (i) $\dfrac{6}{(-25)}$ - $\dfrac{4}{(-15)}$

We have,

$\Rightarrow \dfrac{6}{\left( -25 \right)}-\dfrac{4}{\left( -15 \right)}$

$\Rightarrow -\dfrac{6}{25}+\dfrac{4}{15}$

$\Rightarrow \dfrac{4}{15}-\dfrac{6}{25}$

$\Rightarrow \dfrac{20-18}{75}$

$\Rightarrow \dfrac{2}{75}$

Hence, this is the answer.

Substract: $\dfrac{-4}{15}$ from $\dfrac{-7}{15}$

$\quad \frac { -7 }{ 15 } -\frac { -4 }{ 15 } \\ =\frac { -7-\left( -4 \right) }{ 15 } \\ =\frac { -7+4 }{ 15 } \\ =\frac { -3 }{ 15 } \\ =\frac { -1 }{ 5 }$

Substract: $\dfrac{-3}{8}$ from $\dfrac{5}{8}$

$\quad \frac { 5 }{ 8 } -\frac { -3 }{ 8 } \\ =\frac { 5-\left( -3 \right) }{ 8 } \\ =\frac { 8 }{ 8 } \\ =1$

Solve: $4+\dfrac{7}{8}.$

Given

$4+\dfrac{7}{8}$

$=\dfrac{32+7}{8}$

$=\dfrac{39}{8}$

Substract: $\dfrac{2}{5}$ from $\dfrac{3}{5}$

$\quad \frac { 3 }{ 5 } -\frac { 2 }{ 5 } \\ =\frac { 3-2 }{ 5 } \\ =\frac { 1 }{ 5 }$

Solve: $2\dfrac{2}{3}+3\dfrac{1}{2}.$

Given

$2\dfrac{2}{3}+3\dfrac{1}{2}$

$=\dfrac{8}{3}+\dfrac{7}{2}$

$=\dfrac{16+21}{6}$

$=\dfrac{37}{6}$

Solve: $\dfrac{7}{10}+\dfrac{2}{5}+\dfrac{3}{2}.$

$\dfrac{7}{10}+\dfrac{2}{5}+\dfrac{3}{2}$

$=\dfrac{7}{10}+\dfrac{4+15}{10}$

$=\dfrac{7}{10}+\dfrac{19}{10}$

$=\dfrac{7+19}{10}$

$=\dfrac{26}{10}$

Simi's book is $\frac{3}{5}$ kg and Soma's book is $\frac{3}{8}$ kg in weight. Whose book is heavier and by how much?

Simi's book is heavier and by 9/40 kg .

The value of $\left( \cfrac { 3 }{ 5 } +\cfrac { 4 }{ 7 } \right) -\cfrac { 3 }{ 7 }$

Given

$\left(\dfrac{3}{5}+\dfrac{4}{7}\right)-\dfrac{3}{7}$

$=\left(\dfrac{21+20}{35}\right)-\dfrac{3}{7}$

$=\dfrac{41}{35}-\dfrac{3}{7}$

$=\dfrac{41-15}{35}$

$=\dfrac{26}{35}$

Find out the reciprocal of $\left( \frac { - 1 } { 8 } \right) ^ { 6 } .$

$\left (\dfrac {-1}{8}\right)^6=\dfrac {1}{8^6}=$ reciprocal

$=8^6$

What should be added to $\dfrac{-7}{12}$ as to get $\dfrac{9}{16}$

We have,

Let us suppose;

we add 'x' to

$-\dfrac{7}{12}$

$x+\left(-\dfrac{7}{12}\right)=\dfrac{9}{16}$

$x-\dfrac{7}{12}=\dfrac{9}{16}$

$x-\dfrac{9}{16}+\dfrac{7}{12}$

Now take "LCM".

$x=\dfrac{55}{48}$

Hence, this is answer.

1195654_1242830_ans_244c0acb10ce4b3fa96f1eaef335cbff.jpg

Solve $1 + \frac{1}{8} - \frac{{27}}{8}$

$1+ \dfrac18-\dfrac{27} {8}$

$=1- \dfrac{26}{8}$

$=-\dfrac{18}{8}$

Substract: $\dfrac{-3}{7}$ from $\dfrac{5}{4}$

$\quad \frac { 5 }{ 4 } -\frac { -3 }{ 7 } \\ =\frac { 5\times 7-\left( -3\times 4 \right) }{ 4\times 7 } \\ =\frac { 35-\left( -12 \right) }{ 28 } \\ =\frac { 35+12 }{ 28 } \\ =\frac { 47 }{ 28 }$

$2\dfrac{3}{11}+4\dfrac{1}{9}+\dfrac{1}{3}$

Consider the given expression.

$2\dfrac{3}{11}+4\dfrac{1}{9}+\dfrac{1}{3}$

$=\dfrac{25}{11}+\dfrac{37}{9}+\dfrac{1}{3}$

$=\dfrac{25}{11}+\dfrac{37}{9}+\dfrac{3}{9}$

$=\dfrac{25}{11}+\dfrac{40}{9}$

$=\dfrac{25\times 9+40 \times 11}{11\times 9}$

$=\dfrac{225+440}{99}$

$=\dfrac{665}{99}$

$=6\dfrac{71}{99}$

Hence, this is the answer.

Mixed numbers always represent a -----------------.

Mixed numbers always represent a fraction.

Eg :

$2 \dfrac{5}{3}$ is a fraction.

Simplify: $\dfrac{8}{-15}+\dfrac{4}{-3}$

$\cfrac { -8 }{ 15 } -\cfrac { 4 }{ 3 } =\cfrac { -8-4\left( 5 \right) }{ 15 } =\cfrac { -28 }{ 15 }$

Evaluate
$\dfrac{1}{2}+6\dfrac{2}{7}+\dfrac{2}{5}=$

We have,

$\dfrac{1}{2}+\dfrac{2}{5}+6\dfrac{2}{7}$

$=\dfrac{503}{70}$

$=7\dfrac{13}{70}$

$=7.185714$

Hence, this is answer.

1197647_1242943_ans_7fe3e076fa53421b86ed120fc678cc58.png

Write the following surd in simplest from .
$\dfrac {5}{9} \sqrt {45}$

1338764_1264350_ans_5d4dd2636d2d4781bd1be5f3dc75de71.jpg

Express 1 day as a fraction of 1 week.

1338089_1290858_ans_6b37bc9c2b29425aab04eba32f123614.PNG

Solve:
$3\frac{2}{5} - 2\frac {1}{4}$

$3\dfrac{2}{5}-2\dfrac{1}{4}=\dfrac{17}{5}-\dfrac{9}{4}=\dfrac{68-45}{20}=\dfrac{23}{20}$

Convert into mixed fractions.
(a) $\frac{3}{2}$
(b) $\frac{24}{5}$

$a)\frac{3}{2} = 1\frac{1}{2}$

$b)\frac{24}{5}=4\frac{4}{5}$

1130964_1247055_ans_822f07d0aabc41b284be335d38b08735.jpg

Solve: $\dfrac{4}{9} + \dfrac{2}{7}$

For the

$1^{st}$ fraction, since

$9\times 7=63$

$\dfrac{4}{9}=\dfrac{4\times7}{9\times7}=\dfrac{28}{63}$

likewise for the

$2^{nd}$ fraction,since

$7\times 9=63$

$\dfrac{2}{7}=\dfrac{2\times9}{7\times9}=\dfrac{18}{63}$

Add the two fraction

$\dfrac{28}{63}=\dfrac{28+18}{63}=\dfrac{46}{63}$

Solve: $\dfrac{5}{6} - \dfrac{2}{9}$

1117398_1278041_ans_c5b00a9f787a49229b725b57bd78d992.JPG

Solve: $\dfrac{3}{4} - \dfrac{1}{2}$

1117397_1278027_ans_0850677e6a3245f0a3f8599d28217f96.JPG

Express 20 seconds as a fraction of 1 minute.

1338091_1290864_ans_dbdfbffe6ada45bd86de34a94705c62b.PNG

Solve: $8\dfrac{3}{10} - 4\dfrac{4}{5}$

1117402_1278057_ans_0a08c13a7e4f4126bd3c86c362d2fdb9.JPG

Draw number lines and locate the point on them :
$\dfrac {2}{5},\dfrac {3}{5},\dfrac {8}{5},\dfrac {4}{5}$

Given points are located on number line.
1235385_1298227_ans_67804873ece44d8e8fbdee09998a5ef9.png

$\dfrac{2}{3}$ + $\dfrac{2}{4}$

We have,

$\dfrac{2}{3} + \dfrac{2}{4}$

$= \dfrac{{8 + 6}}{{12}}$

$= \dfrac{{14}}{{12}}$

Subtract $\dfrac {-5}{6}$ from $\dfrac {1}{3}$ .

Since the denominators are different we need to take LCM
The LCM of

$6$ and

$3$ is

$6$

$\therefore$ By multiplying both numerator and denominator with

$1$ we get,

$\dfrac {-5}{6} = \dfrac {-5\times 1}{6\times 1} = \dfrac {-5}{6}$
And by multiplying both numerator and denominator with

$2$ we get,

$\dfrac {1}{3} = \dfrac {1\times 2}{3\times 2} = \dfrac {2}{6}$
Now by subtracting

$\dfrac {1}{3} - \dfrac {-5}{6} = \dfrac {2}{6} - \dfrac {-5}{6}$
Since we have the same denominators we can subtract directly,

$\dfrac {2 - (-5)}{6}\Rightarrow \dfrac {7}{6}$ .

Solve: $2\dfrac{3}{4}+4\dfrac{2}{3}+1$

$\begin{array}{l} 2\dfrac { 3 }{ 4 } +4\dfrac { 2 }{ 3 } +1 \\ Now, \\ =\dfrac { { 11 } }{ 4 } +\dfrac { { 14 } }{ 3 } +1 \\ =\dfrac { { 33+56+12} }{ { 12 } } \\ Hence, \\ =\dfrac { { 101 } }{ { 12 } } \end{array}$

Wrote the following as mixed fraction
$\dfrac { 7 }{ 2 } ,\dfrac { 8 }{ 5 }$

The mixed fraction of:-

$\begin{array}{l} \left( i \right) \dfrac { 7 }{ 2 } is\, \, 3\dfrac { 1 }{ 2 } \\ \left( { ii } \right) \dfrac { 8 }{ 5 } \, is\, \, 1\dfrac { 3 }{ 5 } \end{array}$

simplify :
$2\dfrac{3}{5}+1\dfrac{3}{10}-3\dfrac{2}{15}$

$2 \cfrac{3}{5} + 1 \cfrac{3}{10} - 3 \cfrac{2}{15}$

$= \cfrac{10 + 3}{5} + \cfrac{10 + 3}{10} - \cfrac{45 + 2}{15}$

$= \cfrac{13}{5} + \cfrac{13}{10} - \cfrac{47}{15]}$

$= \cfrac{78 + 39 - 94}{30} = \cfrac{23}{30}$

Express $10$ months as a fraction of $1$ year.

$1$ year

$=12$ months

$1$ month

$=\dfrac{1}{12}$ years

$10$ month

$=\dfrac{10}{12}$ yr

$=\dfrac{5}{6}$ yr.

Javed was given $\dfrac{5}{7}$ of a basket of oranges. What fraction of oranges was left in the basket.

Let the remaining fraction be

$x$

$\dfrac{5}{7}+x=1$ (1 is whole)

$\Rightarrow x=1-\dfrac{5}{7}=\dfrac{2}{7}$

Subtract: $\dfrac {-18}{11}$ from $\dfrac {1}{1}$ .

Since the denominators are different we need to take LCM
The LCM of

$11$ and

$1$ is

$11$

$\therefore$ By multiplying both numerator and denominator with

$1$ we get,

$\dfrac {-18}{11} = \dfrac {-18 \times 1}{11\times 1} = \dfrac {-18}{11}$
And by multiplying both numerator and denominator with

$11$ we get,

$\dfrac {1}{1} = \dfrac {1\times 11}{1\times 11} = \dfrac {11}{11}$
Now by subtracting

$\dfrac {1}{1} - \dfrac {-18}{11} = \dfrac {11}{11} - \dfrac {-18}{11}$
Since we have the same denominators we can subtract directly,

$\dfrac {11 - (-18)}{11} = \dfrac {11 + 18}{11}\Rightarrow \dfrac {29}{11}$ .

$\dfrac { 9 }{ 11 } -\dfrac { 4 }{ 15 }$

$\dfrac{9}{11}$ -

$\dfrac{4}{15}$

$\dfrac{{9\times15}-{4\times11}}{11\times15}$

$\dfrac{91}{165}$

$\dfrac{1}{22}+\dfrac{21}{22}$ .

$\dfrac{1}{22}+\dfrac{21}{22}$

$\dfrac{21+1}{22}$

$\dfrac{22}{22}$

$=1$

Simplify: $\dfrac{6}{9}-\dfrac{3}{18}$

$\frac{6}{9}-\frac{3}{18}= \frac{2}{3}-\frac{1}{6}=\frac{4}{6}-\frac{1}{6}=\frac{3}{6}=\frac{1}{2}$
1194231_1366045_ans_c995f342f9f14102baad43c89a028e94.jpg

Reduce the following fractions to simplest from:

(i) $\dfrac{48}{60}$

(ii) $\dfrac{150}{60}$

1) In

$\dfrac{48}{60},$

Dividing the numerator and denominator by

$12,$

We get

$\dfrac{48}{60} = \dfrac{4}{5}$

2) In

$\dfrac{150}{60},$

Dividing the numerator and denominator by

$30,$

we get

$\dfrac{150}{60} = \dfrac{5}{2}$

Solve:
(a) $\dfrac { 2 }{ 3 } +\dfrac { 1 }{ 7 }$ (b) $\dfrac { 3 }{ 10 } +\dfrac { 7 }{ 15 }$ (c) $\dfrac { 4 }{ 9 } +\dfrac { 2 }{ 7 }$ (d) $\dfrac { 5 }{ 7 } +\dfrac { 1 }{ 3 }$

(e) $\dfrac { 4 }{ 5 } +\dfrac { 2 }{ 3 }$ (f) $\dfrac { 3 }{ 4 } -\dfrac { 1 }{ 3 }$ (g) $\dfrac { 5 }{ 6 } -\dfrac { 1 }{ 3 }$

$(a) \dfrac{2}{3} + \dfrac{1}{7} = \dfrac{17}{21}$

$(b) \dfrac{3}{10} + \dfrac{7}{15} = \dfrac{45+70}{150} = \dfrac{115}{150} =\dfrac{23}{30}$

$(c) \dfrac{4}{9} + \dfrac{2}{7} = \dfrac{28+18}{63} =\dfrac{46}{63}$

$(d) \dfrac{5}{7} + \dfrac{1}{3} = \dfrac{22}{21}$

$(e) \dfrac{4}{5} + \dfrac{2}{3} = \dfrac{22}{15}$

$(f) \dfrac{3}{4} - \dfrac{1}{3} = \dfrac{5}{12}$

$(g) \dfrac{5}{6} - \dfrac{1}{3} = \dfrac{9}{18} =\dfrac{1}{2}$

Solve
$\dfrac {-8}{19}+\dfrac {(-2)}{57}$

$\frac{-8}{19}+(\frac{-2}{57})$

$\Rightarrow \frac{-8}{19}\div \frac{2}{57}=\frac{-24-2}{57}$

$\Rightarrow \frac{-26}{57}$

1212599_1364603_ans_59d5b32332004a048a0125d3cd9f09f1.JPG

Solve : $\dfrac{3}{4} - \dfrac{1}{3}$

$\dfrac{3}{4}-\dfrac{1}{3}$

$\dfrac{3\times3}{4\times3}-\dfrac{1\times4}{3\times4}$

$\dfrac{9}{12}-\dfrac{4}{12}$

$\dfrac{9-4}{12}=\dfrac{5}{12}$

Represent each of the following number on the number line:
$\dfrac {1}{3}$ .

We know that

$1/3$ is greater than

$0$ and lesser than

$1$ .

$\therefore$ it lies between

$0$ and

$1$ . It can be represented on number line as.
1602759_1366040_ans_f5d2354262e4449f904c4ebca20db53d.jpg

Simplify: $\dfrac{16}{39}+\dfrac{9}{-26}$

We have

$\dfrac{16}{39}+\dfrac{9}{-26}=\dfrac{16}{39}+\dfrac{-9}{26}$

Now, the L.C.M of

$39$ and

$26$ is

$78$

$\therefore$ Rewriting

$\dfrac{16}{39}$ and

$\dfrac{-9}{26}$ in such a way that they have the same denominator

$78$

$\dfrac{16}{39}=\dfrac{16\times 2}{39\times 2}=\dfrac{32}{78}$

$\dfrac{-9}{26}=\dfrac{-9\times 3}{26\times 3}=\dfrac{-27}{78}$

$\therefore \dfrac{16}{39}+\dfrac{-9}{26}=\dfrac{32}{78}+\dfrac{-27}{78}=\dfrac{32-27}{78}=\dfrac{5}{78}$

Subtract the following fraction and reduce to the lowest term $\dfrac{7}{15}$ from $\dfrac{13}{20}$

Given

$\frac{7}{15}$ form

$\frac{13}{20}$

now we have P=

$\frac{13}{20}-\frac{7}{15}$

Taking LCM

$P=\frac{13\times 3-7\times 4}{60}=$

$\frac{39-28}{60}$

$P=\frac{11}{60}$

1211574_1362741_ans_685b440ee0b1423585fb74302de8a110.JPG

Add: $\dfrac{3}{4}{a}^{2}b,\,\dfrac{2}{3}{a}^{2}b$

Let

$a=\dfrac{3}{4}{a}^{2}b+\dfrac{2}{3}{a}^{2}b$

$=\left(\dfrac{3}{4}+\dfrac{2}{3}\right){a}^{2}b$

$=\left(\dfrac{3}{4}\times \dfrac{3}{3}+\dfrac{2}{3}\times \dfrac{4}{4}\right){a}^{2}b$

$=\left(\dfrac{9}{12}+\dfrac{8}{12}\right){a}^{2}b$

$=\dfrac{9+8}{12}{a}^{2}b$

$a=\dfrac{17}{12}{a}^{2}b$

$\dfrac{3}{4}{a}^{2}b+\dfrac{2}{3}{a}^{2}b=\dfrac{17}{12}{a}^{2}b$

Add the following
$\dfrac {4}{7}\ and\ \dfrac {5}{7}$

$\dfrac{4}{5}+\dfrac{5}{7}$

$=\dfrac{4+5}{7}$

$=\dfrac{9}{7}$

$=1\dfrac{2}{7}$

Add: $\dfrac{3}{5}x \ \ ,\,\dfrac{2}{3}x \ \ \ \ ,\dfrac{-4}{5}x$

Let

$a=\dfrac{3}{5}x+\dfrac{2}{3}x-\dfrac{4}{5}x$

$=\left(\dfrac{3}{5}x-\dfrac{4}{5}x\right)+\dfrac{2}{3}x$

$=\dfrac{3-4}{5}x+\dfrac{2}{3}x$

$=\dfrac{-1}{5}x+\dfrac{2}{3}x$

$=\left(\dfrac{-1}{5}+\dfrac{2}{3}\right)x$

$=\dfrac{-3+10}{15}x$

$a=\dfrac{7}{15}x$

$\dfrac{3}{5}x+\dfrac{2}{3}x-\dfrac{4}{5}x=\dfrac{7}{15}x$

$5\dfrac {1}{3}-3\dfrac {1}{2}=$ _______

$5\dfrac{1}{3}-3\dfrac{1}{2}$

$=\dfrac{16}{3}-\dfrac{7}{2}$

$=\dfrac{32-21}{6}$

$=\dfrac{11}{6}$

$=1\dfrac{5}{6}$

Add: $\dfrac{5}{-11}$ and $\dfrac{8}{11}$

$\dfrac{5}{-11}+\dfrac{8}{11}$

$=\dfrac{-5}{11}+\dfrac{8}{11}$

$=\dfrac{-5+8}{11}=\dfrac{3}{11}$

Simplify :
$6+\{ \frac { 4 }{ 3 } +(\frac { 3 }{ 4 } -\frac { 1 }{ 3 } )\}$

$6+\{ \frac { 4 }{ 3 } +(\frac { 3 }{ 4 } -\frac { 1 }{ 3 } )\}$

$=6+\dfrac{4}{3} +\dfrac{3}{4} -\dfrac{1}{3}$

$=\dfrac{72+16+9-4}{12}$ =

$\dfrac{93}{12}$

Raju plucked two papayas. One of them weighed $\dfrac{1}{4}$ kg, and the other weighed $\dfrac{1}{2}$ kg, so what is their total weight.

Weight of first papaya

$=\dfrac{1}{4}$ kg

Weight of second papaya

$=\dfrac{1}{2}$ kg

Total weight

$=\dfrac{1}{4}+\dfrac{1}{2}$ kgs

$=\dfrac{1+2}{4}=\dfrac{3}{4}$ kg

$\therefore$ total weight

$=\dfrac{3}{4}$ kg

$=\dfrac{3}{4}\times 1000$ gms

$=750$ gms

Solve:
$\dfrac {-8}{19}+\dfrac {(-2)}{57}$

Solve

$\frac{-8}{19} + \frac{(-2)}{57}$

on taking LCM

$\frac{-8}{19} \frac{-2}{57}$

$\Rightarrow \frac{-24 -2}{57} = \frac{-26}{57}$ Ans

1215378_1368788_ans_2f33f3023b994ad7bb2f1e154f55f06a.jpg

Add : $\dfrac{5}{6}$ & $\dfrac{13}{15}$

$\dfrac{5}{6}+\dfrac{13}{15}=\dfrac{75+78}{90} = \dfrac{153}{90}$

Add the following.
$\dfrac { -5 }{ 3 } ,\dfrac { -4 }{ 3 } ,\dfrac { -3 }{ 3 }$

$\dfrac { -5 }{ 3 } +\dfrac { -4 }{ 3 } +\dfrac { -3 }{ 3 }$

$=\dfrac{-5-4-3}{3}$

$=\dfrac{-12}{3}$

$=-4$

Multiply and write your answer in the lowest form.
$\dfrac{4}{5} \times 2$

We've,

$\dfrac{4}{5} \times 2$

$=$

$\dfrac{8}{5}$

This is the required lowest form.

Find the interval between $7:15$ am and $12:15$ pm.

The interval between

$7:15$ am and

$12:15$ pm is

$12:15-7:15=5$ hrs.

The sum of two fraction is $5\cfrac { 3 }{ 9 }$ . If one of the fractions is $2\cfrac { 3 }{ 4 }$ , find the other fraction.

Let the other fraction be

$x$

Given one of the fraction is

$2\dfrac{3}{4}=2+\dfrac{3}{4}=\dfrac{8+3}{4}=\dfrac{11}{4}$

Given Sum of two fractions is

$5\dfrac{3}{9}=5+\dfrac{1}{3}=\dfrac{15+1}{3}=\dfrac{16}{3}$

$\implies x+\dfrac{11}{4}=\dfrac{16}{3}$

$\implies x=\dfrac{16}{3}-\dfrac{11}{4}=\dfrac{64-33}{12}=\dfrac{31}{12}=2\dfrac{7}{12}$

Therefore the other fraction is

$2\dfrac{7}{12}$

Convert $4\dfrac{5}{6}$ in to improper fraction

$4\dfrac{5}{6}\\4+\dfrac{5}{6}\\\dfrac{24}{6}+\dfrac{5}{6}\\\dfrac{24+5}{6}\\\dfrac{29}{6}$

Simplify:
$\dfrac {8}{10}+\dfrac {9}{10}$

$\dfrac{8}{10}+\dfrac{9}{10}=\dfrac{8+9}{10}=\dfrac{17}{10}$

Add the following.

$a)\dfrac{3}{24}+\dfrac{11}{24}$

$b)\dfrac{3}{10}+\dfrac{2}{10}+\dfrac{7}{10}$

$a)\dfrac{3}{24}+\dfrac{11}{24}$

$=\dfrac{3+11}{24}$

$=\dfrac{14}{24}=\dfrac{7}{12}$

$b)\dfrac{3}{10}+\dfrac{2}{10}+\dfrac{7}{10}$

$=\dfrac{3+2+7}{10}$

$=\dfrac{12}{10}=\dfrac{6}{5}=1\dfrac{1}{5}$

Simplify:
$3\dfrac {1}{4}+4\dfrac {5}{6}$

Now,

$3\dfrac {1}{4}+4\dfrac {5}{6}$

$=\dfrac{13}{4}+\dfrac{29}{6}$

$=\dfrac{39+58}{12}$

$=\dfrac{97}{12}$ .

Convert $3\dfrac{5}{6}$ into an improper fraction.

$3\dfrac{5}{6}$ into an improper fraction =

$\dfrac{(6 \times 3)+5}{6} = \dfrac{23}{6}$ .

Find the sum of $9\dfrac { 3 }{ 8 } +10\dfrac { 5 }{ 8 }$

$9\dfrac{3}{8}+10\dfrac{5}{8}$

$=\dfrac{75}{8}+\dfrac{85}{8}$

$=\dfrac{75+85}{8}=\dfrac{160}{8}=20$

Add.
$\dfrac {1}{3}+\dfrac {3}{4}+\dfrac {2}{5}$

$\dfrac{1}{3} + \dfrac{3}{4} + \dfrac{2}{5}$

$= \dfrac{(4\times 5)+(3\times 3\times 5)+(2\times 3\times 4)}{3\times 4 \times 5}$

$= \dfrac{20+ 45 + 24}{60}$

$= \dfrac{89}{60}$

Add $\dfrac { 3 }{ 7 }$ and $\dfrac { 2 }{ 3 }$ .

$\dfrac{3}{7}+\dfrac{2}{3}$

$=\dfrac{3\times 3}{7\times 3}+\dfrac{2\times 7}{3\times 7}$

$=\dfrac{9+14}{21}=\dfrac{23}{21}$

What number should be added to $\dfrac { -6 }{ 11 }$ so as to get $\dfrac { 4 }{ 9 }$ ?

Let the number to be added is

$x$ .

$x+\left(\dfrac{-6}{11}\right)=\dfrac{4}{9}$

$x=\dfrac{4}{9}+\dfrac{6}{11}$ {LCM

$(9,11)=99$ }

$x=\dfrac{4\times 11+6\times 9}{99}$

$x=\dfrac{44+54}{99}=\dfrac{98}{99}$ .

1199949_1445446_ans_0be663ef45fa4138a185916f4e9f996c.jpg

Find the sum of $\dfrac { 1 } { 3 } + \dfrac { 1 } { 4 }$ .

Sum is

$\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{4+3}{12}=\dfrac{7}{12}$

Solve: $6{1 \over 3} + 2{1 \over 3}$

$6\frac{1}{3}+2\frac{1}{3}= (6+2)[\frac{1}{3}+\frac{1}{3}]$

$= 8[\frac{2}{3}]=8\frac{2}{3}$

1214714_1431600_ans_1c54a38e10a247c1b39883eb4587d6d1.jpg

Compare: $\dfrac { 3 }{ 7 }$ and $\dfrac { 2 }{ 3 }$

$\dfrac{3}{7}$ and

$\dfrac{2}{3}$

L.C.M of

$7,3=21$

$\dfrac{3}{7}\times \dfrac{3}{3}=\dfrac{9}{21}$

and

$\dfrac{2}{3}\times \dfrac{7}{7}=\dfrac{14}{21}$

$14>9$

$\Rightarrow \dfrac{14}{21}>\dfrac{9}{21}$

$\therefore \dfrac{2}{3}>\dfrac{3}{7}$

The sum of two numbers is $11$ and the sum of their reciprocals is $\dfrac{{11}}{{28}}$ Find the numbers.

Given sum of two numbers is

$11$

$x+y=11$ _____ (1)

sum of their reciprocals is

$\dfrac{11}{28}$ .

$\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{11}{48}$

$\dfrac{y+x}{xy}=\dfrac{11}{48}$

$28(x+y)= 11xy$ _______ (2)

$28(11)= 11.xy$

$xy=28$

Now

$(x-y)^{2}=x^{2}+y^{2}-2xy\\$

$(x-y)^{2}=(x+y)^{2}-4xy\\$

$(x-y)^{2}=(11)^{2}-4\times28\\$

$(x-y)^{2}=121-112\\$

$(x-y)^{2}=9\\$

$x-y=\sqrt{9}\\$

$x-y=3$ _______ (3)

On adding (1) & (3)

$x+y=1$

$x-y=3$

_________

$2x=14\\$

$x=\dfrac{14}{2}\\$

$x=7\\$

putting

$x=7$ in equation _____ (1)

$x+y=11\\$

$7 +y=11\\$

$y=11-7\\$

$y=4\\$

$\therefore$ The two numbers are

$(x,y)=(7,4)$

simplify:
$1 \frac{1}{2}+\frac{4}{9}+13 \frac{1}{2}$

$1\frac{1}{2}=1+\dfrac{1}{2}=\dfrac{2+1}{2}=\dfrac{3}{2}$

$13\frac{1}{2}=13+\frac{1}{2}=\dfrac{26+1}{2}=\dfrac{27}{2}$

$1\frac{1}{2}+\frac{4}{9}+13\frac{1}{2}=\dfrac{3}{2}+\dfrac{4}{9}+\dfrac{27}{2}=\dfrac{27+8+243}{18}=\dfrac{278}{18}=\dfrac{139}{9}=15\dfrac{4}{9}$

Find :
$\dfrac { 2 }{ 5 } +\dfrac { 3 }{ 4 } +\dfrac { 1 }{ 4 }$

$\dfrac{2}{5}^{\times 4}+\dfrac{3}{4}^{\times 5}+\dfrac{1}{4}^{\times 5}$

$\dfrac{8+15+5}{20}=\dfrac{28}{20}=\dfrac{7}{5}$ .

1199847_1444784_ans_30836f6f78204df49347c29a569125c6.jpg

Add $\dfrac { 16 }{ 5 } ,\dfrac { 3 }{ 5 }$ .

Solution :-

$=\dfrac{16}{5}+\dfrac{3}{5}$

$=\dfrac{16+3}{5}$

$=\dfrac{19}{5}$

$=3\dfrac{4}{5}$ (Ans)

What should be added to $(\frac{1}{3}+\frac{3}{5})$ to get $\frac{-6}{20}?$

Let the number should be added be

$x$

Given

$x$ should be added to

$\bigg(\dfrac{1}{3}+\dfrac{3}{5}\bigg)$ to get

$\dfrac{-6}{20}$

$\implies x+\dfrac{1}{3}+\dfrac{3}{5}=-\dfrac{3}{10}$

$\implies x=-\dfrac{3}{10}-\dfrac{1}{3}-\dfrac{3}{5}=\dfrac{-9-10-18}{30}=-\dfrac{37}{30}$

So the number is

$\dfrac{-37}{30}$

Write the following in order of size, starting with the smallest
$0.239$ $\sqrt{0.057}$ $\cfrac{11}{46}$

$\dfrac{11}{46}$ =

$0.23913$

$\sqrt{0.057}$ will be less than 0.239 hence ascending order is

$\sqrt{0.057}$

$0.239$

$\cfrac{11}{46}$

Convert into proper fractions: $8\dfrac { 3 }{ 4 }$

$8\dfrac { 3 }{ 4 }$

$= \dfrac{{ ( 4 \times 8 ) + 3 }}{ 4}$

$=\dfrac{ { 32 + 3 }}{4}$

$=\dfrac{ 35}{ 4}$

Add
$5 \frac{1}{15}+5+\frac{11}{10}$

1316808_1538310_ans_30dddf2143714e2f9f983bbf167e3a1b.jpeg

$100+50+5+\frac{1}{100}$

Given the sum of

$100,50,5,\frac{1}{100}$

$\implies 100+50+5+\dfrac{1}{100}=100+50+5+0.01=155.01$

Write each of the following as decimals.
$20 + 9 + \dfrac {4}{10} + \dfrac {1}{100}$

$20 + 9 + \dfrac {4}{10} + \dfrac {1}{100} = 29 + 0.4 + 0.01 = 29.41$ .

Solve $\displaystyle 5\frac { 2 }{ 7 } -\frac { 8 }{ 9 } +\left( \frac { -3 }{ 14 } \right) -21\frac { 3 }{ 19 } .$

$5\dfrac { 2 }{ 7 } -\dfrac { 8 }{ 9 } +\left( \dfrac { -3 }{ 14 } \right) -21\dfrac { 3 }{ 19 }$

$=\dfrac { 37 }{ 7 } -\dfrac { 8 }{ 9 } -\dfrac { 3 }{ 14 } -\dfrac { 402 }{ 19 }$

$=\dfrac { 12654-2128-513-50652 }{ 2394 }$

$=\dfrac { -40639 }{ 2394 }$

$=-16\dfrac { 2335 }{ 2394 }$

Evaluate $\displaystyle \frac{{ - 12}}{5} + \frac{{31}}{{10}} + \frac{{11}}{{ - 15}} + \frac{{ - 7}}{{ - 18}}$

$\frac{-12}{5}+\frac{31}{10}+\frac{11}{-15}+(\frac{-7}{-18})$

$=\frac{-12\times 2}{5\times 2}+\frac{31}{10}-\frac{11}{15}+\frac{7}{18}$

$=\frac{31-24}{10}+\frac{7\times 5}{18\times 5}-\frac{11\times 6}{15\times 6}$

$=\frac{7}{10}+\frac{35-66}{90}$

$=\frac{7}{10}-\frac{31}{90}$

$=\frac{7\times 9}{10\times 9}-\frac{31}{90}=\frac{63}{90}-\frac{31}{90}$

$=\frac{32}{90}$

$=\frac{32}{90}=\frac{16}{25}$

1217541_1507478_ans_b0387b07395047b8adb37df42c32a55e.jpg

Simplify: $\dfrac{3}{5} - \dfrac{7}{{20}}$

Now,

$\dfrac{3}{5}-\dfrac{7}{20}$

$=\dfrac{12-7}{20}$

$=\dfrac{5}{20}$

$=\dfrac{1}{4}$ .

Evaluate $\dfrac{1}{9}+\dfrac{2}{9}$ .

$\dfrac{1}{9} + \dfrac{2}{9} = \dfrac{3}{9} = \dfrac{1}{3} = 0.3333$

Find the difference:
$1 / 2 - 3 / 8$

Given

$1 / 2 - 3 / 8$
(Refer image)
LCM of 2 and 8

$=(2\times 2\times 2)=8$
Now,

$1 / 2 =(1\times 4)/(2\times 4)=4 / 8$

$3 / 8=(3\times 1)/(8\times 1)= 3 / 8$

$\therefore 1 / 2-3 / 8=4 / 8 - 3/ 8$

$=(4 - 3)/8$

$=1 / 8$
Hence,

$1 / 2 - 3 / 8 =1 / 8$
1630928_1777285_ans_cb4e8d69bcd741acb1022cad705353a2.JPG

Find the difference:
$3\dfrac{5}{8}-2\dfrac{5}{12}$

Give

$3\dfrac{5}{8}-2{5}{12}$
(Refer image 8.1)
LCM of 8 and 12

$=(2\times 2\times 2\times 3)=24$
Now,

$29 / 8= (29\times 3)/(8\times 3)=87/24$

$29/12=(29\times 2)/(12\times 2)=58/24$

$\therefore 29 / 8 - 29 / 12=87 / 24 - 58 / 24$

$(87 - 58) / 24$

$= 29 / 24$
(Refer image 8.2)

$=1\dfrac{5}{24}$
Hence,

$3\dfrac{5}{8}-2\dfrac{5}{12}=1\dfrac{5}{24}$
1630933_1777349_ans_46a0391953e240078cedba9e368190a3.JPG

Find the difference:
$2\dfrac{3}{10}=1\dfrac{7}{15}$

Given

$2\dfrac{3}{10}=1\dfrac{7}{15}$

$23 / 10 - 22 / 15$
(Refer image)
LCM of 10 and 15

$= (5\times 3\times 2)=30$
Now,

$23 / 10=(23\times 3)/(10\times 3)=69/30$

$22/15=(22\times 2)/(15\times 2)=44/30$

$23/10 - 22/15 = 69/30 - 44/30$

$=(69 - 44)/ 30$

$= 25 / 30$

$= 5 / 6 \, \, \, \,(multiplication \, \,by \, \,5)$
Hence,

$2\dfrac{3}{10}-1\dfrac{7}{15}=5 / 6$
1630935_1777388_ans_f26ec864bd80456fa1e493c97adbf5eb.JPG

Find the difference:
$\dfrac{7 }{ 12} - \dfrac{5 }{ 12}$

To find the difference:

$\dfrac{7 }{ 12} - \dfrac{5 }{ 12}$

Now,

$\dfrac{7 }{ 12} - \dfrac{5 }{ 12}$

$=\dfrac{7 -5}{ 12}$

$=\dfrac{2}{ 12}$

$=\dfrac{1}{ 6}$

Hence, the answer is

$\dfrac{1}{ 6}$ .

Find the difference:
$2\dfrac{7}{9}-1\dfrac{8}{15}$

Given

$2\dfrac{7}{9}-1\dfrac{8}{15}$

$25 / 9 - 23 / 15$
(Refer image 7.1)
LCM of 9 and 15

$=(3\times 3\times 5)=45$
Now,

$25 / 9=(25\times 5)/(9\times 5)=125 / 45$

$23 / 15 = (23\times 3)/(15\times 3)=69 /45$

$\therefore 25/ 9 - 23 / 15=125 / 45 - 69 / 45$

$(125 - 69)/45$

$56 / 45$
(Refer image 7.2)

$=1\dfrac{11}{45}$
Hence,

$2\dfrac{7}{9}-1\dfrac{8}{15}=1\dfrac{11}{45}$
1630932_1777316_ans_6386b2cf6ccf4ebb9060291dcd63a9a0.JPG

Find the difference:
$\dfrac{5}{6}-\dfrac{4}{9}$

It is known that

$\dfrac{a}{b}-\dfrac{c}{d}=\dfrac{(a\times d)-(b\times c)}{b\times d}$

Thus,

$\dfrac{5}{6}-\dfrac{4}{9}=\dfrac{(5\times 9)-(4\times 6)}{6\times 9}$

$=\dfrac{45-24}{54}$

$=\dfrac{7}{18}$

Without using you calculator, work out $1\cfrac{7}{12}+\cfrac{13}{20}$
You must show all your working and give your answer as a mixed number in its simplest form.

$\cfrac{95}{60}+\cfrac{13\times3}{20\times3})$

$\cfrac{95}{60}+\cfrac{39}{60})$

$\cfrac{67}{30})$

Find the difference:
$5 / 8 1 / 8$

We have

$\dfrac {5} {8} \, \,– \, \dfrac 1 8$

$\dfrac {5} {8} \, \,– \, \dfrac 1 8= \dfrac {(5 \,- \,1)} {8}$

$=\dfrac 4 8$

$= \dfrac 2 4$

$= \dfrac 1 2$

Hence,

$\dfrac 5 8 \,– \dfrac 1 8 = \dfrac 1 2$

Find the difference:
$4\dfrac{3}{7}-2\dfrac{4}{7}$

Given

$4\dfrac{3}{7}-2\dfrac{4}{7}$

$31 / 7 -18 / 7$

$31 / 7 - 18 / 7=(31 - 18)/ 7$

$=13 / 7$
(Refer image)

$=1\dfrac{6}{7}$
1630917_1777265_ans_3345a8709a6a416e9295bab07c2b3a9d.JPG

Write down the numerator and the denominator of the fraction given below:
$6/11$

Numerator of

$6 / 11$ is

$6$ Denominator of

$6 / 11$ is

$11$

Write down the numerator and the denominator of the fraction given below:
$8/15$

Numerator of

$8 / 15$ is

$8$ Denominator of

$8 / 15$ is

$15$

Find the difference:
$6\dfrac{2}{3}-3\dfrac{3}{4}$

Given

$6\dfrac{2}{3}-3\dfrac{3}{4}$

$20/3-15/4$
(Refer image 10.1)
LCM of 3 and 4

$=(2\times 2\times 3)=12$
Now,

$20 /3=(20\times 4)/(3\times 4)=80 /12$

$15/4=(15\times 3)/(4\times 3)=45/12$

$\therefore 20/3-15/4=80/12-45/12$

$=(80-45)/12$

$=35/12$
(Refer image 10.2)

$=2\dfrac{11}{12}$
Hence,

$6\dfrac{2}{3}-3\dfrac{3}{4}=2\dfrac{11}{12}$

1630943_1777469_ans_726f0e6e1af14c73a08dd44c0472e49e.JPG

Find the difference:
$7-5\dfrac{2}{3}$

Given

$7-5\dfrac{2}{3}$

$7-17/3$
LCM of

$3$ and

$1$

$= 3$

$7/1 =(7\times 3)/(1 \times 3)=21/3$

$17 / 3 = 17 / 3$
Now,

$7 -17 / 3 = (21-17)/3$

$=4/3$
(Refer image)

$=1\dfrac{1}{3}$
1630948_1777484_ans_607b9eda863642088f4e92c4b5e66636.JPG

Write down the numerator and the denominator of the fraction given below:
$4/9$

Numerator of

$4 / 9$ is

$4$

Denominator of

$4 / 9$ is

$9$

Write a fraction for the following:
Three - tenths

The fraction for three - tenths is

$\dfrac3{10}.$

Solve: $2+\dfrac{11}{15} - \dfrac{5}{9}$

To solve:

$2+\dfrac{11}{15} - \dfrac{5}{9}$

$15=3\times 5$

and,

$9=3\times3$

So, LCM of the denominators

$15$ and

$9$

$=3\times3\times5$

$=45$

Now,

$2+\dfrac{11}{15} - \dfrac{5}{9}$

$=\dfrac{2\times45+11\times3-5\times5}{45}$

$=\dfrac{90+33-25}{45}$

$=\dfrac{98}{45}$

$=2\dfrac{8}{45}$

Hence, the required solution is

$2\dfrac{8}{45}$ .

Find the difference:
$10-6\dfrac{3}{8}$

Given

$10 - 6\dfrac{3}{8}$

$10 - 51 / 8$
LCM of

$8 = 8$
Now,

$10 =(10\times 8) = 80$

$51 / 8 = 51 / 8$

$\therefore 10-51 / 8=(80-51) /8$

$=29 / 8$
(Refer image)

$3\dfrac{5}{8}$
Hence,

$10-6\dfrac{3}{3}=3\dfrac{5}{8}$
1630966_1777495_ans_93d816570350472fad7bda79cf1854b9.JPG

Find the difference:
$5/8 + 3/4 - 7/12$

Given

$5/8 + 3/4 - 7/12$
(Refer image)
LCM of 4, 8 and 12

$=(2\times 2\times 2\times 3)=24$

$5 / 8=(5\times 3)/(8\times 3)=15 / 24$ (by dividing 24/ 8=3)

$3 / 4=(3\times 6)/(4\times 6)=9/12$ (by dividing 24/ 4=6)

$7 / 12=(7\times 2)/(12\times 2)=14 / 24$ (by dividing 24/12 =2)
Now,

$5 / 8 + 3 / 4 -7 / 12=(15+18-14)/24$

$=(33-14)/24$

$=19/24$
Hence,

$5/8 + 3/4 - 7/12=19/24$
1630997_1777573_ans_fbe49b3055f645dda634139cb9e119bb.JPG

Write a fraction for the following:
Four- sevenths

The fraction for four - sevenths is

$4 / 7$

Write down the fraction in which
numerator $= 7$ , denominator $= 16$

Fraction of numerator as

$7$ and denominator as

$16$ is

$\dfrac{7}{16}$ .

Find the sum:
$\dfrac{5}{8} +\dfrac{ 1}{8}$

Given

$\dfrac{5}{8} +\dfrac{ 1}{8}=\dfrac{5+1}{8}$

$=\dfrac{6}{8}$

$=\dfrac{3}{4}$

Write down the numerator and the denominator of the fraction given below:
$5/1$

Numerator of

$5 / 1$ is

$5$ Denominator of

$5 / 1$ is

$1$

Write down the fraction in which
numerator $= 8$ , denominator $= 15$

Given, numerator

$= 8$ , denominator

$= 15$

Hence, the fraction is

$\dfrac{8}{15}$

Write down the fraction in which
numerator $= 5$ , denominator $= 12$

Fraction of numerator

$= 5$ , denominator

$=12$ is

$5/12$

Write down the fraction in which
numerator $= 3$ , denominator $= 8$

Fraction of numerator as

$3$ and denominator

$8$ is

$\dfrac{3}{8}$ .

Find the sum:
$\dfrac{4}{9} + \dfrac{8}{9}$

Given

$\dfrac{4}{9} + \dfrac{8}{9}$

$=\dfrac{4+8}{9}$ (on taking

$9$ as L.C.M of the denominator)

$=\dfrac{12}{9}$

$=\dfrac{4}{3}$ (on dividing both numerator and denominator by

$3$ )

$\therefore \dfrac{4}{9} + \dfrac{8}{9}=\dfrac{4}{3}$

$=1\dfrac{1}{3}$

Write down the numerator and the denominator of the fraction given below:
$12/17$

Numerator of

$12 / 17$ is

$12$ Denominator of

$12 / 17$ is

$17$

Which of the following are improper fractions?
$3 / 2, 5 / 6, 9 / 4, 8 / 8, 3, 27 / 16, 23 / 31, 19 / 18, 10 / 13, 26 / 26$

A fraction whose numerator is greater than or equal to its denominator is called an improper fraction
Here,

$3 / 2, 9 / 4, 8 / 8, 3, 27 / 16, 19 / 18$ and

$26 / 26$ are improper fractions.

Which of the following are proper fractions?
$1/2,3/5,10/7,7/4,2,15 /8,16/16,10/11,23/ 10$

A fraction whose numerator is less than its denominator is called a proper fraction.
Here,

$1 / 2, 3/5$ and

$10 / 11$ are proper fractions.

Find the sum:
$1\dfrac{3}{5}+2\dfrac{4}{5}$

$1\dfrac{3}{5}+2\dfrac{4}{5}=\dfrac{8}{5}+\dfrac{14}{5}$

$\Rightarrow \dfrac{8}{5}+\dfrac{14}{5}=\dfrac{22}{5}=4\dfrac{2}{5}$
1631272_1778354_ans_48021ae5ff1c4712af83881cf629520a.JPG

Find the sum:
$3\dfrac{2}{3}+1\dfrac{5}{6}+2$

Given

$3\dfrac{2}{3}+1\dfrac{5}{6}+2$

$11/3+11/6+2$
(Refer image 10.1)
LCM of 3 and 6

$=(3\times2)=6$

$11/3+11/6+2=(11\times2)/(3\times2)+(11\times1)/(6\times1)+(2\times6)$

$=(22+11+12)/6$

$=45/6$

$15/2$
(Refer image 10.2)

$=7\dfrac{1}{2}$
1631322_1778364_ans_06171164b91744a88f996b05e685dd94.JPG

Find the sum:
$\dfrac{2}{9} + \dfrac{5}{6}$

Given

$\dfrac{2}{9} + \dfrac{5}{6}$

We need to find the L.C.M of

$9,6$

(Refer to Image

$4.1$ ) L.C.M of

$9$ and

$6$ is

$=3\times3\times 2=18$

So,

$\dfrac{2}{9}=\dfrac{2\times 2}{9 \times 2}=\dfrac{4}{18}$

and

$\dfrac{5}{6}=\dfrac{5 \times 3}{6 \times 3}=1\dfrac{5}{18}$

Now adding:

$\dfrac{2}{9}+\dfrac{5}{6}=\dfrac{4+15}{18}$

$=\dfrac{19}{18}$

$=1 \dfrac{1}{19}$

Find the sum:
$3\dfrac{1}{8}+1\dfrac{5}{12}$

Given

$3\dfrac{1}{8}+1\dfrac{5}{12}$

$=25/8+17/12$
(Refer image 8.1)
LCM of 8 and 12

$=(2\times2\times2\times3)=24$

$25/8 + 17/12=(25\times3)/(8\times3)+(17\times2)/(12\times2)$

$=75/24 + 34/24$

$=109/24$
(Refer image 8.2)

$=4\dfrac{13}{24}$
1631315_1778359_ans_86bd9e6b24c54f829c9626792884353c.JPG

Find the sum:
$2\dfrac{3}{4}+5\dfrac{5}{6}$

Given

$2\dfrac{3}{4}+5\dfrac{5}{6}$

$11/4 + 35/6$
(Refer image 7.1)
LCM of 4 and 6

$=(2\times2\times3)=12$

$11/4+35/6=(11\times3)/(4\times3)+(35\times2)/(6\times2)$

$=33/12 + 70/12$

$=103/12$
(Refer image 7.2)

$=8\dfrac{7}{12}$
1631311_1778358_ans_c59812e7d3104d4db75a7521d3047a59.JPG

Find the sum:
$\dfrac{4}{15} + \dfrac{17}{20}$

Given

$\dfrac{4}{15} + \dfrac{17}{20}$

Factors of

$15 = 3\times 5$

Factors of

$20 = 2\times 2\times 5$

$\therefore$ LCM of 15 and 20

$=(5\times3\times2\times2)=60$

$\dfrac{4}{15} + \dfrac{17}{20}=\dfrac{(4\times4)}{(15\times4)}+\dfrac{(17\times3)}{(20\times3)}$

$=\dfrac{16}{60} + \dfrac{51}{60}$

$=\dfrac{67}{60}$

$=1\dfrac{7}{60}$

Find the sum:
$9\dfrac{7}{10}+3\dfrac{8}{15}$

Given

$9\dfrac{7}{10}+3\dfrac{8}{15}$

$27/10+53/15$
(Refer image 9.1)
LCM of 10 and 15

$=(5\times3\times2)=30$

$27/10+53/15=(27\times3)/(10\times3)/(10\times3)/(53\times2)/(15\times2)$

$=81/30+106/30$

$=187/30$
(Refer image 9.2)

$=6\dfrac{7}{30}$
1631317_1778360_ans_5d9ba4c0660b472b8629a249d4c77c0e.JPG

Find the sum:
$\dfrac{7}{12} + \dfrac{9}{16}$

Given

$\dfrac{7}{12} + \dfrac{9}{16}$

Prime factorisation of

$12$ and

$16$ to find LCM

$12=2\times 2\times 3=2^2\times 3$

$16=2\times 2\times 2\times 2=2^4$

$\therefore$ LCM of

$12$ and

$16$

$=(2\times2\times2\times2\times3)=48$

$\implies \dfrac{7}{12} + \dfrac{9}{16}=\dfrac{(7\times4)+(9\times3)}{48}$

$=\dfrac{28+27}{48}$

$=\dfrac{55}{48}$

$=1\dfrac{7}{48}$

Write down the numerator and the denominator of the following rational number:
$\cfrac5{-8}$

$\cfrac pq$ , numerator is

$p$ and denominator is

$q.$

On comparing fraction

$\cfrac5{-8}$ with above we get that, numerator is

$5$ and the denominator is

$-8.$

Write six improper fractions with numerator $13.$

In an improper fraction, the numerator is always greater than the denominator.

Examples of improper fractions with numerator

$13$ are

$\dfrac{13}{2},\ \dfrac{13}{4},\ \dfrac{13}{5},\ \dfrac{13}{8},\ \dfrac{13}{9},\ \dfrac{13}{11},...$

Write six improper fractions with denominator $5$

$6 / 5, 7 / 5, 8 / 5, 9 / 5, 11 / 5, 12 / 5$ are improper fractions with denominator

$5$

Write the following fractions in number form :
(i) One - sixth
(ii) Three - eleventh,
(iii) Seven - fortieth
(iv) Thirteen - one hundred twenty fifth

(i) One - sixth

$= \dfrac{1}{6}$

(ii) Three - eleventh

$= \dfrac{3}{11}$

(iii) Seven - fortieth

$=\dfrac{7}{40}$

(iv) Thirteen - one hundred twenty fifth

$= \dfrac{13}{125}$

Simplify the following :
$(i) 1 \dfrac{2}{3} + 2 \dfrac{1}{2} + \dfrac{3}{4}$
$(ii) 3 \dfrac{2}{9} + 2 \dfrac{1}{3} +2 \dfrac{7}{12}$
$(iii) \dfrac{7}{12} + \dfrac{8}{9} - \dfrac{5}{6}$
$(iv) 1 \dfrac{3}{25} + \dfrac{7}{20} - \dfrac{2}{5}$
$(v) 1 \dfrac{13}{14} - 2 \dfrac{5}{6} + 1\dfrac{6}{7}$

$(i) 1 \dfrac{2}{3} + 2 \dfrac{1}{2} + \dfrac{3}{4}$

L.C.M. of

$3,2$ and

$4 = 12$

$= \dfrac{5}{3} + \dfrac{5}{2} + \dfrac{3}{4} = \dfrac{5 \times 4 + 5 \times 6 + 3 \times 3}{12}$

$= \dfrac{20 + 30 + 9 }{12} = \dfrac{59}{12} = 4 \dfrac{11}{12}$

$(ii) 3\dfrac{2}{9} + 2\dfrac{1}{3} + 2\dfrac{7}{12}$

$L.C.M. \, of\, 9,3,12 = 36$

$= \dfrac{29}{9} + \dfrac{7}{3} + \dfrac{31}{12}$

$= \dfrac{29 \times 4 + 7 \times 12 + 31 \times 3}{36}$

$= \dfrac{116 + 84 + 93}{36} = \dfrac{293}{36} = 8\dfrac{5}{36}$

$(iii) \dfrac{7}{12} + \dfrac{8}{9} - \dfrac{5}{6}$

$L.C.M \, of \, 12, 9,6 = 36$

$= \dfrac{7 \times 3 + 8 \times 4 - 5 \times 6 }{36}$

$= \dfrac{7 \times 3 + 8 \times 4 - 5 \times 6}{36}$

$= \dfrac{53 - 30}{36} = \dfrac{23}{36}$

$(iv) 1 \dfrac{3}{25} + \dfrac{7}{20} - \dfrac{2}{5}$

$L.C.M. \, 25,20, 5 = 100$

$=\dfrac{28}{25} + \dfrac{7}{20} - \dfrac{2}{5}$

$= \dfrac{28\times 4 + 7 \times 5 - 2 \times 20}{100}$

$=\dfrac{112 + 35 - 40}{100}$

$= \dfrac{147 - 40}{100} = \dfrac{107}{100} = 1\dfrac{7}{100}$

$(v) 1 \dfrac{13}{14} - 2\dfrac{5}{6} + 1 \dfrac{6}{7}$

$L.C.M. of \, 14,6,7 = 42$

$= \dfrac{27}{14} - \dfrac{17}{6} + \dfrac{13}{7}$

$= \dfrac{27 \times 3 - 17 \times 7 + 13 \times 6}{42}$

$=\dfrac{81 - 119 + 78}{42}$

$= \dfrac{159 - 119}{42} = \dfrac{40}{42} = \dfrac{20}{21}$

Work out the following :
$(i)\ \dfrac{2}{3} + \dfrac{3}{4}$
$(ii)\ \dfrac{5}{7} - \dfrac{4}{9}$
$(iii)\ \dfrac{1}{2} + \dfrac{3}{5}$
$(iv)\ 1 \dfrac{4}{9} + 3 \dfrac{5}{12}$

$(i)\ \dfrac{2}{3} + \dfrac{3}{4}$

$(L.C.M$ of

$7 \,$ and

$9 = 63)$

$= \dfrac{\left(2 \times 4\right) + \left(3 \times 3\right)}{12} = \dfrac{8 + 9}{12}$

$= \dfrac{17}{12} = 1 \dfrac{5}{12}$

$(ii)\ \dfrac{5}{7} - \dfrac{4}{9}$

$(L.C.M$ of

$\, 7\,$ and

$\, 9 = 63)$

$= \dfrac{\left(5 \times 9 \right) -\left(7 \times 4\right)}{63} = \dfrac{45 - 28}{63} = \dfrac{17}{63}$

$(iii)\ \dfrac{1}{2} + \dfrac{3}{5}$

$(L.C.M.$ of

$\,2$ and

$\, 5 = 10)$

$= \dfrac{\left(1 \times 5\right) + \left(3 \times 2 \right)}{10} = \dfrac{5 + 6}{10}=\dfrac{11}{10} = 1 \dfrac{1}{10}$

$(iv)\ 1 \dfrac{4}{9} + 3 \dfrac{5}{12}$

$\Rightarrow \dfrac{13}{9} + \dfrac{41}{12}$

$(L.C.M. \,$ of

$\, 9 \,$ and

$\, 12 = 36)$

$= \dfrac{\left(13 \times 4\right) + \left(41 \times 3\right)}{36} = \dfrac{52 + 123}{36}$

$= \dfrac{175}{36} = 4 \dfrac{31}{36}$

Convert $2435\ m$ to $km$ and express the result as a mixed fraction.

We know that

$1\ km = 1000\ m$

$\Rightarrow 1\ m = \dfrac {1}{1000}\ km$

$\Rightarrow 2435\ m = \dfrac {1}{1000} \times 2435\ km$

$\Rightarrow 2435\ m = \dfrac {2435}{1000}\ km = \dfrac{487}{200}\ km$

$\therefore 2435\ m=2 \dfrac {87}{200} \ km$

1643667_1788597_ans_58f49a01fc0d46bda5e0cb3a7daa6500.png

Find in the missing fractions :
$(i) \dfrac{7}{10} - \Box = \dfrac{3}{10}$
$(ii) \Box + \dfrac{5}{27} = \dfrac{12}{27}$
$(iii) \Box - \dfrac{5}{7} = \dfrac{2}{7}$ .

$(i)\ \dfrac{7}{10} - \Box = \dfrac{3}{10}$

Let the missing fraction be

$x$ .

$\Rightarrow \dfrac{7}{10} - x = \dfrac{3}{10}$

$\Rightarrow -x = \dfrac{3}{10} - \dfrac{7}{10}$

$\Rightarrow -x = \dfrac{-4}{10}$

$\Rightarrow x = \dfrac{4}{10}$

$(ii)\ \Box + \dfrac {5}{27} = \dfrac{12}{27}$

Let the missing fraction be

$x$

$\Rightarrow x + \dfrac{5}{27} = \dfrac{12}{27}$

$\Rightarrow x = \dfrac{12}{27} - \dfrac{5}{27}$

$\Rightarrow x = \dfrac{7}{27}$

$(iii)\ \Box - \dfrac{5}{7} = \dfrac{2}{7}$

Let the missing fraction be

$x$

$\Rightarrow x - \dfrac{5}{7} = \dfrac{2}{7}$

$\Rightarrow x = \dfrac{2}{7} + \dfrac{5}{7}$

$\Rightarrow x = \dfrac{7}{7} = 1$

Simplify: $[8(5/6)]-[3(3/8)]+[1(7/12)]$

First we convert each mixed fraction into an improper fraction.

$\Rightarrow [8(5/6)]=(53/6)$
also

$\Rightarrow [3(3/8)]=(27/8)$

and

$\Rightarrow [1(7/12)]=(19/12)$

Then,

$(53/6)-(27-8)-(19/12)$
LCM of

$6,8,12=24$
Now, let us change each of the given fraction into an equivalent fraction having

$24$ as the denominator.

$=[(53/6)\times(4/4)]=(212/24)$
also

$=[(27/8)\times (3/3)]=(81/24)$

and

$=[(19/12)\times(2/2)]=(38/24)$

Now,

$=(212/24)-(81/24)-(38/24)$

$=[(212-81+38)/24]$

$=[(250-81/24)]$

$=(169/24)$

$=[7(1/24)]$

Fill in the blanks.
$7 \dfrac{2}{5} + ........ = 12$

$7 \dfrac{2}{5} +x= 12$

$x=12-\dfrac{37}{5}$

$x=\dfrac{60-37}{5}$

$x=\dfrac{23}{5}=4\dfrac{4}{5}$

$\therefore7 \dfrac{2}{5} + 4\dfrac{3}{5} = 12$

Fill in the blanks
A proper fraction lies between $0$ and .......

A proper fraction is a fraction where the numerator is less than the denominator.

So a proper fraction lies between –1 and 0 and between 0 and 1.

Fill in the blanks.
A mixed fraction can be converted into ......... fraction.

A mixed fraction can be converted into an improper fraction.

Sarita bought $\dfrac{2}{5}$ metre of ribbon and Lalita $\dfrac{3}{4}$ metre of ribbon.
What is the total length of the ribbon they bought?

Ribbon bought by Sarita

$= \dfrac{2}{5} m$
Ribbon bought by Laiita

$= \dfrac{3}{4} m$

$\therefore$ Total length of the ribbon they bought

$=\dfrac{2}{5} m + \dfrac{3}{4} m = \left(\dfrac{2}{5} + \dfrac{3}{4}\right)m$

$\left[L.C.M. \left(5,4\right) = 20\right]$

$= \left(\dfrac{2 \times 4 + 3 \times 5}{20}\right)m$

$=\left(\dfrac{8}{20} + \dfrac{15}{20}\right) m = \dfrac{8 + 15}{20} m$

$=\dfrac{23}{20} m = 1 \dfrac{3}{20} m$

The weight of three packets are $2 \dfrac{3}{4} kg$ . $3 \dfrac{1}{3} kg.$ and $5\dfrac{2}{5} kg.$ Find total weight of all the three packets.

Weigth of 1st packet

$= 2\dfrac{3}{4}$
Weigth of 2nd packet

$= 3\dfrac{1}{3}$
Weigth of 3rd packet

$= 5\dfrac{2}{5}$

$\therefore$ Total weight

$= 2\dfrac{3}{4} + 3\dfrac{1}{3} + 5\dfrac{2}{5} = \dfrac{11}{4} + \dfrac{10}{3} + \dfrac{27}{5}$

$\left(\because L.C.M \, 4,3,5 = 60 \right)$

$= \dfrac{11 \times 15 + 10 \times 20 + 27 \times 12}{60}$

$=\dfrac{165 + 200 + 324}{60}$

$= \dfrac{689}{60} = 11\dfrac{29}{60}kg$

Evaluate :
$3 \dfrac{5}{12} - 1 \dfrac{2}{3}$

$3 \dfrac{5}{12} - 1 \dfrac{2}{3}$
It can be written as

$= \dfrac{12 \times 3 + 5}{12} - \dfrac{3 \times 1 + 2}{3}$

$= \dfrac{41}{12} - \dfrac{5}{3}$

LCM of

$12$ and

$3$ is

$12$

$= \dfrac{41 \times 1}{12 \times 1} - \dfrac{5 \times 4}{3 \times 4}$

By further calculation

$= \dfrac{41 - 20}{12}$

$= \dfrac{21}{12}$

$= \dfrac{7}{4}$

$= 1 \dfrac{3}{4}$

The reciprocal of an improper fraction is an
improper fraction.

False, the reciprocal of an improper fraction is

a proper fraction.

Evaluate :
$3 \dfrac{5}{12} + 1 \dfrac{2}{3}$

$3 \dfrac{5}{12} + 1 \dfrac{2}{3}$
It can be written as

$= \dfrac{12 \times 3 + 5}{12} + \dfrac{3 \times 1 + 2}{3}$

$= \dfrac{41}{12} + \dfrac{5}{3}$

LCM of

$12$ and

$3$ is

$12$

$= \dfrac{41 \times 1}{12 \times 1} + \dfrac{5 \times 4}{3 \times 4}$

By further calculation

$= \dfrac{41}{12} + \dfrac{20}{12}$

$= \dfrac{41 + 20}{12}$

$= \dfrac{61}{12}$

$= 5 \dfrac{1}{12}$

The reciprocal of a proper fraction is a proper fraction.

False, the reciprocal of a proper fraction is an
improper fraction.

Enter 1 if it is true else enter 0.
$\dfrac{7}{6}$

Expressing a rational number into standard form means there is no common factor, other than 1, available in its numerator and denominator and its denominator is a positive integer.

The correct answer is 1

1887765_1802384_ans_70a9a1e24a814b828b1ee924ddf6e1c2.jpeg

Enter 1 if true else enter 0.
The rational number $\dfrac{-81}{15}$ reduced to simplest form is $\dfrac{-7}{13}$ .

The correct answer is 0

1887671_1802381_ans_4da27cdfed60489590a7e50c422d9b53.jpeg

Subtract $\left(2/7 - 5/21 \right)$ from the sum of 3/4, 5/7 and 7/12.

We can write it as

$\left (3/4 + 5/7 + 7/12 \right ) - \left (2/7 - 5/21 \right )$
LCM of 4. 7 and 12 is 84 and 7 and 21 is 21

$= \left[\left (63 + 60 + 49 \right )/84\right] - \left[\left (6 - 5 \right )/21\right]$
By further calculation

$= 172/84 - 1/21$
LCM of 21 and 84 is 84
=

$\left(172 - 4 \right)/84$

$= 168/84$

$= 2$

Simplify : $2 \dfrac{1}{2} - 3 \dfrac{1}{4} + 5\dfrac{5}{6}$

$2 \dfrac{1}{2} - 3 \dfrac{1}{4} + 5\dfrac{5}{6} = \dfrac{5}{2} - \dfrac{13}{4} + \dfrac{35}{6}$

( Changing into like fractions )

$= \dfrac{5 \times 6}{2 \times 6} - \dfrac{13 \times 3}{4 \times 3} + \dfrac{35 \times 2}{6 \times 2}$

$\left( LCM \, of \, 2, 4, 6 = 12\right)$

$= \dfrac{30 - 39 + 70}{12} = \dfrac{61}{12} = 5 \dfrac{1}{12}$

Put the $(\checkmark),$ wherever applicable

Number

Natural Number

Whole Number

Integer

Fraction

Rational Number

$-19\dfrac{3}{4}$

As we know that
Integer are $-2,-1,0,1,2,3$
Whole numbers are $0,1,2,3,$
Natural numbers are $1,2,3,4...$
Rational numbers are $\dfrac{1}{2},\dfrac{3}{5},\dfrac{7}{2}$

Hence according to the number systems,
$-19\dfrac{3}{4}=\dfrac{-79}{4}$ - Rational number

Number	Natural Number	Whole Number	Integer	Fraction	Rational Number
$-19\dfrac{3}{4}$					$\checkmark$

Evaluate :
$\dfrac{7}{26} + 2 + \dfrac{-11}{13}$

$\dfrac{7}{26} + 2 + \dfrac{-11}{13}$
It can be written as

$= \dfrac{7}{26} + \dfrac{2}{1} + \dfrac{-11}{13}$
Here LCM of

$26$ and

$13$ is

$26$
So we get

$= \dfrac{7 \times 1}{26 \times 1} + \dfrac{2 \times 26}{1 \times 26} - \dfrac{11 \times 2}{13 \times 2}$
By further calculation

$= \dfrac{7 + 52 - 22}{26}$
So we get

$= \dfrac{59 - 22}{26}$

$= \dfrac{37}{26}$

$= 1 \dfrac{11}{26}$
1747171_1826137_ans_9bd14556dff14ab7bd82e2460fcb2028.png

Evaluate :
$\dfrac{-5}{9} + \dfrac{-7}{12} + \dfrac{11}{18}$

$\dfrac{-5}{9} + \dfrac{-7}{12} + \dfrac{11}{18}$
Here LCM of

$9 , 12$ and

$18$ is

$36$
So we get

$= \dfrac{-5 \times 4}{9 \times -4} - \dfrac{7 \times 3}{12 \times 3} + \dfrac{11 \times 2}{18 \times 2}$
By further calculation

$= \dfrac{-20 - 21 + 22}{36}$
So we get

$= \dfrac{-41 + 22}{36}$

$= \dfrac{-19}{36}$
1747114_1826133_ans_feb2a9e906b540c6b39e39a690c4c370.png

Evaluate :
$\dfrac{7}{-27} + \dfrac{11}{18}$

$\dfrac{7}{-27} + \dfrac{11}{18}$
Here LCM of

$27$ and

$18$ is

$54$
So we get

$= \dfrac{7 \times 2}{-27 \times 2} + \dfrac{11 \times 3}{18 \times 3}$
By further calculation

$= \dfrac{-14 + 33}{54}$

$= \dfrac{19}{54}$
1739695_1826130_ans_a4b0aece39784e4b8b5b24ee0ced9b24.png

Evaluate :
$\dfrac{9}{-16} + \dfrac{-5}{-12}$

$\dfrac{9}{-16} + \dfrac{-5}{-12}$
It can be written as

$= \dfrac{9}{-16} + \dfrac{5}{12}$
Here LCM of

$16$ and

$12$ is

$48$
So we get

$= \dfrac{9 \times 3}{-16 \times 3} + \dfrac{5 \times 4}{12 \times 4}$
By further calculation

$= \dfrac{-27 + 20}{48}$

$= \dfrac{-7}{48}$
1739722_1826132_ans_4dcb5eeb7d2b4878a0430c382062c05e.png

Reduce to a single fraction:
$2 \dfrac{1}{2} + 2 \dfrac{1}{3} - 1 \dfrac{1}{4}$

$2 \dfrac{1}{2} + 2 \dfrac{1}{3} - 1 \dfrac{1}{4}$
It can be written as

$= 5/2 + 7/3 - 5/4$
Here the LCM of

$2, 3$ and

$4$ is

$12$

$= \left (5 \times 6 \right )/ \left (2 \times 6 \right ) - \left (7 \times 4 \right )/ \left (3 \times 4 \right ) - \left (5 \times 3 \right )/ \left (4 \times 3 \right )$
By further calculation

$= 30/12 + 28/12 - 15/12$
So we get
=

$\left (30 + 28 - 15 \right )/12$
=

$\left (58 - 15 \right )/12$

$= 43/12$

$= 3 \dfrac{7}{12}$

Evaluate :
$\dfrac{-3}{8} + \dfrac{-5}{12}$

$\dfrac{-3}{8} + \dfrac{-5}{12}$
Here LCM of

$8$ and

$12$ is

$24$
So we get

$= \dfrac{-3 \times 3}{8 \times 3} + \dfrac{-5 \times 2}{12 \times 2}$
By further calculation

$= \dfrac{-9 - 10}{24}$

$= \dfrac{-19}{24}$
1739704_1826131_ans_a7f0b083923c4f03881b111ebab311a8.png

Evaluate :
$\dfrac{-2}{3} + \dfrac{3}{4}$

$\dfrac{-2}{3} + \dfrac{3}{4}$
Here the LCM of

$3$ and

$4$ is

$12$
So we get

$= \dfrac{-2 \times 4}{3 \times 4} + \dfrac{3 \times 3}{4 \times 3}$
By further calculation

$= \dfrac{-8 + 9}{12}$

$= \dfrac{1}{12}$
1739686_1826129_ans_c1a9fb9a03f7430dabc0a0f91ba32b09.png

Evaluate :
$\dfrac{-5}{7} - \left (\dfrac{-3}{8} \right )$

$\dfrac{-5}{7} - \left (\dfrac{-3}{8} \right )$
It can be written as
Here LCM of

$7$ and

$8$ is

$56$
So we get

$= \dfrac{-5 \times 8}{7 \times 8} + \dfrac{3 \times 7}{8 \times 7}$
By further calculation

$= \dfrac{-40 + 21}{56}$

$= \dfrac{-19}{56}$
1747155_1826136_ans_210a829b7eeb450ebb56135ba3e21a57.png

Evaluate :
$\dfrac{7}{-26} + \dfrac{16}{39}$

$\dfrac{7}{-26} + \dfrac{16}{39}$
Here LCM of

$26$ and

$39$ is

$78$
So we get

$= \dfrac{-7 \times 3}{26 \times 3} + \dfrac{16 \times 2}{39 \times 2}$
By further calculation

$= \dfrac{-21 + 32}{78}$

$= \dfrac{11}{78}$
1747120_1826134_ans_54b06635e1eb419abfe9e99e3986ab1a.png

Reduce to a single fraction:
1/2 + 2/3

$1/2 + 2/3$
Here the LCM of

$2$ and

$3$ is

$6$

$= \left (1 \times 3 \right )/ \left (2 \times 3 \right ) + \left (2 \times 2 \right )/ \left (3 \times 2 \right )$
By further calculation

$= 3/6 + 4/6$
=

$\left (3 + 4 \right )/6$
So we get

$= 7/6$

$= 1 \dfrac{1}{6}$

Reduce to a single fraction:
$2 \dfrac{5}{8} - 2 \dfrac{1}{6} + 4 \dfrac{3}{4}$

$2 \dfrac{5}{8} - 2 \dfrac{1}{6} + 4 \dfrac{3}{4}$
It can be written as

$= 21/8 - 13/6 + 18/4$
Here the LCM of

$8, 6$ and

$4$ is

$24$

$= \left (21 \times 3 \right )/ \left (8 \times 3 \right ) - \left (13 \times 4 \right )/ \left (6 \times 4 \right ) - \left (19 \times 6 \right )/ \left (4 \times 6 \right )$
By further calculation

$= 63/24 - 52/24 + 114/24$
So we get
=

$\left (63 - 52 + 114 \right )/24$
=

$\left (117 - 52 \right )/12$

$= 125/24$

$= 5 \dfrac{5}{24}$

Subtract:
$3/7$ and $3/7$

$3/7$ and

$3/7$
It can be written as

$= 3/7 - \left(-3/7\right)$

$= 3/7 + 3/7$
So we get

$= \left(3 + 3\right)/7$

$= 6/7$

Subtract:
$2$ form $2/3$

$2$ form

$2/3$
It can be written as

$= 2/3 - 2/1$
LCM of

$3$ and

$1$ is

$3$

$= 2/3 - \left(2 \times 3 \right)/3$

$= 2/3 - 6/3$
So we get

$= \left(2 - 6 \right)/3$

$= -4/3$

Evaluate :
$\dfrac{5}{9} + \dfrac{-7}{6}$

$\dfrac{5}{9} + \dfrac{-7}{6}$

$\dfrac{5}{9} + \dfrac{-7}{6} Taking L. C.M$
2 9 , 6
3 9 3
3 3 1
1 , 1

$\therefore LCM of 9 and 6 = 2 \times 3 \times 3 = 18$

$\dfrac{5 \times 2}{9 \times2} - \dfrac{7 \times 3}{6 \times 3}$
(

$\therefore LCM of 9 and 6 = 18$ )

$= \dfrac{10 - 21}{18} = \dfrac{-11}{8}$

Subtract:
$1/8$ from $5/8$

$1/8$ from

$5/8$
It can be written as

$= 5/8 - 1/8$

$= \left(5 - 1 \right)/8$
So we get

$= 4/8$

$= 1/2$

Subtract:
$-4/7$ from $-6/11$

$-4/7$ from

$-6/11$
It can be written as

$= -6/11 - \left(- 4/7 \right)$

$= -6/11 + 4/7$
LCM of 7 and 11 is 77

$= \left(-6 \times7 \right)/\left(11 \times 7 \right) - \left(4 \times 11 \right)/\left(7 \times 11 \right)$
By further calculation

$= -42/77 + 44/77$
So we get

$= \left(-42 + 44\right)/77$

$= 2/77$

Subtract:
$-2/5$ and $2/5$

$-2/5$ and

$2/5$
It can be written as

$= 2/5 - \left(-2/5\right)$

$= 2/5 + 2/5$
So we get

$= \left(2 + 2\right)/5$

$= 4/5$

Subtract:
$0$ and $-4/5$

$0$ and

$-4/5$
It can be written as

$= -4/5 - 0$

$= - 4/5$

What number should be added to $8 \dfrac{2}{3}$ to get $12 \dfrac{5}{6}$ ?

To find the fraction we must subtract

$8 \dfrac{2}{3}$ from

$12 \dfrac{5}{6}$
So the required number

$=12 \dfrac{5}{6}$

$-8 \dfrac{2}{3}$
It can be written as

$= 77/6 - 26/3$
Here the LCM of 3 and 6 is 6

$= \left(77 \times 1 \right)/\left(6 \times 1 \right) - \left(26 \times 2 \right)/\left(3 \times 2 \right)$
By further calculation

$= \left(77 - 52\right)/6$

$= 25/6$

$= 4 \dfrac{1}{6}$

What should be subtracted from $8 \dfrac{3}{4}$ to get $2 \dfrac{2}{3}$ ?

Required number =

$8 \dfrac{3}{4}$

$-2 \dfrac{2}{3}$
It can be written as

$= 35/4 - 8/3$
LCM of 4 and 3 is 12

$= \left(35 \times 3 \right)/\left(4 \times 3 \right) - \left(8 \times 4 \right)/\left(3 \times 4 \right)$
By further calculation

$= \left(105 - 32\right)/12$

$= 73/12$

$= 6 \dfrac{1}{12}$

Evaluate :
$\dfrac{-4}{9} - \dfrac{2}{-3}$

$\dfrac{-4}{9} - \dfrac{2}{-3}$
Taking LCM

$3| 9,3$

$5| 3,1$

$|1,1$
(∴LCM of 3 and 9=9)

$\dfrac{-4}{9} - \dfrac{2}{-3} = \dfrac{-4 \times 1}{9 \times1} - \dfrac{\left ( -2 \times3 \right )}{3 \times 3}$

$= \dfrac{-4 + 6}{9} = \dfrac{2}{9}$

Evaluate :
$\dfrac{-5}{18} - \dfrac{-2}{9}$

Taking $LCM$
$2 |18,9$
$3 |9,9$
$3| 3,3$
$|1,1$

$∴LCM of 9 and 18=2\times3\times3=18$

$\dfrac{-5}{18} - \dfrac{-2}{9} = \dfrac{-5 \times 1 }{18 \times 1} - \dfrac{\left ( -2 \times 2 \right )}{9 \times 2}$

$= \dfrac{-5 + 4}{18}$

$= \dfrac{-1}{18}$

Subtract :
$\dfrac{-5}{8} From \dfrac{-13}{16}$

Subtraction

$\dfrac{-5}{8} From \dfrac{-13}{16}$ Taking LCM

$2 |8,16$

$2 |4,8$

$2 |2,4$

$2 |1,2$

$|1,1$
∴LCM of

$8$ and

$16=16$

$\dfrac{-13}{16} - \left ( \dfrac{-5}{8} \right ) = \dfrac{-13 \times 1}{16 \times1} + \dfrac{5 \times 2}{8 \times2}$

$= \dfrac{-13 + 10}{16} = \dfrac{-3}{16}$

Subtract :
$\dfrac{1}{4} From \dfrac{-3}{8}$

Subtracting

$\dfrac{1}{4} From \dfrac{-3}{8}$
Taking

$LCM$

$2 |4,8$

$2 |2,4$

$2 |1,2$

$|1,1$

$\therefore LCM\ of 4 , 8 = 2 \times 2 \times 2 = 8$

$\dfrac{-3}{8} - \dfrac{1}{4} = \dfrac{-3 \times1}{8 \times 1} - \dfrac{1 \times 2}{4 \times 2}$

$= \dfrac{-3-2}{8} = \dfrac{-5}{8}$

Evaluate :
$\dfrac{2}{3} - \dfrac{4}{5}$

$\dfrac{2}{3} - \dfrac{4}{5}$
Taking LCM

$3 |\underline{3,5}$

$5 |\underline{1,5}$

$|1,1$
∴LCM of

$3$ and

$5=15$

$\dfrac{2}{3} - \dfrac{4}{5} = \dfrac{2 \times 5}{3 \times5} - \dfrac{4 \times 3}{5 \times 3}$

$= \dfrac{10 - 12}{15} = \dfrac{-2}{15}$

Evaluate :
$- 1 - \dfrac{4}{9}$

$- 1 - \dfrac{4}{9} = \dfrac{-1 \times9}{1 \times 9} - \dfrac{4 \times1}{9 \times1}$

$= \dfrac{-9-4}{9} = \dfrac{-13}{9}$

Subtract :
$\dfrac{4}{9} From \dfrac{-5}{9}$

Subtraction

$\dfrac{4}{9} From \dfrac{-5}{9}$

$\dfrac{-5}{9} - \dfrac{4}{9} = \dfrac{-5- 4}{9} = \dfrac{-9}{9} = -1$

Evaluate :
$\dfrac{-2}{7} - \dfrac{3}{-14}$

$\dfrac{-2}{7} - \dfrac{3}{-14}$
Taking LCM

$2 |2,7$

$7 |7,7$

$|1,1$

$\dfrac{-2}{7} - \dfrac{3}{-14} = \dfrac{-2 \times 2}{7 \times 2} - \dfrac{\left ( -3 \times 1\right )}{14 \times 1}$

$=\dfrac{-4 + 3}{14} = \dfrac{-1}{14}$

Subtract :
$\dfrac{-8}{11} From \dfrac{4}{11}$

Subtracting

$\dfrac{-8}{11} From \dfrac{4}{11}$

$\dfrac{4}{11} - \left ( \dfrac{-8}{11} \right ) = \dfrac{4 + 8}{11}$

$= \dfrac{12}{11} = 1 \dfrac{1}{11}$

Evaluate :
$\dfrac{5}{21} - \dfrac{-13}{42}$

Taking LCM

$2 |21,42$

$3 |21,21$

$7 |7,7$

$|1,1$

$∴LCM of 21, 42=2\times3\times7=42$

$= \dfrac{5 \times 2}{21 \times 2} -\dfrac{\left ( -13\times1 \right )}{42 \times 1}$

$= \dfrac{10 + 13}{42} = \dfrac{23}{42}$

Subtract :
$\dfrac{-9}{22} From \dfrac{5}{33}$

Subtracting

$\dfrac{-9}{22} From \dfrac{5}{33}$

Taking $LCM$
$2 |22,33$
$3 |11,33$
$11 |1,11$
$|1,1$

$∴LCM of 22 and 33=2 \times3 \times 11 = 66$

$\dfrac{5}{33} - \left ( \dfrac{-9}{22} \right ) = \dfrac{5 \times 2}{33 \times 2} + \dfrac{9 \times 3}{22 \times 3}$

$= \dfrac{10 + 27 }{66} = \dfrac{37}{66}$

Solve:
$\frac{4}{9} + \frac{2}{7}$

$= \dfrac{(4 \times 7) + (2 \times 9)}{63}$ (Taking 63 as L.C.M.)

$= \dfrac{28 + 18}{63} = \dfrac{46}{63}$

Solve:
$\displaystyle \frac{3}{4} - \frac{1}{3}$

$\displaystyle \frac{3}{4} - \frac{1}{3}= \frac{(3 \times 3) - (1 \times 4)}{12} = \frac{9 - 4}{12} = \frac{5}{12}$

On the basis of given number line, answer the following.

A is

$\dfrac{5}{6}$

B is

$\dfrac{6}{6}$

C is

$\dfrac{7}{6}$

D is

$\dfrac{8}{6}$

Consider the following number line and match each point to its value.

A is

$\dfrac{2}{4}=\dfrac{1}{2}$

B is

$\dfrac{5}{4}$

C is

$\dfrac{7}{4}$

The given number line represents the fraction ____.

the number is

$\dfrac{7}{4}$

Subtract: $\dfrac {3}{4}$ from $\dfrac {1}{3}$ .

Since the denominators are different we need to take LCM
The LCM of

$4$ and

$3$ is

$12$

$\therefore$ By multiplying both numerator and denominator with

$3$ we get,

$\dfrac {3}{4} = \dfrac {3\times 3}{4\times 3} = \dfrac {9}{12}$

And by multiplying both numerator and denominator with

$4$ we get,

$\dfrac {1}{3} = \dfrac {1\times 4}{3\times 4} = \dfrac {4}{12}$

Now by subtracting

$\dfrac {1}{3} - \dfrac {3}{4} = \dfrac {4}{12} - \dfrac {9}{12}$
Since we have the same denominators we can subtract directly,

$\dfrac {4 - 9}{12}\Rightarrow \dfrac {-5}{12}$ .

$\dfrac{-32}{13}$ from 2

1608654_564860_ans_c29561dacc1a4ae283a18b3e5d75824b.jpg

Solve the following:
$\cfrac { 7 }{ 9 } +\cfrac { 1 }{ 3 }$ =K. Find 9K

Given,

$\dfrac{7}{9} + \dfrac{1}{3} = K$

By taking LCM,

We can write,

$\dfrac{21 + 9}{27} = K$

$\dfrac{30}{27} = K$

So,

$K = \dfrac{10}{9}$

The value of

$9K =9\times\dfrac{10}{9}$

$9K = 10$

Fill in the missing fraction.
$\displaystyle \square -\frac{3}{21}=\frac{5}{21}$ .

Let the missing fraction be

$x$

According to the given condition,

$x-\dfrac{3}{21}=\dfrac{5}{21}$

$x=\dfrac{3}{21}+\dfrac{5}{21}$

$x=\dfrac{5+3}{21}\\\ \ \ =\dfrac{8}{21}$

$\Rightarrow x=\dfrac{8}{21}$

Convert the following mixed fraction into improper fraction.
$\displaystyle 3\frac{2}{8}$ .

Given, mixed fraction

$=3\dfrac{2}{8}$ .

$3\dfrac{2}{8}=\dfrac{8\times 3+2}{8}=\dfrac{26}{8}$ .

$-\dfrac{7}{5}$

1290699_564897_ans_b8cbfee29eba482ab03a16d11eea839e.jpg

Fill in the missing fraction.
$\displaystyle \square +\frac{5}{27}=\frac{12}{27}$ .

Let the missing fraction =

$x$

$x+\dfrac{5}{27}=\dfrac{12}{27}$

$x=\dfrac{12-5}{27}=\dfrac{7}{27}$

$x=\dfrac{7}{27}$

If x is a proper fraction, show that $\displaystyle\frac{x}{1-x^2}-\frac{x^3}{1-x^6}+\frac{x^5}{1-x^{10}}-.......=\displaystyle \frac{x}{1+x^2}+\frac{x^3}{1+x^6}+\frac{x^5}{1+x^{10}}+.....$

$\cfrac { x }{ 1+{ x }^{ 2 } } =x{ \left( 1-{ x }^{ 2 } \right) }^{ -1 }$

$=x\left( 1+{ x }^{ 2 }+{ x }^{ 4 }+.... \right)$

$=x+{ x }^{ 3 }+{ x }^{ 5 }+{ x }^{ 7 }+.......$

$\therefore \cfrac { { x }^{ 3 } }{ 1-{ x }^{ 6 } } ={ x }^{ 3 }{ \left( 1-{ x }^{ 6 } \right) }^{ -1 }$

$={ x }^{ 3 }+{ x }^{ 9 }+{ x }^{ 15 }+{ x }^{ 21 }+....$

Similarly;

$\cfrac { x }{ 1-{ x }^{ 2 } } -\cfrac { { x }^{ 3 } }{ 1-{ x }^{ 6 } } +\cfrac { { x }^{ 5 } }{ 1-{ x }^{ 10 } } -\cfrac { { x }^{ 7 } }{ 1-{ x }^{ 14 } } +....$

$=x+{ x }^{ 3 }+{ x }^{ 5 }+{ x }^{ 7 }+.....$

$-{ x }^{ 3 }-{ x }^{ 9 }-{ x }^{ 15 }-{ x }^{ 21 }-....$

$+{ x }^{ 5 }+{ x }^{ 15 }+{ x }^{ 25 }+{ x }^{ 35 }+....$

$-{ x }^{ 7 }-{ x }^{ 21 }-{ x }^{ 35 }-{ x }^{ 19 }-....$

....................

By adding vertical columns;-

$=x\left( 1-{ x }^{ 2 }+{ x }^{ 4 }-{ x }^{ 6 }.... \right) +{ x }^{ 3 }\left( 1-{ x }^{ 6 }+{ x }^{ 12 }-{ x }^{ 18 }+.... \right) +{ x }^{ 5 }\left( 1-{ x }^{ 10 }+{ x }^{ 20 }-{ x }^{ 30 }+.... \right)$

$=x{ \left( 1+{ x }^{ 2 } \right) }^{ -1 }+{ x }^{ 3 }{ \left( 1+{ x }^{ 6 } \right) }^{ -1 }+{ x }^{ 5 }{ \left( 1+{ x }^{ 10 } \right) }^{ -1 }+{ x }^{ 7 }{ \left( 1+{ x }^{ 14 } \right) }^{ -1 }+....$

$=\cfrac { x }{ 1+{ x }^{ 2 } } +\cfrac { { x }^{ 3 } }{ 1+{ x }^{ 6 } } +\cfrac { { x }^{ 5 } }{ 1+{ x }^{ 10 } } +\cfrac { { x }^{ 7 } }{ 1+{ x }^{ 14 } } +......$

Solve the following:
$1-\cfrac { 4 }{ 7 } =K.$ Find $14K$

Given,

$1 - \dfrac{4}{7} = K$

By taking LCM,

We can write,

$\dfrac{7 - 4}{7} = K$

$\dfrac{3}{7} = K$

So,

$K = \dfrac{3}{7}$

The value of

$14K = 14\times\dfrac{3}{7}$

$14K = 3$ x

$2 = 6$

$14 K = 6$

Convert the following improper fraction into mixed fraction.
$\displaystyle\frac{11}{2}$ .

Given improper fraction is

$\dfrac{11}{2}$ .

$\ \ \ 2)11(5$

$-10$

_______

$\ \ \ 1$

_______

We know that, a mixed fraction is formed as

$\text{Quotient}\dfrac{\text{Remainder}}{\text{Divisor}}$

$\Rightarrow 5\dfrac{1}{2}$ .

Hence,

$\dfrac{11}{2}=5\dfrac{1}{2}$ .

Solve the following:
$2+\cfrac { 3 }{ 4 }$ = K. Find 4K

Given,

$2 + \dfrac{3}{4} = K$

By taking LCM,

We can write,

$\dfrac{8 + 3}{4} = K$

$\dfrac{11}{4} = K$

So,

$K =\dfrac{11}{4}$

The value of

$4K = 4\times\dfrac{11}{4}$

$4K = 11$

Represent the following mixed infinite decimal periodic fractions as common fractions:
$\displaystyle\, (2^{3/2} + 27y^{3/5}) : \left ( \left ( \frac{1}{2} \right )^{-1/2} + 3 y^{1/5} \right )$

$(2^{3/2}+27 y^{3/5}):\bigg(\bigg(\dfrac{1}{2}\bigg)^{-1/2}+3{y}^{1/5}\bigg)=\dfrac{(2^{1/2})^{3}+(3 y^{1/5})^{3}}{(2^{1/2}+3 y^{1/5})}=2+9 y^{2/5}-2^{1/2}.3 y^{1/5}$

$\dfrac{3}{7}+\bigg(\dfrac{-6}{11}\bigg)+\bigg(\dfrac{-8}{21}\bigg)+\bigg(\dfrac{5}{22}\bigg)$ = $\bigg(\dfrac{-125}{462}\bigg)$ .If true enter $1$ else $0$ .

$\dfrac{3}{7}+\left(\dfrac{-6}{11}\right)+\left(\dfrac{-8}{4}\right)+\left(\dfrac{5}{22}\right)$

$\dfrac{3}{7}-\dfrac{6}{11}-\dfrac{8}{21}+\dfrac{5}{22}$

$\dfrac{198-252-176+105}{1\times 11\times 3\times 2}$

$=\dfrac{-125}{462}$

$\therefore$ its true the answer is

$1$

Represent the following mixed infinite decimal periodic fractions as common fractions:
$\displaystyle\, \frac{\frac{x - y}{\sqrt{x} - \sqrt{y}} - \frac{x - y}{\sqrt{x} + \sqrt{y}}}{\frac{\sqrt{x} - \sqrt{y}}{x - y} + \frac{\sqrt{x} + \sqrt{y}}{x - y}} \cdot \frac{2 \sqrt{xy}}{y -x}$

$\dfrac{x-y}{\sqrt{x}-\sqrt{y}}-\dfrac{x-y}{\sqrt{x}+\sqrt{y}}=(x-y)\dfrac{2\sqrt{y}}{x-y}=2\sqrt{y}$

$\dfrac{\sqrt{x}-\sqrt{y}}{x-y}+\dfrac{\sqrt{x}+\sqrt{y}}{x-y}=\dfrac{2\sqrt{x}}{x-y}$

$\dfrac{\dfrac{x-y}{\sqrt{x}-\sqrt{y}}-\dfrac{x-y}{\sqrt{x}+\sqrt{y}}}{\dfrac{\sqrt{x}-\sqrt{y}}{x-y}+\dfrac{\sqrt{x}+\sqrt{y}}{x-y}}.\dfrac{2\sqrt{x}\sqrt{y}}{y-x}=-2{y}$

Represent the following mixed infinite decimal periodic fractions as common fractions:
Simplify the following expressions.
$\displaystyle\left ( \left ( \frac{y}{y - x} \right )^{-2} - \frac{(x + y)^2 - 4xy}{x^2 -xy}\right ) \frac{x^4}{x^2 y^2 - y^4}$

The given equation can be written as

$[({ \dfrac { x-y }{ y } ) }^{ 2 }-\dfrac { { (x-y) }^{ 2 } }{ x(x-y) } ]\dfrac { { x }^{ 4 } }{ { y }^{ 2 }({ x }^{ 2 }{ -y }^{ 2 }) }$ [since

$(x+y)^2-4xy=(x-y)^2$ ]

$\Rightarrow [\dfrac { (x-y)({ x }^{ 2 }{ -xy-y }^{ 2 }) }{ x{ y }^{ 2 } } ][\dfrac { { x }^{ 4 } }{ { y }^{ 2 }({ x }^{ 2 }-{ y }^{ 2 }) } ]=\dfrac { { x }^{ 3 }({ x }^{ 2 }-xy-{ y }^{ 2 }) }{ { y }^{ 4 }(x+y) }$

$\Rightarrow \dfrac { { x }^{ 3 }({ x }^{ 2 }-xy-{ y }^{ 2 }) }{ { y }^{ 4 }(x+y) } =\dfrac { { x }^{ 5 }-{ x }^{ 4 }y-{ x }^{ 2 }{ y }^{ 2 } }{ x{ y }^{ 4 }+{ y }^{ 5 } }$

Using suitable rearrangement and find the sum,
$- 5 + {7 \over {10}} + ( - 3) + {3 \over 7} + \left( {{{ - 4} \over 5}} \right) + {5 \over {14}}$

$-5+\dfrac{7}{10}+(-3)+\dfrac{3}{7}+\bigg(\dfrac{-4}{5}\bigg)+\dfrac{5}{14}=-5-3+\dfrac{7-8}{10}+\dfrac{6+5}{14}=2-\dfrac{1}{10}+\dfrac{11}{14}=\dfrac{140-7+55}{70}=\dfrac{188}{70}=\dfrac{94}{35}$

What is the value of $2 + {1 \over {2 + {2 \over {5 + {2 \over 3}}}}}?$

$5+\dfrac{2}{3}=\dfrac{17}{3}$

$2+\dfrac{2}{\dfrac{17}{3}}=2+\dfrac{6}{17}=\dfrac{40}{17}$

$2+\dfrac{1}{\dfrac{40}{17}}=\dfrac{97}{17}$

$2+\dfrac{1}{2+\dfrac{2}{5+\dfrac{2}{3}}}=\dfrac{97}{17}$

If 1 is added in the numerator and denominator, then fraction becomes ${\dfrac 3 4}$ and if we substract 1 from both numerator and denominator then becomes ${\dfrac 2 3}$ , then find the fraction.

Let fraction be

$\cfrac{x}{y},$ so

$\cfrac{x+1}{y+1}=\cfrac{3}{4}\Rightarrow4x+1=3y+3$

$\Rightarrow4x-3y=-1$ ...... (i)

And also

$\cfrac{x-1}{y-1}=\cfrac{2}{3}\Rightarrow3x-3=2y-2$

$\Rightarrow3x-2y=1$ ...... (ii)

On solving equations (i) and (ii), we get

$x=5$ and

$y=7$

Thus, the fraction is

$\cfrac{5}{7}$

Find the multiplicative inverse of the following.
(i) $-13$ (ii) $\dfrac{-13}{19}$ (iii) $\dfrac{1}{5}$

(i) The multiplicative inverse of

$-13$ is

$-\dfrac{1}{13}$ . Since

$(-13).\dfrac{1}{-13}=1$ (Multiplicative identity)

(ii) The multiplicative inverse of

$\dfrac{-13}{19}$ is

$-\dfrac{19}{13}$ . Since

$\dfrac{(-13)}{19}.\dfrac{19}{-13}=1$ (Multiplicative identity)

(iii)The multiplicative inverse of

$\dfrac{1}{5}$ is

$5$ . Since

$\dfrac{1}{5}.5=1$ (Multiplicative identity)

Subtract 3 kg 765 g from 9 kg 401g

$9$ kg

$401$ g

$=9.401$ kg

$3$ kg

$765$ g

$=3.765$ kg

$9.401-3.765=5.636=5$ kg

$636$ g

Represent the rational numbers $\dfrac{1 }{ 2},\dfrac{ - 3} {4},\dfrac{3 }{ 2}$ and $\dfrac{5 }{ 8}$ on the number line.

1432549_1015362_ans_0c7ac3faf1a04c70bab86fda2f966f8c.png

What should be added to $\dfrac{4} {5}$ to get $\dfrac{49}{30}?$

Let

$x$ should be added

$\implies \dfrac{4}{5}+x=\dfrac{49}{30}\implies x=\dfrac{49}{30}-\dfrac{4}{5}=\dfrac{25}{30}=\dfrac{5}{6}$

Add the following rational numbers:
(a) $\dfrac{2}{7}$ and $\dfrac{3}{7}$
(b) $\dfrac{-4}{11}$ and $\dfrac{7}{11}$

$(a)\ \dfrac27+\dfrac37$

$\Rightarrow \dfrac{2+3}{7}$

$=\dfrac{5}{7}$

$(b)\ \dfrac{-4}{11}+ \dfrac{7}{11}$

$\Rightarrow\dfrac{-4+7}{11}$

$= \dfrac{3}{11}$

The value of $\left ({\dfrac {3} {5}}\right )$ + $\left (\dfrac {1} {2}\right )$ is

$\left (\dfrac35+\dfrac12\right )$

Take LCM of 2,5

$=LCM(2,5)=10$

$=\dfrac{(3\times2)+(1\times5)}{10}$

$=\dfrac{6+5}{10}$

$=\dfrac{11}{10}\ or\ 1.1$

Simplify :
$8-4\dfrac{1}{2}-2\dfrac{1}{4}$

1310900_1036633_ans_dc2bf0c7e3454ccfa6a1cfd14ce8aa5f.jpeg

Subtract:
$2\dfrac {3}{5}$ from $-3/7$ .

$2\cfrac { 3 }{ 5 }$ from

$\cfrac { -3 }{ 7 }$

$\cfrac { -3 }{ 7 } -\cfrac { 13 }{ 5 } =-\left( \cfrac { 15+91 }{ 35 } \right) =\cfrac { -106 }{ 35 }$

A bag contains $20$ balls of which $3$ balls are green, $10$ balls are red and the remaining are blue. What fraction of the ball represent red, blue and green balls?

Among

$20$ balls

$3$ balls are green,

$10$ balls are red,

$7$ balls are blue.

$\dfrac{3}{20}$ part of total balls are green balls,

$\dfrac{10}{20}$ part of total balls are red balls and

$\dfrac{7}{20}$ part of total balls are blue balls.

$\displaystyle 2\dfrac{3}{11}+1\dfrac{3}{77}$ .

We can write mixed function as below

$a\dfrac{b}{c}=\dfrac{a.c+b}{c}=a+\dfrac{b}{c}----(1)$

$\rightarrow$ We are given,

$2\dfrac{3}{11}+1\dfrac{3}{77}$

$=2+\dfrac{3}{11}+1+\dfrac{3}{77}$

$=3+\left(\dfrac{3}{11}+\dfrac{3}{77}\right)$

$=3+\left(\dfrac{3.7}{77}+\dfrac{3}{77}\right)$ ( Take LCM of

$11$ &

$77$ )

$=3+\left(\dfrac{21+3}{77}\right)$

$=3+\dfrac{24}{77}$

$=3\dfrac{24}{77}$ ( as question

$(1)$ )

So,

$\boxed{2\dfrac{3}{11}+1\dfrac{3}{77}=3\dfrac{24}{77}}$

Dhriti,s house is $\dfrac{9}{10}\ km$ from her school. She walked some distance and then took a bus for $\dfrac{1}{2}\ km$ to reach the school. How far did she walk ?

given distance to school

$=\dfrac{9}{10}$ km

she traveled in bus for

$\dfrac 12$ km

therefore the distance she walked

$=\dfrac{9}{10}-0.5=0.4km$

Convert $\displaystyle 3\dfrac{920}{1331}$ into improper fraction.

$3 \dfrac{920}{1331}$

$= 3+\dfrac{920}{1331} =\dfrac{1331 \times 3+ 920}{1331} = \dfrac{3993+ 920}{1331}$

$=\dfrac{4913}{1331}$

Express the following as mixed fraction
$\dfrac{17}{7}$ .

$\dfrac{17}{7}$ as mixed function

$=2\dfrac{3}{7}.$

Hence, the answer is

$2\dfrac{3}{7}.$

If the numerator of a fraction is multiplied by $2$ and its denominator is increasing by $1$ , it becomes $1$ .However if the numerator is increasing by $4$ and denominator is multiplied by $2$ , the fraction becomes $\dfrac {1}{2}$ . Find the fraction.

Let the fraction be

$\dfrac{x}{y}$

case 1

$\dfrac{{x \times 2}}{{y + 1}} = 1$

$\Rightarrow 2x = y + 1$

$\Rightarrow y = 2x - 1 - \left( 1 \right)$

case 2

$\dfrac{{x + 4}}{{y \times 2}} = \dfrac{1}{2}$

$\Rightarrow x + 4 = y$

$\Rightarrow y = x - 4$

from eq 1 &2

$x + 4 = 2x - 1$

$x = 5$

from eq 2

$y = x + 4 = 5 + 4 = 9$

Hence the required fraction is

$\dfrac{5}{9}$

A rectangle sheet of paper is $12\dfrac {1}{2}\ cm$ long and $10\dfrac {2}{3}\ cm$ wide Find the perimeter of the rectangle.

Consider the diagram shown below.

Given:

Length of the rectangle $=12\dfrac{1}{2}\ cm$

Breadth of the triangle $=10\dfrac{2}{3}\ cm$

Calculate the perimeter of the rectangle.

Perimeter $=2\times$ (Length $+$ Breadth)

$=2\times \left[ \left( 12\dfrac{1}{2} \right)+\left( 10\dfrac{2}{3} \right) \right]$

$=2\times \left[ \left( 12+\dfrac{1}{2} \right)+\left( 10+\dfrac{2}{3} \right) \right]$

$=2\times \left[ \left( 12+10 \right)+\left( \dfrac{1}{2}+\dfrac{2}{3} \right) \right]$

$=2\times \left[ 22+\left( \dfrac{3+4}{6} \right) \right]$

$=2\times \left[ 22+\dfrac{7}{6} \right]$

$=44+\dfrac{7}{3}$

$=44+2\dfrac{1}{3}$

$=46\dfrac{1}{3}\ cm$

Hence, the perimeter of the rectangle is

$46\dfrac{1}{3}\ cm$ .
1031370_1057529_ans_6aef7e91c0bb4dadabb501b9d1bb2b05.PNG

Find the area of rectangular park which is $18\dfrac{3}{5}$ m long and $8\dfrac{2}{3}$ m breadth.

The length of park is

$l=18\dfrac{3}{5}\ m$

The breadth of park is

$b=8\dfrac{2}{3}\ m$

We can write mixed fraction

as below:

$a\dfrac{b}{c}=a+\dfrac{b}{c}=\dfrac{ac+b}{c}$

Area of park

$=A=l\times b$

$A=\left(18\dfrac{3}{5}\right)\times \left(8\dfrac{2}{3}\right)$

$=\left(18+\dfrac{3}{5}\right)\times \left(8+\dfrac{2}{3}\right)$

$=18\times 8+18\times \dfrac{2}{3}+\dfrac{3}{5}\times 8+\dfrac{3}{5}\times \dfrac{2}{3}$

$=144+12+\dfrac{24}{5}+\dfrac{2}{5}$

$A=156+\dfrac{26}{5}=156+5+\dfrac{1}{5}=161\dfrac{1}{5}$

$A=161\dfrac{1}{5}$

Express the following as mixed fraction
$\dfrac{11}{5}$ .

$\dfrac{11}{5}$ as mixed function

$=2\dfrac{1}{5}.$

Hence, the answer is

$2\dfrac{1}{5}.$

How many pieces of $\dfrac{3}{2}$ m each can be cut from a lace measuring 15 m?

Given,

The length of lace $=15\,m$ and length of piece $=\dfrac{3}{2}\,m$

Now we know that ,

$number\ of\ lace$ $=\dfrac{length\ of\ lace}{length\ of\ Piece\,}$

$=\dfrac{15}{\dfrac{3}{2}}$

$=\dfrac{15\times 2}{3}$

$=\dfrac{30}{3}$

$=10$

Hence, $10$ pieces of $\dfrac32\ m$ each can be cut from a lace measuring $15\ m$ .

Evaluate the following:
$\dfrac {3}{10}+\dfrac {2}{5}$
$2\dfrac {3}{4}-\dfrac {1}{4}$

(a)Taking LCM in

$\dfrac{3}{{10}} + \dfrac{2}{5}$ .

$\dfrac{3}{{10}} + \dfrac{2}{5} = \dfrac{{3 + 2\left( 2 \right)}}{{10}}$

$= \dfrac{7}{{10}}$

(b)Simplifying

$2\dfrac{3}{4} - \dfrac{1}{4}$ .

$2\dfrac{3}{4} - \dfrac{1}{4} = \dfrac{{11}}{4} - \dfrac{1}{4}$

$= \dfrac{{11 - 1}}{4}$

$= \dfrac{{10}}{4}$

$= \dfrac{5}{2}$

Sneha can complete a work in $36$ days. Rashmi can complete the same work in $24$ days. If both work together, then they will be able to complete the work approximately.

Work done by Sneha in one day

$=\cfrac { 1 }{ 36 }$

Work done by Rashmi in one day

$=\cfrac { 1 }{ 24 }$

Total work done by both in one day

$=\cfrac { 1 }{ 36 } +\cfrac { 1 }{ 24 }$

$=\cfrac { 1 }{ 12 } \left( \cfrac { 1 }{ 3 } +\cfrac { 1 }{ 2 } \right)$

$=\cfrac { 5 }{ 72 }$

$\therefore$ Time taken by both to complete the work

$=\cfrac { 72 }{ 5 }$ days

$=14\cfrac { 2 }{ 5 }$ days

Evaluate $\left[ \dfrac { 5 }{ 4 } +\dfrac { -1 }{ 2 } \right] +\dfrac { -3 }{ 2 } \dfrac { -5 }{ 4 } +\left( \dfrac { 1 }{ 2 } +\dfrac { -3 }{ 2 } \right)$

$\left[ \cfrac{5}{4} + \cfrac{-1}{2} \right] + \cfrac{-3}{2} \cfrac{-5}{4} + \left( \cfrac{1}{2} + \cfrac{-3}{2} \right)$

$= \left[ \cfrac{5 + \left( -2 \right)}{4} \right] + \cfrac{\left( -3 \right) \times \left( -5 \right)}{8} + \left( \cfrac{1 + \left( -3 \right)}{2} \right)$

$= \cfrac{3}{4} + \cfrac{15}{8} + \cfrac{-2}{2}$

$= \cfrac{3}{4} + \cfrac{15}{8} - 1$

$= \cfrac{6 + 15 - 8}{8}$

$= \cfrac{1}{8}$

Hence the required answer is

$\cfrac{1}{8}$ .

Solve $\dfrac {5}{6}+\dfrac {(-3)}{21}$

Simplifying $\dfrac{5}{6} + \left( {\dfrac{{ - 3}}{{21}}} \right)$ .

$\dfrac{5}{6} + \left( {\dfrac{{ - 3}}{{21}}} \right)$

$= \dfrac{5}{6} - \dfrac{1}{7}$

$= \dfrac{{5\left( 7 \right) - 1\left( 6 \right)}}{{42}}$

$= \dfrac{{35 - 6}}{{42}}$

$= \dfrac{{29}}{{42}}$

Simplify $\left(2\dfrac{4}{5}-\dfrac{7}{10}\right)$

$\left(2\dfrac 45-\dfrac 7{10}\right)\\\dfrac {14}5-\dfrac 7{10}\\\dfrac{28-7}{10}=\dfrac {21}{10}=2.1$

Simplify $3\dfrac{4}{7}\times 2\dfrac{2}{5}\times 1\dfrac{3}{4}$ .

$3 \dfrac{4}{7} \times 2 \dfrac{2}{5} \times 1 \dfrac{3}{4}$

$=\dfrac{21 +4}{7} \times \dfrac{10+2}{5} \times \dfrac{4+3}{4}$

$=\dfrac{25}{7} \times \dfrac{12}{5} \times \dfrac{7}{4}= 15$

Convert into single fraction
$\dfrac{25}{11}+\dfrac{80}{77}$ .

$\dfrac{25}{11}+\dfrac{80}{77}$

$=$

$\dfrac{25\times 7+80}{77}=\dfrac{175+80}{77}=\dfrac{255}{77}$

What should be subtracted from $8\dfrac{3}{10}$ to get $2\dfrac{2}{3}$ ?

Let $x$ be the number that is to be subtracted from $8\cfrac{3}{{10}}$ to get $2\cfrac{2}{3}$ .

$8\cfrac{3}{{10}} - x = 2\cfrac{2}{3}$

$\cfrac{{21}}{{10}} - x = \cfrac{8}{3}$

$x = \cfrac{{21}}{{10}} - \cfrac{8}{3}$

$x = \cfrac{{21\left( 3 \right) - 8\left( {10} \right)}}{{30}}$

$x = \cfrac{{ - 17}}{{30}}$

Therefore, $- \cfrac{{17}}{{30}}$ is to be subtracted from $8\cfrac{3}{{10}}$ to get $2\cfrac{2}{3}$ .

Honey baked a cake. She used $4\cfrac{5}{8}$ cups of flour and $3\cfrac{1}{8}$ cups of sugar. How much more flour than sugar did Honey use?

Extra flour used = Flour used - Sugar used

$= 4\dfrac{5}{8}-3\dfrac{1}{8}$

$= \dfrac{37}{8}-\dfrac{25}{8}=\dfrac{12}{8}$

$= \dfrac{3}{2} cups$

$= 1\dfrac{1}{2}cups.$

Subtract $\dfrac{3}{4}$ from $\dfrac{1}{3}$ .

1329537_1080902_ans_0900f94d3e414e51beea22cc0bea57d8.jpg

Subtract $-7$ from $\dfrac{-4}{7}$ .

1329542_1080905_ans_6f4e73c16c654b9ead45ea4c243cd310.jpg

Bunty walked $1\cfrac{5}{8}km$ . Then he took a bus and travelled another $3\cfrac{1}{6}km$ . After that he took a taxi and travelled $\cfrac{11}{4}km$ . How much distance did he travel in all?

Total Distance

$= 1 \dfrac{5}{8}+3\dfrac{1}{6}+\dfrac{11}{4}$

$=\dfrac{13\times 3+19\times 4+11\times 6}{24}$

$=\dfrac{181}{24}km$

$=7\dfrac{13}{24} km$

Solve : $\dfrac{7}{10} + \dfrac{2}{5} + \dfrac{3}{2}$

Given expression:

$\dfrac{7}{10}+\dfrac{2}{5}+\dfrac{3}{2}=\dfrac{7+4+15}{10}$

$=\dfrac{26}{10}=\dfrac{13}{5}$

The sum of a fraction and its reciprocal is $\dfrac{13}{6}$ . Find the fraction if its numerator is $1$ less than the denominator.

Suppose that the dfraction $\dfrac{x}{y}$

According to the given question,

$\dfrac{x}{y}+\dfrac{y}{x}=\dfrac{13}{6}$

$\Rightarrow {{x}^{2}}+{{y}^{2}}=\dfrac{13}{6}xy$

$\Rightarrow 6{{x}^{2}}+6{{y}^{2}}=13xy$

$\Rightarrow 6{{x}^{2}}+6{{y}^{2}}-13xy=0\,\,\,\,\,\,\,\,.......\,\,\,\left( 1 \right)$

Now, According to the given question,

$x=y-1\,\,\,\,........\,\,\,\,\left( 2 \right)$

Put the value of y by equation (2) in (1) and we get,

$6{{\left( y-1 \right)}^{2}}+6{{y}^{2}}-13\left( y-1 \right)y=0$

$\Rightarrow 6\left( {{y}^{2}}+{{1}^{2}}-2y \right)+6{{y}^{2}}-13{{y}^{2}}+13y=0$

$\Rightarrow 6{{y}^{2}}+6-12y+6{{y}^{2}}-13{{y}^{2}}+13y=0$

$\Rightarrow -{{y}^{2}}+y+6=0$

$\Rightarrow {{y}^{2}}-y-6=0$

$\Rightarrow {{y}^{2}}-\left( 3-2 \right)y-6=0$

$\Rightarrow {{y}^{2}}-3y+2y-6=0$

$\Rightarrow y\left( y-3 \right)+2\left( y-3 \right)=0$

$\Rightarrow \left( y-3 \right)\left( y+2 \right)=0$

$\Rightarrow y-3=0\,\,\,\,\,\,\,\,\,\,\,\,y+2=0$

$\Rightarrow y=3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y=-2$

Put the value of y in equation (2) and we get,

$x=y-1\,\,\,\,........\,\,\,\,\left( 2 \right)$

$x=3-1=2$

Then, the fraction is

$=\dfrac{x}{y}=\dfrac{2}{3}$

Hence, this is the answer.

Subtract $\dfrac{-32}{13}$ from $2$ .

1329539_1080904_ans_55df0403eff9415192554f5aaed04253.jpg

Draw number lines and locate the points on them:
$\dfrac{1}{8}, \dfrac{2}{8}, \dfrac{3}{8}, \dfrac{7}{8}$

Divide space between 0 to 1 in 8 equal parts.
1437056_1075999_ans_ff2cca3e16294da88c906e73c50de110.JPG

Evaluate $2+\dfrac{1}{1-\dfrac{1}{1-\dfrac{1}{3}}}$

$=2+\dfrac{1}{1-\dfrac{1}{1-\dfrac{1}{3}}}$

$=2+\dfrac{1}{1-\dfrac{1}{\dfrac{3-1}{3}}}$

$=2+\dfrac{1}{1-\dfrac{1}{\dfrac{2}{3}}}$

$=2+\dfrac{1}{1-\dfrac{3}{2}}$

$=2+\dfrac{1}{\dfrac{2-3}{2}}$

$=2+\dfrac{1}{\dfrac{-1}{2}}$

$=2+\dfrac{2}{-1}$

$=2+(-2)=0$

Solve : $3 + \dfrac{6}{14} + 4 \dfrac{1}{4}$

$3+\dfrac{6}{14}+4\dfrac 14=3+\dfrac{6}{14}+\dfrac{17}{4}$

$=\dfrac{42+12+119}{28}=\dfrac{173}{28}$

Write the fractions representing each of the shaded regions. Are all these fractions equivalent?

$(i)$ For the first circle,

Total number of parts

$=2$

Number of shaded parts

$=1$

Fraction for the shaded region

$=\dfrac{1}{2}$

$(ii)$ For the second circle,

Total number of parts

$=4$

Number of shaded parts

$=2$

Fraction for the shaded region

$=\dfrac{2}{4}=\dfrac{1}{2}$

$(iii)$ For the third circle,

Total number of parts

$=6$

Number of shaded parts

$=3$

Fraction for the shaded region

$=\dfrac{3}{6}=\dfrac{1}{2}$

$(iv)$ For the fourth circle,

Total number of parts

$=8$

Number of shaded parts

$=4$

Fraction for the shaded region

$=\dfrac{4}{8}=\dfrac{1}{2}$

So, the fractions for the shaded region are same for all the cases.

Hence, all these fractions are equivalent.

2120402_1129281_ans_18e53af5d2ed43b78b38adc3b273069b.png

Solve : $8 \dfrac{7}{10} - 3 \dfrac{5}{11}$

Given:

$8\dfrac{7}{10}-3\dfrac{5}{11}=\dfrac{87}{10}-\dfrac{38}{11}$

$=\dfrac{87\times 11-38\times 10}{110}$

$=\dfrac{957-380}{110}$

$=\dfrac{577}{110}=5\dfrac{27}{110}$

Solve
$\dfrac{-7}{10} + \dfrac{13}{15} + \dfrac{27}{20}$

1087841_1114343_ans_6d2123da9b22489abe491bb64813a4ab.png

Solve:
$\dfrac { 5 }{ 12 } +\dfrac { 7 }{ 16 }$ = ?

Find the least common denominator or LCM of the two denominators: LCM of 12 and 16 is 48

$For\quad the\quad 1st\quad fraction,since\quad 12\ast 4=48,$

$\dfrac { 5 }{ 12 } =\dfrac { 5\times 4 }{ 12\times 4 } =\dfrac { 20 }{ 48 }$

$Likewise,for\quad the\quad 2nd\quad \quad fraction,since\quad 16×3=48,$

$\dfrac { 7 }{ 16 } =\dfrac { 7×3 }{ 16×3 } =\dfrac { 21 }{ 48 }$

$Add\quad the\quad two\quad fractions:\dfrac { 20 }{ 48 } +\dfrac { 21 }{ 48 } =\dfrac { 20+21 }{ 48 } =\dfrac { 41 }{ 48 }$

Simplify: $\dfrac { 8 }{ 5 } +\left[ \dfrac { 3 }{ 4 } +\dfrac { 5 }{ 6 } \right]$

$\cfrac { 8 }{ 5 } +\left[ \cfrac { 3 }{ 4 } +\cfrac { 5 }{ 6 } \right] \Rightarrow \cfrac { 8 }{ 5 } +\left[ \cfrac { 9+10 }{ 12 } \right] =\cfrac { 8 }{ 5 } +\cfrac { 19 }{ 12 } =\cfrac { 96+95 }{ 60 } =\cfrac { 191 }{ 60 } =3\cfrac { 11 }{ 60 }$

Simplify: $\dfrac{4}{5}+\dfrac{1}{5}$

Given,

$\dfrac{4}{5} + \dfrac{1}{5}$

$\dfrac{4 + 1}{5}$

$\dfrac{5}{5}$

= 1

$4+\dfrac{1}{2}+3\dfrac{1}{2}+9\dfrac{6}{7}$ .

1149101_1113002_ans_9d17ecce34e84f828aeee12cf304de6f.jpg

Add $2\dfrac { 5 }{ 6 } +1\dfrac { 11 }{ 12 }$

The mixed fraction convert as

$\dfrac{17}{6}+\dfrac{23}{12}$

$=\dfrac{17\times 2+23}{12}$

$=\dfrac{34+23}{12}$

$=\dfrac{57}{12}$

$=\dfrac {19}{4}$

Solve:
$4\dfrac{3}{8}-\dfrac{25}{18}$ .

$4\dfrac {3}{8}-\dfrac {25}{18}$

$\dfrac {35}{8}-\dfrac {25}{18}$

$L.C.M.\ 1118$

$2\times 2\times 2\times 3\times 3$

$=72$

$\dfrac {(35 \times 9)-(25\times 4)}{72}$

$\dfrac {5(7\times 9-5\times 4)}{72}$

$5\dfrac {(63-20)}{72}$

$\dfrac {5\times 43}{72}$

$=\dfrac {215}{72}$

Sarita bought $\dfrac{2}{5}$ meter of ribbon and Lalita $\dfrac{3}{4}$ meter of ribbon. What is the total length of the ribbon they bought?

Length of ribbon Sarita bought $= \dfrac25$ meter

Length of ribbon Lalita bought $= \dfrac34$ meter

Total length of ribbon $= \dfrac25 + \dfrac34$

LCM of $5$ & $4$ is $20$

$\dfrac {4\times 2}{5 \times 4} + \dfrac {3\times 5}{4\times 5}$

$= \dfrac {8}{20} + \dfrac {15}{20}$

$= \dfrac {23} {20}$

Thus,they bought $\dfrac {23}{20}$ meters of ribbon together.

Simplify:

$\dfrac{8}{{ - 13}} + \dfrac{6}{{ - 3}} - \dfrac{1}{2}$

Given,

$\Rightarrow \dfrac{8}{-13}+\dfrac{6}{-3}-\dfrac{1}{2}$

$\Rightarrow \dfrac{-8}{13}+\dfrac{-6}{3}-\dfrac{1}{2}$

$\Rightarrow \dfrac{-48-156-39}{78}$

$\Rightarrow \dfrac{-243}{78}\Rightarrow \dfrac{-81}{26}$

$\Rightarrow \dfrac{-81}{26}$ (Ans)

From a $20\ m$ roll of cloth, three pieces of lengths $2\dfrac{5}{8}m, 3\dfrac{1}{3}m$ and $3\dfrac{2}{5}m$ are cut out. What length of the cloth is left?

Length of cloth is

$20m$

$3$ pices with are:-

$2\dfrac {5}{6}m, 3\dfrac {1}{3}m, 3\dfrac {2}{5}$

$\Rightarrow \ \dfrac {21}{6}m, \dfrac {10}{3}m, \dfrac {17}{5}m$

cloth left is

$\Rightarrow \$

$\Rightarrow \ 20-\left [\dfrac {21}{8}+\dfrac {10}{3}+\dfrac {17}{5}\right]$

$\Rightarrow \ 20-\dfrac {(21\times 15)+10\times 40+17(24)}{120}$

$20-\dfrac {1123}{120}$

$=\dfrac {2400-1123}{120} \Rightarrow \ \dfrac {1277}{120}$

$\Rightarrow \ \boxed {1\dfrac {77}{120}m}$ remaining cloth

Solve
$4 \frac { 1 } { 8 } - 8 \frac { 1 } { 3 }$

$= 4\frac{1}{8} - 8\frac{1}{3}$

$\Rightarrow \frac{{33}}{8} - \frac{{25}}{3}$

$\Rightarrow \frac{{99 - 200}}{{24}}$

$\Rightarrow \frac{{ - 101}}{{24}} \Rightarrow - 4\frac{5}{{24}}$

Rakaesh had $1$ rope of $7\dfrac{1}{4}$ length and another $6\dfrac{3}{7}$ meters length. How much length of rope did rakesh have in all?

Total Length of Rope

$=7\dfrac14+6\dfrac37=\dfrac{29}{4}+\dfrac{45}{7}=\dfrac{29*7+45*4}{28}=\dfrac{383}{28}=13\dfrac{19}{28}$

Find the value of the following
$\dfrac{3}{5}+\dfrac{7}{10}+\left ( \dfrac{-8}{12} \right )+\dfrac{4}{3}$

Given

$=\cfrac{3}{5}+\cfrac{7}{10}-\cfrac{8}{12}+\cfrac{4}{3}$

$=\cfrac{6+7}{10}+\cfrac{16+8}{12}$

$=\cfrac{13}{10}+\cfrac{8}{12}$

$=\cfrac{(13\times 12)+80}{120}$

$=\cfrac{236}{120}$

$=\cfrac{59}{30}$

Find the sum of following.
(a) $\dfrac{3}{7},\dfrac{5}{7}$ (b) $\dfrac{4}{5},\dfrac{3}{6}$

1100460_1176041_ans_8e74412ae5244dd8a4a84d6fca272928.jpg

Change the following groups of fractions to like fractions:

(i) $\dfrac{1}{3},\dfrac{2}{5},\dfrac{3}{4},\dfrac{1}{6}$

(ii) $\dfrac{5}{6},\dfrac{7}{8},\dfrac{11}{12},\dfrac{3}{10}$

(iii) $\dfrac{2}{7},\dfrac{7}{8},\dfrac{5}{14},\dfrac{9}{16}$

$(i)\dfrac{1}{3},\dfrac{2}{5},\dfrac{3}{4},\dfrac{1}{6}$

LCM

$\left(3,4,5,6\right)=60$

$=\dfrac{1\times 20}{3\times 20},\dfrac{2\times 12}{5\times 12},\dfrac{3\times 15}{4\times 15},\dfrac{1\times 10}{6\times 10}$

$=\dfrac{20}{60},\dfrac{24}{60},\dfrac{45}{60},\dfrac{10}{60}$

$(ii)\dfrac{5}{6},\dfrac{7}{8},\dfrac{11}{12},\dfrac{3}{10}$

LCM

$\left(6,8,10,12\right)=120$

$=\dfrac{5\times 20}{6\times 20},\dfrac{7\times 15}{8\times 15},\dfrac{11\times 10}{12\times 10},\dfrac{3\times 12}{10\times 12}$

$=\dfrac{100}{120},\dfrac{105}{120},\dfrac{110}{120},\dfrac{36}{120}$

$(iii)\dfrac{2}{7},\dfrac{7}{8},\dfrac{5}{14},\dfrac{9}{16}$

LCM

$\left(7,8,14,16\right)=112$

$=\dfrac{2\times 16}{7\times 16},\dfrac{7\times 14}{8\times 14},\dfrac{5\times 8}{14\times 8},\dfrac{9\times 7}{16\times 7}$

$=\dfrac{32}{112},\dfrac{98}{112},\dfrac{40}{112},\dfrac{63}{112}$

From a rope $10 \dfrac {1}{2} m$ long, $4 \dfrac {5}{8} m$ is cut off. Find the length of the remaining rope.

Initial length of rope

$=10^{1/2}m=\dfrac{21}{2}m$

Length cut

$=4\dfrac{5}{8}m=\dfrac{31}{8}m$

Length of remaining

$=\dfrac{21}{2}-\dfrac{37}{8}$

$=\left(\dfrac{84}{8}-\dfrac{37}{8}\right)$

$=\left(\dfrac{47}{8}\right)m$

$\boxed {Length\ of\ remaining\ rope =5\dfrac{7}{8}m}$

What should be added to $- \dfrac{{17}}{{15}}$ to get $- \dfrac{{8}}{{21}}.$

Let

$x$ be added to

$\dfrac{-17}{15}$ to get

$\dfrac{-8}{21}.$

So,

$\Rightarrow \dfrac{-17}{15} + x$ =

$\dfrac{-8}{21}$

$\Rightarrow x = \dfrac{-8}{21} - \left(\dfrac{-17}{15}\right)$

$\Rightarrow x = \dfrac{-8}{21} + \dfrac{17}{15}$

$\Rightarrow x = \dfrac{(-8)(5) + (17)(7)}{105}$

$\Rightarrow x = \dfrac{(-40) + 119}{105}$

$\Rightarrow x = \dfrac{79}{105}$

Hence,

$\dfrac{79}{105}$ should be added to

$\dfrac{-17}{15}$ to get

$\dfrac{-8}{21}$

Solve:
${\left(6\dfrac {4}{11}\right)}^{2}-{\left(4\dfrac {7}{11}\right)}^{2}$

$\left( 6\dfrac{4}{11}\right)^2-\left( 4\dfrac{7}{11}\right)^2$

$=\left( \dfrac{70}{11}\right)^2-\left( \dfrac{51}{11}\right)^2$

$=\dfrac{\left(70\right)^2-\left( 51\right)^2}{\left( 11\right)^2}$

$=\dfrac{\left( 70+51\right)\left( 70-51\right)}{\left(11\right)^2}$

$\left( a^2-b^2=\left(a+b\right) \left( a-b\right)\right)$

$=\dfrac{121\left( 70-51\right)}{121}$

$\left( \left(11\right)^2=121\right)$

$=19$

Hence, the answer is

$19.$

Solve:
$\dfrac{1}{2} + \dfrac{1}{3}$

$\dfrac{1}{2}+\dfrac{1}{3}$

Taking LCM of 2 & 3 i.e, 6

$\dfrac{1\times 3}{2\times 3}+\dfrac{1\times 2}{3\times 2}$

$= \dfrac{3+2}{6}$

$= \dfrac{5}{6}$

Solve $\dfrac{2}{3}+\dfrac{1}{7}$

For the

$1^{st}$ fraction, since

$3\times 7=21$

$\dfrac{2}{3}=\dfrac{2\times7}{3\times7}=\dfrac{14}{21}$

Likewise, for the

$2{nd}$ fraction, since

$7\times3=21$

$\dfrac{1}{7}=\dfrac{1\times 3}{7\times3}=\dfrac{3}{21}$

Add the two fraction

$=\dfrac{14}{21}+\dfrac{3}{21}=\dfrac{14+3}{21}=\dfrac{17}{21}$

$=\dfrac{2}{3}+\dfrac{1}{7} is \dfrac{17}{21}$

solve : ${(5 + 2\sqrt 6 )^{x2 - 3}} + {(5 - 2\sqrt 6 )^{x2 - 3}} = 10$

$(5+2\sqrt{6})^{x^2-3}+(5-2\sqrt{6})^{x^2-3}=10$

Let

$5+2\sqrt{6}=A$ then

$5-2\sqrt{6}=\dfrac{1}{A}$

and let

$x^2-3=z$

$\therefore A^z+\dfrac{1}{A^z}=10$

$A^{2z}-10A^z+1=0$

Now let

$A^z=y$

Let

$y^2-10y+1=0$

$\Rightarrow y=\dfrac{10\pm \sqrt{100-4}}{2}$

$=\dfrac{10\pm \sqrt{96}}{2}$

$=\dfrac{5\pm 2\sqrt{6}}{0}$

$\therefore A^z=5\pm 2\sqrt{6}$

and

$(5+2\sqrt{6})^{x^2-3}(5+2\sqrt{6})^1$

$\Rightarrow x^2-3=1$

$\Rightarrow x=4$

$\Rightarrow x=\pm 2$ .

$\dfrac{2}{3}+\left(\dfrac{-4}{5}\right)+\dfrac{1}{4}+\left(\dfrac{-2}{3}\right)+\left(\dfrac{-11}{5}\right)$

$\cfrac { 2 }{ 3 } +\left( \cfrac { -4 }{ 5 } \right) +\cfrac { 1 }{ 4 } +\left( \cfrac { -2 }{ 3 } \right) +\left( \cfrac { -11 }{ 5 } \right)$

$=\cfrac { 2 }{ 3 } +\cfrac { 1 }{ 4 } -\cfrac { 2 }{ 3 } -\cfrac { 15 }{ 5 }$

$=\cfrac { 1 }{ 4 } -3=-\cfrac { 11 }{ 4 }$

Solve:
$\dfrac{3}{5} + \dfrac{1}{4}$

$\dfrac{3}{5}+\dfrac{1}{4}$

Taking LCM of 5 & 4 i.e,

$20$

$\Rightarrow \dfrac{3\times 4+1\times 5}{20} = \dfrac{12+5}{20}$

$= \dfrac{17}{20}$

Solve:
$\dfrac{1}{5} + \dfrac{1}{6}$

$\dfrac{1}{5}+\dfrac{1}{6}$

Taking LCM of

$5 \ \& \ 6$ i.e,

$30$

$\Rightarrow \dfrac{1\times 6}{5\times 6}+\dfrac{1\times 5}{6\times 5}$

$= \dfrac{6+5}{30} = \dfrac{11}{30}$

Simplify:
$4 \dfrac { 1 } { 3 } - 2 \dfrac { 2 } { 3 }$

Given:

$4\dfrac{1}{3}-2\dfrac{2}{3}$

$=\dfrac{3\times 4+1}3{-\dfrac{3\times 2+2}{3}}$

$=\dfrac{13}{3}-\dfrac{8}{3}=\dfrac{5}{3}$ .

Find the value of the equation

$\displaystyle 20 + 3 + {1 \over {10}} + {3 \over {100}} + {4 \over {1000}}$

Given:-

$=20+3 + \dfrac{1}{10} +\dfrac{3}{100} + \dfrac{4}{1000}$

$=23+0.1+0.03+0.004$

$\Rightarrow 23.1+0.03+0.004$

$\Rightarrow 23.13+0.004$

$\Rightarrow 23.134$

Replace and $mnsq^2$ in each of the following by the correct number :
(a) $\dfrac{2}{7}=\dfrac{8}{\Box }$ (b) $\dfrac{5}{8}=\dfrac{10}{\Box }$

(c) $\dfrac{3}{5}=\dfrac{\Box}{20}$ (d) $\dfrac{45}{60}=\dfrac{15}{\Box}$

(e) $\dfrac{18}{24}=\dfrac{\Box}{4}$

$a)\dfrac{2}{7}=\dfrac{8}{\Box}$

cross multiplying

$2\times{\Box}=7\times 8$

$\Box=\dfrac{7\times 8}{2}$

$\Box=7\times4$

$\Box=28$

$\dfrac{2}{7}=\dfrac{8}{28}$

$b)\dfrac{5}{8}=\dfrac{10}{\Box}$

$5\times\Box=8\times10$

$\Box=\dfrac{8\times10}{5}$

$\Box=8\times 2$

$\Box=16$

$\dfrac{5}{8}=\dfrac{10}{16}$

$c)\dfrac{3}{5}=\dfrac{\Box}{20}$

cross multiplying

$20\times 3=\Box\times 5$

$5\times\Box=3\times20$

$\Box=\dfrac{20\times3}{5}$

$\Box=4\times3$

$\Box=12$

$d)\dfrac{45}{60}=\dfrac{15}{\Box}$

cross multiplying

$45\times\Box=15\times60$

$\Box=\dfrac{15\times60}{45}$

$=\dfrac{60/3}{3/3}$

$\Box=20$

$e)\dfrac{18}{24}=\dfrac{\Box}{4}$

cross multiplying

$18\times4=\Box\times24$

$\Box=18\times4$

$\Box=\dfrac{18\times4}{24}$

$\Box=\dfrac{18/2}{6/2}$

$\Box=\dfrac{18/2}{3/3}$

$\Box=3$

Convert into improper fraction :
$2\dfrac{3}{4}$

Convert into improper fraction

$2\dfrac {3}{4}$

denominator will be same for fixed and improper multiply denominator with

$'2'$ and add to numerator

$\dfrac {4(2)+3}{4}=\dfrac {8+3}{4}=\dfrac {11}{4}$

$4\dfrac{1}{2} + 2\dfrac{5}{7} \times \dfrac{7}{{19}} - \dfrac{7}{{19}} \div 2$

$4\dfrac{1}{2}+2\dfrac{5}{7}\times \dfrac{7}{9}-\dfrac{7}{19}\div 2$

Apply BODMAS

$\Rightarrow \dfrac{9}{2}+\dfrac{19}{7}\times \dfrac{7}{9}-\dfrac{7}{19}\times 2\dfrac{1}{2}$

$\Rightarrow \dfrac{9}{2}+\dfrac{19}{9}-\dfrac{7}{38}$

LCM of

$2, 9, 38=342$

$\Rightarrow \dfrac{171\times 9+19\times 38-7\times 18}{342}$

$\Rightarrow \dfrac{2135}{342}$ .

1368966_1235813_ans_c1f4526ac4804ee0b5d4cfb3967928e1.JPG

Add the following rational numbers-
$\dfrac{3}{4},\dfrac{-2}{5}$

$\dfrac{3}{4}, \dfrac{-2}{5}$

$\dfrac{3}{4} + \left(\dfrac{-2}{5} \right)$

$= \dfrac{3 \times 5 + (-2) 4}{4 \times 5}$

$= \dfrac{15 - 8}{20}$

$= \dfrac{7}{20}$

Add the following rational numbers-
$\dfrac{3}{7},\dfrac{13}{17}$

$\dfrac{3}{7} , \dfrac{13}{17}$

$\dfrac{3}{7} + \dfrac{13}{17} = \dfrac{3 \times 17 + 13 \times 7}{7 \times 17}$

$= \dfrac{51 + 91}{119} = \dfrac{142}{119}$

$= \dfrac{142}{119}$

Convert the following into improper fraction:
$5\dfrac{3}{4}$

$5 \dfrac{3}{4}$

$=\dfrac{5(4)+3}{4}$

$=\dfrac{20+3}{4}$

$=\dfrac{23}{4}$

Convert into improper fraction :
$3\dfrac{2}{5}$

Convert into improper fraction

$3 \dfrac{2}{5}$

$=\dfrac{5(3)+2}{5}$

$=\dfrac{15+2}{5}$

$=\dfrac{17}{5}$

Add the following rational numbers-
$\dfrac{-7}{9},\dfrac{-3}{4}$

$\dfrac{-7}{9} , \dfrac{-3}{4}$

$\dfrac{-7}{9} + \left(\dfrac{-3}{4} \right)$

$= \dfrac{-7(4) + (-3) (9)}{9 \times 4}$

$= \dfrac{-28 - 27}{36}$

$= \dfrac{-55}{36}$

Find the value of $\left[ {{1^3} + {2^3} + {3^3} + {8^2}} \right] -\displaystyle {5 \over {2.}}$

Given:

$[1^3+2^3+3^3+8^2]-\dfrac{5}{2}$

$=[1+8+27+64]-\dfrac{5}{2}$

$=100-\dfrac{5}{2}$

$=\dfrac{200-5}{2}$

$=\dfrac{195}{2}$ .

Solve the following equation :-
$\dfrac {2}{3}+\dfrac {5}{3}$

$\dfrac{2}{3}+\dfrac{5}{3}$

Bases are equal (denominator)

So add numerator

$\dfrac{2+5}{3}=\dfrac{7}{3}.$

Find the value of 4 + $\dfrac{1}{2} -\dfrac{1}{2}+ 2 +\dfrac{1}{2}\left(2\dfrac{1}{2}\right)$

$4+\dfrac{1}{2}-\dfrac{1}{2}+2+\dfrac{1}{2}\left(\dfrac{5}{2}\right)$

$=4+2+\dfrac{5}{4}=6+\dfrac{5}{4}=\dfrac{29}{4}$

Shubham painted $\dfrac{2}{3}$ of the wall space in his room. His sister Madhavi helps and painted $\dfrac{1}{3}$ of the wall space. How much did they paint together.

Fraction of the wall painted by Shubham =

$\dfrac{2}{3}$

Fraction of the wall painted by Madhavi =

$\dfrac{1}{3}$

Together painted =

$\dfrac{2}{3}+\dfrac{1}{3}=1$

Hence, they painted the complete wall.

Naina was given $1\dfrac { 1 } { 2 }$ piece of cake and Najma was given $1\dfrac { 1 } { 3 }$ piece of cake. Find the total amount of cake was given to both of them.

$1\dfrac{1}{2}=\dfrac{3}{2}$

$1\dfrac{1}{3}=\dfrac{4}{3}$

Total amount of cake

$=\dfrac{3}{2}+\dfrac{4}{3}=\dfrac{9+8}{6}=\dfrac{17}{6}$

$\Rightarrow 2\dfrac{5}{6}$

$2\dfrac{5}{6}$ is the total amount of cake given to both of them.

Find the proportionality constant between $3:4$ and $2:5$

$\dfrac{3}{4}=k\times \dfrac{2}{5}$

$k=\dfrac{3\times 5}{4\times 2}=1\dfrac{15}{8}$ .

Show $\dfrac {2}{6},\ \dfrac {4}{6},\ \dfrac {8}{6}$ and $\dfrac {6}{6}$ on the number line.

$\dfrac{2}{6},\dfrac{4}{6},\dfrac{8}{6}$ and

$\dfrac{6}{6}$

$\rightarrow$ Arrange in ascending order

$\dfrac{2}{6}<\dfrac{4}{6}<\dfrac{6}{6}<\dfrac{8}{6}$

1339145_1266838_ans_0a2a792f0733499fac9744f84c23ac93.PNG

Simplify
$\dfrac{4}{7}-\dfrac{8}{9}-\dfrac{13}{7}+\dfrac{17}{21}$ .

We have,

$\dfrac{4}{7}+\left(-\dfrac{8}{9}\right)+\left(-\dfrac{13}{7}\right)+\dfrac{17}{21}$

$=\dfrac{4}{7}-\dfrac{8}{9}-\dfrac{13}{7}+\dfrac{17}{21}$

$=\dfrac{(4\times 9)-(8\times 7)-(13\times 9)+(17\times 3)}{63}$

$=\dfrac{36-56-117+51}{63}$

$=\dfrac{87-173}{63}$

$=\dfrac{-86}{63}$

$=-1.36507$

Hence, this is answer.

1197632_1242939_ans_235de498a79340f080c7fd9bf50e1a99.png

$\dfrac {8}{2^{2}+1}+\dfrac {16}{2^{6}+1}+\dfrac {32}{2^{14}+1}+\dfrac {64}{2^{14}+1}+\dfrac {128}{2^{18}+1}=$ ?

We have,

$\begin{array}{l} \frac { 8 }{ { { 2^{ 2 } }+1 } } +\frac { { 16 } }{ { { 2^{ 6 } }+1 } } +\frac { { 32 } }{ { { 2^{ 14 } }+1 } } +\frac { { 64 } }{ { { 2^{ 14 } }+1 } } +\frac { { 128 } }{ { { 2^{ 18 } }+1 } } \\ \frac { { 32 } }{ { { 2^{ 14 } }+1 } } +\frac { { 64 } }{ { { 2^{ 14 } }+1 } } :\frac { { 96 } }{ { 16385 } } \\ \frac { 8 }{ { { 2^{ 2 } }+1 } } +\frac { { 16 } }{ { { 2^{ 6 } }+1 } } +\frac { { 96 } }{ { 16385 } } +\frac { { 128 } }{ { { 2^{ 18 } }+1 } } \\ { 2^{ 2 } }+1=5 \\ =\frac { 8 }{ 5 } +\frac { { 16 } }{ { { 2^{ 6 } }+1 } } +\frac { { 96 } }{ { 16385 } } +\frac { { 128 } }{ { { 2^{ 18 } }+1 } } \\ { 2^{ 6 } }+1=65 \\ =\frac { 8 }{ 5 } +\frac { { 16 } }{ { 65 } } +\frac { { 96 } }{ { 16385 } } +\frac { { 128 } }{ { { 2^{ 18 } }+1 } } \\ Least\, \, common\, \, multiplier\, \, of\, \, 5,65,16385:859049165 \\ =\frac { { 1374478664 } }{ { 859049165 } } +\frac { { 211458256 } }{ { 859049165 } } +\frac { { 5033184 } }{ { 859049165 } } +\frac { { 419456 } }{ { 859049165 } } \\ Since\, \, the\, deno\min ator\, \, are\, \, equal\, \, ,combine\, \, the\, \, fraction: \\ \frac { a }{ b } \ne \frac { b }{ c } =\frac { { a\ne b } }{ c } \\ =\frac { { 1374478664+211458256+5033184 } }{ { 859049165 } } \\ =\frac { { 1591389560 } }{ { 859049165 } } \\ =\frac { { 318277912 } }{ { 171809833 } } \end{array}$

If $1 \dfrac { 1 } { 4 } : 2 \dfrac { 1 } { 3 } = p : q$ and $q : r = 4 \dfrac { 1 } { 2 } : 5 \dfrac { 1 } { 4 } ,$ find $p : r$

$1\dfrac{1}{4}:2\dfrac{1}{3}=p;q$

$\&$

$q:r=4\dfrac{1}{2}:5\dfrac{1}{4}$

$\dfrac{5}{4}:\dfrac{7}{3}=p:q$

$\&$

$q:r=\dfrac{9}{2}:\dfrac{21}{4}$

$\Rightarrow \dfrac{p}{q}=\dfrac{5}{4}\times \dfrac{3}{7}=\dfrac{15}{25}$

$\Rightarrow \dfrac{q}{r}=\dfrac{9}{2}\times \dfrac{9}{2}\times \dfrac{4}{21}=\dfrac{6}{7}$

$\Rightarrow p:r=\dfrac{p}{r}=\dfrac{p}{q}\times \dfrac{q}{r}$

$=\dfrac{15}{28}\times \dfrac{6}{7}$

$=\dfrac{15\times 3}{14\times 7}$

$=\dfrac{45}{98}$

Hence, the answer is

$\dfrac{45}{98}.$

Compare the fractions.
$\dfrac{3}{8}\square \dfrac{2}{4}$ .

$\frac{3}{8},\frac{2}{4}$ LCM

$\Rightarrow = 2\times 2\times 2 = 8$

$\frac{3,4}{8}$

$\Rightarrow \frac{3}{8}< \frac{4}{8} \Rightarrow \frac{3}{8}< \frac{2}{4}$

1179294_1304912_ans_2ac0262ca49a498a88af67d4e4df3ef8.jpg

Rahul studied science for $\dfrac{1}{3}$ hour and English for $\dfrac{1}{3}$ hour? How long did he study?

We have,

Rahul studied science for $=\dfrac{1}{3}\,hour=\dfrac{1}{3}\times 60=20\,\min .$

English for $=\dfrac{1}{3}\,hour=\dfrac{1}{3}\times 60=20\,\min .$

Total time $=20+20=40\,\min .$

Or $\dfrac{40}{60}hr\,=\dfrac{2}{3}\,hr.$

Hence, this is the answer.

Compare the fractions.
$\dfrac{4}{5}\square \dfrac{4}{8}$ .

$\frac{4}{5},\frac{4}{8}$ LCM

$= 2\times 2\times 2\times 5 = 40$

$\frac{32, 20}{40}$

$\Rightarrow \frac{32}{40} > \frac{20}{40} \Rightarrow \frac{4}{5} > \frac{4}{8}$

1179295_1304920_ans_d99f8a92f9974fe8b93dd2a93b59f16d.jpg

Compare the fractions.
$\dfrac{1}{4}\square \dfrac{1}{2}$ .

$\frac{1}{4},\frac{1}{2}$ LCM

$= 2\times 2 = 4$

$\frac{1,1}{4} \Rightarrow \frac{1}{4}< \frac{2}{4} \rightarrow \frac{1}{4}< \frac{1}{2}$

1179297_1304924_ans_5fd9680a1d4a4e91bce8cd829e46af6f.jpg

Simplify:
$1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{7}{5}}}$

$\begin{array}{l} We\, have \\ 1+\dfrac { 1 }{ { 1+\dfrac { 1 }{ { 1+\dfrac { 7 }{ 5 } } } } } \\ =1+\dfrac { 1 }{ { 1+\dfrac { 1 }{ { \dfrac { { 5+7 } }{ 5 } } } } } \\ =1+\dfrac { 1 }{ { 1+\dfrac { 5 }{ { 12 } } } } \\ =1+\dfrac { 1 }{ { \dfrac { { 12+5 } }{ { 12 } } } } \\ =1+\dfrac { { 12 } }{ { 17 } } \\ =\dfrac { { 17+12 } }{ { 17 } } \\ =\dfrac { { 29 } }{ { 17 } } \end{array}$

$\left( -\dfrac { 2 }{ 3 } \right) ^{ 4 }-\left( -\dfrac { 3 }{ 4 } \right) ^{ 3 }$

We have,

${{\left( -\dfrac{2}{3} \right)}^{4}}-{{\left( -\dfrac{3}{4} \right)}^{3}}$

$=\dfrac{16}{81}-\left( -\dfrac{27}{64} \right)$

$=\dfrac{16}{81}+\dfrac{27}{64}$

$=\dfrac{16\times 64+27\times 81}{81\times 64}$

$=\dfrac{1024+2187}{5184}$

$=\dfrac{3211}{5184}$

Hence, this is the answer.

Verify
$\dfrac { 2 }{ 5 } +\dfrac { 1 }{ 3 } =\dfrac { 1 }{ 3 } +\dfrac { 2 }{ 5 }$

We have,

$\dfrac{2}{5}+\dfrac{1}{3}=\dfrac{1}{3}+\dfrac{2}{5}$

$\Rightarrow \dfrac{6+5}{15}=\dfrac{5+6}{15}$

$\Rightarrow \dfrac{11}{15}=\dfrac{11}{15}$

Hence, this is the answer.

Solve
$5-\left(\dfrac {8}{11}-3\dfrac {3}{11}\right)$

$\begin{array}{l} 5-\left( { \dfrac { 8 }{ { 11 } } -3\dfrac { 3 }{ { 11 } } } \right) \\ =5-\left( { \dfrac { 8 }{ { 11 } } -\dfrac { { 36 } }{ { 11 } } } \right) \\ =5-\left( { \dfrac { { -28 } }{ { 11 } } } \right) \\ =5+\dfrac { { 28 } }{ { 11 } } \\= \dfrac { { 55+28 } }{ { 11 } } \\ =\dfrac { { 83 } }{ { 11 } } \end{array}$

Simplify : $\left[ \dfrac { 5 } { 6 } - \dfrac { 2 } { 12 } \right] - \left[ \dfrac { 6 } { 12 } - \dfrac { 1 } { 3 } \right]$

$\begin{array}{l} We\, have \\ \left[ { \dfrac { 5 }{ 6 } -\dfrac { 2 }{ { 12 } } } \right] -\left[ { \dfrac { 6 }{ { 12 } } -\dfrac { 1 }{ 3 } } \right] \\ =\left[ { \dfrac { 5 }{ 6 } -\dfrac { 1 }{ 6 } } \right] -\left[ { \dfrac { 1 }{ 2 } -\dfrac { 1 }{ 3 } } \right] \\ =\dfrac { 5 }{ 6 } -\dfrac { 1 }{ 6 } -\dfrac { 1 }{ 2 } +\dfrac { 1 }{ 3 } \\ =\dfrac { { 5-1-3+2 } }{ 6 } \\ =\dfrac { 3 }{ 6 } \\ =\dfrac { 1 }{ 2 } \\ which\, is\, the\, required\, asnwer. \end{array}$

Find the value of c if $\dfrac1c=\dfrac23+\dfrac35$

$\dfrac1c=\dfrac23+\dfrac35\\\dfrac1c=\dfrac{10+9}{15}\\\dfrac1c=\dfrac{19}{15}\\c=\dfrac{15}{19}$

$[3\frac { 1 }{ 4 } \div \{ 1\frac { 1 }{ 4 } -\frac { 1 }{ 2 } (2\frac { 1 }{ 2 } -{ \frac { 1 }{ 4 } -\frac { 1 }{ 6 } } )\} ]$

$[3\dfrac{1}{4} \div (1\dfrac{1}{4}-\dfrac{1}{2}(2\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{6}))]$

$=3\dfrac{1}{4} \div [\dfrac{5}{4}-\dfrac{1}{2}(\dfrac{5}{2}-\dfrac{1}{4}-\dfrac{1}{6})]$

$=\dfrac{13}{4} \div [\dfrac{5}{4}-\dfrac{5}{4}+\dfrac{1}{8}+\dfrac{1}{12}]$

$=\dfrac{13}{4} \div [\dfrac{1}{8}+\dfrac{1}{12}]$

$=\dfrac{13}{4} \div \dfrac{5}{24}$

$=\dfrac{13}{4} \times \dfrac{24}{5}$

$=\dfrac{78}{5}$

Represent the following number on the number line:
$\dfrac {2}{7}$

We need to find the position of the number 2/7 on the number line.

Steps of making the number:

1. Draw a number line between

$-1$ and

$1$ .

2.Each interval of the number line is divided into

$7$ parts.

3. Marked the numbers on the number line are

$-1,...,-2/7, -1/7, 0, 1/7, 2/7,..., 1$

The number

$\dfrac{2}{7}$ is less than

$1$ and more than

$0$ .

Now, mark the point

$\dfrac{2}{7}$ by A.

1602769_1366056_ans_71db001bcd264064849aacc5c797990f.png

Solve:
$\dfrac{-5}{10}+\dfrac{6}{13}+8$

LCM

$\left(10,13,1\right)=130$

$\dfrac{-5}{10}+\dfrac{6}{13}+\dfrac{8}{1}$

$=\dfrac{-5}{10}\times\dfrac{13}{13}+\dfrac{6}{13}\times\dfrac{10}{10}+\dfrac{8}{1}\times\dfrac{130}{130}$

$=\dfrac{-65+60+1040‬}{130}=\dfrac{1055}{130}=\dfrac{211}{26}$

Simplify: $\dfrac{5}{3}+\dfrac{11}{2}+\dfrac{-9}{4}+\dfrac{-8}{3}+\dfrac{-7}{2}$

$\dfrac{5}{3}+\dfrac{11}{2}+\dfrac{-9}{4}+\dfrac{-8}{3}+\dfrac{-7}{2}$

$=\left(\dfrac{5}{3}+\dfrac{-8}{3}\right)+\left(\dfrac{11}{2}+\dfrac{-7}{2}\right)+\dfrac{-9}{4}$

$=\dfrac{5-8}{3}+\dfrac{11-7}{2}+\dfrac{-9}{4}$

$=\dfrac{-3}{3}+\dfrac{4}{2}+\dfrac{-9}{4}$

$=-1+2+\dfrac{-9}{4}$

$=1+\dfrac{-9}{4}$

$=\dfrac{4-9}{4}=\dfrac{-5}{4}$

A person earns $Rs. 100$ in a day. If he spents $Rs. 14\dfrac {2}{7}$ on food and $Rs. 30\dfrac {2}{3}$ on petrol. How much did he save on that day?

Given
A person's earning in a day

$= Rs. 100$
Money spent on food

$= Rs. 14\dfrac {2}{7} = Rs. 100/ 7$
Money spent on petrol

$= Rs. 30\dfrac {2}{3} = Rs. 92/3$
The savings of a person is calculated as follows:
Savings

$= Rs. 100 - \left \{(100/ 7 + 92/ 3)\right \}$

$= Rs. 100 - \left \{(300 + 644) / 21\right \}$
On further calculation, we get

$= Rs. 100 - (944/ 21)$

$= (2100 - 944) / 21$

$= Rs. 1156/21$

$= Rs. 55\dfrac {1}{21}$
Hence, a person saved Rs.

$55\dfrac {1}{21}$ on that day.

Simplify : $\dfrac{2}{3}+3\dfrac{1}{6}+4\dfrac{2}{9}+2\dfrac{5}{18}$

We have,

$\dfrac { 2 }{ 3 } +3\dfrac { 1 }{ 6 } +4\dfrac { 2 }{ 9 } +2\dfrac { 5 }{ 18 }$

$\Rightarrow \dfrac { 2 }{ 3 } +\dfrac { 19 }{ 6 } +\dfrac { 38 }{ 9 } +\dfrac { 41 }{ 18 }$

$\Rightarrow \dfrac { \left( 2\times 6 \right) +\left( 19\times 3 \right) +\left( 38\times 2 \right) +41 }{ 18 }$

$\Rightarrow \dfrac { 186 }{ 18 } =\dfrac { 31 }{ 3 } =\boxed { 10\dfrac { 1 }{ 3 } }$

Hence, this is the answer.

Subtract $\dfrac{-3}{8}$ from $\dfrac{-5}{7}$

The additive inverse of

$\dfrac{-3}{8}$ is

$\dfrac{3}{8}$

$\therefore \dfrac{-5}{7}-\left(\dfrac{-3}{8}\right)$

$=\dfrac{-5}{7}+\dfrac{3}{8}$

$=\dfrac{-5\times 8}{7\times 8}+\dfrac{3\times 7}{8\times 7}$

$=\dfrac{-40}{56}+\dfrac{21}{56}$

$=\dfrac{-40+21}{56}$

$=\dfrac{-19}{56}$

Verify the following:
$\dfrac{-5}{8}+\dfrac{3}{5}=\dfrac{3}{5}+\dfrac{-5}{8}$

L.H.S

$=\dfrac{-5}{8}+\dfrac{3}{5}$

$=\dfrac{-5\times 5}{8\times 5}+\dfrac{3\times 8}{5\times 8}$

$=\dfrac{-25}{40}+\dfrac{24}{40}$

$=\dfrac{-25+24}{40}=\dfrac{-1}{40}$

R.H.S

$=\dfrac{3}{5}+\dfrac{-5}{8}$

$=\dfrac{3\times 8}{5\times 8}+\dfrac{-5\times 5}{8\times 5}$

$=\dfrac{24}{40}+\dfrac{-25}{40}$

$=\dfrac{24-25}{40}=\dfrac{-1}{40}$

$\therefore$ L.H.S

$=$ R.H.S

Hence verified

I bought one dozen bananas and ate five of them.What fraction of the total number of bananas was left?

Number of bananas bough

$=1$

Dozen

$=12$

Number of bananas eaten by me

$=5$

Number of bananas left

$=12-5=7$

Fraction

$=\dfrac{7}{12}$

Solve
$\left(\dfrac{1}{2}\right)^{-2}+\left(\dfrac{1}{3}\right)^{-2}+\left(\dfrac{1}{4}\right)^{-2}$

$(\frac{1}{2})^{-2}+(\frac{1}{3})^{-2}+(\frac{1}{4})^{-2}$

$=(\frac{2}{1})^{2}+(\frac{3}{1})^{2}+(\frac{4}{1})^{2}=2^{2}+3^{2}+4^{2}$

=4+9+16=29.

1188263_1343638_ans_55a01a82446d48629152dd8a388507e5.jpg

Simplify: $\dfrac{1}{3}+\dfrac{1}{6}$

$\dfrac{1}{3}+\dfrac{1}{6}$

$=\dfrac{2}{6}+\dfrac{1}{6}$

$=\dfrac{3}{6}=\dfrac{1}{2}$

Anita put $\dfrac{1}{4}$ cup of flour in a bowl.Meena mixed $\dfrac{1}{6}$ cup of flour in it.How much more flour should Arti mix in it equal to $1$ cup of flour?

Anita put

$\dfrac{1}{4}^{th}$ cup of flour in a bowl.

Meena put

$\dfrac{1}{6}^{th}$ cup of flour.

Total quantity of flour mixed

$=\dfrac{1}{4}+\dfrac{1}{6}=\dfrac{6+4}{24}=\dfrac{10}{24}=\dfrac{5}{12}$

Arti mixes flour to get

$1$ cup of flour

$=1-\dfrac{5}{12}=\dfrac{12}{12}-\dfrac{5}{12}=\dfrac{12-5}{12}=\dfrac{7}{12}$

$\therefore\,\dfrac{7}{12}^{th}$ cup of flour is mixed.

Solve:
$\displaystyle \frac{5}{2} - \frac{3}{5} \times \frac{7}{2} + \frac{3}{5} \times \left( {\frac{{ - 2}}{3}} \right)$

Now,

$\displaystyle \frac{5}{2} - \frac{3}{5} \times \frac{7}{2} + \frac{3}{5} \times \left( {\frac{{ - 2}}{3}} \right)$

$=\displaystyle \frac{5}{2} - \frac{21}{10} -\frac{2}{5}$

$=\displaystyle \frac{25-21}{10} -\frac{2}{5}$

$=\displaystyle \frac{4}{10} -\frac{2}{5}$

$=\displaystyle \frac{2}{5} -\frac{2}{5}$

$=0$ .

Steve is planning to bake 3 loaves of bread. Each loaf calls for $5\frac{1}{4}$ cups of flour. He knows he has 20 cups on hand . Will he have enough flour left for a cake recipe that requires $3\frac{3}{4}$ cups?

Let us change the mixed fractions to improper fractions to ease our calculations.

So,

$\,\,5\dfrac{1}{4}=\dfrac{20+1}{4}=\dfrac{21}{4}$

Similarily,

$3\dfrac{3}{4}=\dfrac{12+3}{4}=\dfrac{15}{4}$

Now, Cups of flour required for 3 loaves of bread

$=3\times \dfrac{21}{4}=\dfrac{63}{4}$ cups

Flour left with Steve

$=20-\dfrac{63}{4}=\dfrac{17}{4}$ cups

Flour required for cake recipe

$=\dfrac{15}{4}$ cups

Hence, he will have enough flour for cake recipe.

Simplify $1\dfrac { 2 }{ 75 } -\dfrac { 3 }{ 8 } +\dfrac { 1 }{ 4 }$ .

$1\dfrac{2}{75}-\dfrac{3}{8}+\dfrac{1}{4}$

$=\dfrac{77}{75}-\dfrac{3}{8}+\dfrac{2}{8}$

$=\dfrac{77}{75}+\dfrac{2-3}{8}$

$=\dfrac{77}{75}-\dfrac{1}{8}$

$=\dfrac{616‬-75}{600}$

$=\dfrac{541}{600}$

Which is greater, the sum of $\dfrac {-7}{12}$ and $\dfrac {2}{5}$ of the sum of $\dfrac {-3}{8}$ and $\dfrac {-11}{-12}$ , and by how much?

Let

$x=\dfrac{-7}{12}+\dfrac{2}{5}=\dfrac{-35+24}{60}=\dfrac{-11}{60}; y=\dfrac{-3}{8}-\dfrac{11}{-12}=\dfrac{-3}{8}+\dfrac{11}{12}$

$=\dfrac{-36+88}{96}=\dfrac{52}{96}=\dfrac{13}{24}$

$y > 0, x < 0\Rightarrow y$ is greater by

$y-x=\dfrac{13}{24}+\dfrac{11}{60}=\dfrac{1}{6}\left(\dfrac{13}{4}+\dfrac{11}{10}\right)=\dfrac{130+44}{40\times 6}=\dfrac{174}{40\times 6}=\dfrac{29}{40}$ .

1214412_1395849_ans_449beef66bc247c491410088d0225944.jpg

Show that $\dfrac{3}{7}$ lies between $\dfrac{2}{5}$ and $\dfrac{5}{7}$

$\dfrac{3}{7}$ will lie between

$\dfrac{2}{5}$ and

$\dfrac{5}{7}$ if

$\dfrac{2}{5}>\dfrac{3}{7}>\dfrac{5}{7}$ or

$\dfrac{2}{5}<\dfrac{3}{7}<\dfrac{5}{7}$

Now, comparing

$\dfrac{2}{5},\dfrac{3}{7},\dfrac{5}{7}$

L.C.M of

$5$ and

$7-35$

$\therefore\dfrac{2}{5}=\dfrac{2\times 7}{5\times 7}=\dfrac{14}{35}$

$\dfrac{3}{7}=\dfrac{3\times 5}{7\times 5}=\dfrac{15}{35}$

and

$\dfrac{5}{7}=\dfrac{5\times 5}{7\times 5}=\dfrac{25}{35}$

$\therefore\,\dfrac{14}{35}<\dfrac{15}{35}<\dfrac{25}{35}$

$\dfrac{2}{5}<\dfrac{3}{7}<\dfrac{5}{7}$

$\therefore\,\dfrac{3}{7}$ lies between

$\dfrac{2}{5}$ and

$\dfrac{5}{7}$

Evaluate $1\cfrac{2}{5} - \cfrac{3}{8} + \cfrac{1}{4}$ .

$1\dfrac{2}{5}-\dfrac{3}{8}+\dfrac{1}{4}$

$=\dfrac{7}{5}-\dfrac{3}{8}+\dfrac{2}{8}$

$=\dfrac{7}{5}-\dfrac{1}{8}$

$=\dfrac{56-5}{40}$

$=\dfrac{51}{40}$

Solve: $\dfrac{x+35}{6}-\dfrac{x+7}{9}=\dfrac{x+21}{4}$

$\dfrac{x+35}{6}-\dfrac{x+7}{9}=\dfrac{x+21}{4}$

Factors of

$6=2\times 3$

Factors of

$9=3\times 3$

L.C.M of

$6,\,9=18$

$\Rightarrow \,\dfrac{3\left(x+35\right)}{18}-\dfrac{2\left(x+7\right)}{18}=\dfrac{x+21}{4}$

$\Rightarrow \,\dfrac{3x+105}{18}-\dfrac{2x+14}{18}=\dfrac{x+21}{4}$

$\Rightarrow \,\dfrac{3x+105-2x-14}{18}=\dfrac{x+21}{4}$

$\Rightarrow \,\dfrac{x+91}{18}=\dfrac{x+21}{4}$

$\Rightarrow \,\dfrac{x+91}{9}=\dfrac{x+21}{2}$

$\Rightarrow \,2x+182=9x+189$

$\Rightarrow \,9x-2x=182-189=-7$

$\Rightarrow \,7x=-7$

$\Rightarrow \,x=\dfrac{-7}{7}=-1$

$\therefore x=-1$

Evaluate $2 \dfrac{3}{4}+2 \dfrac{2}{3}$ .

$2\dfrac{3}{4}+2\dfrac{2}{3}$

$=\dfrac{11}{4}+\dfrac{8}{3}$

$=\dfrac{33+32}{12}$

$=\dfrac{65}{12}$

$=5\dfrac{5}{12}$

Simplify :- $\dfrac { -2 }{ 3 } +\dfrac { 5 }{ 7 } + \dfrac { 14 }{ 11 }$

$\dfrac{-2}{3}+\dfrac{5}{7}+\dfrac{14}{11}$

L.C.M of

$3,7,11=231‬$ since LCM of prime numbers is its product.

$=\dfrac{-2\times 77}{231}+\dfrac{5\times 33}{231}+\dfrac{14\times 21}{231}$

$=\dfrac{-154+165+294}{231}$

$=\dfrac{305}{231}$

(a)
$\dfrac{2}{3}$ $\dfrac{4}{3}$
$\dfrac{1}{3}$ $\dfrac{2}{3}$
(b)
$\dfrac{1}{2}$ $\dfrac{1}{3}$
$\dfrac{1}{3}$ $\dfrac{1}{4}$

$a)\dfrac 23+\dfrac 43=\dfrac 63=2\\\dfrac 13+\dfrac 23=\dfrac 33=1\\\dfrac 23-\dfrac 13=\dfrac 13\\\dfrac 43-\dfrac 23=\dfrac 23\\\dfrac 13+\dfrac 23=\dfrac 33=1$

$b)\dfrac 12+\dfrac 13=\dfrac {3+2}6=\dfrac 56\\\dfrac 13+\dfrac 14=\dfrac {4+3}{12}=\dfrac 7{12}\\\dfrac 12-\dfrac 13=\dfrac {3-2}6=\dfrac 16\\\dfrac 13-\dfrac 14=\dfrac {4-3}{12}=\dfrac 1{12}\\\dfrac 16+\dfrac 1{12}=\dfrac {2+1}{12}=\dfrac 14$

Express $\dfrac{-11}{9}$ as a rational number with denominator $-99$

Now we are to express

$\dfrac{-11}{9}$ as a rational number with denominator

$99$ then the given fraction will be

$\dfrac{-11\times -11}{9\times -11}$

$=\dfrac{121}{-99}$ .

This is the required fraction.

Simplify
$\dfrac { 1 }{ 2 } \times\dfrac{1}{4}$

$\dfrac{1}{2}\times\dfrac{1}{4}=\dfrac{1\times 1}{2\times 4}=\dfrac{1}{8}$

Simplify $\dfrac { 1 }{ 2 } -\dfrac { 1 }{ 3 }$ .

$\dfrac{1}{2}-\dfrac{1}{3}$

$=\dfrac{3-2}{6}$

$=\dfrac{1}{6}$

Simplify:
$3\dfrac{1}{5}$ + $\dfrac{6}{8}$

Now,

$3\dfrac{1}{5}$ +

$\dfrac{6}{8}$

$=\dfrac{16}{5}+\dfrac{3}{4}$

$=\dfrac{64+15}{20}$

$=\dfrac{79}{20}$ .

Find the reduced form of ratio $\dfrac {38}{57}$ .

The reduced form of ratio

$\dfrac {38}{57}=\dfrac{2\times 19}{3\times 19}$ is

$\dfrac{2}{3}$ .

Express $\dfrac{-11}{9}$ as a rational number with denominator $-18$

Now we are to express

$\dfrac{-11}{9}$ as a rational number with denominator

$-18$ then the given fraction will be

$\dfrac{-11\times -2}{9\times -2}$

$=\dfrac{22}{-18}$

This is the required fraction.

Show $\dfrac{2}{6}, \dfrac{4}{6}, \dfrac{8}{6}, \dfrac{5}{6}$ and $\dfrac{6}{6}$ on the number line also arrange them in ascending order.

1213049_1411768_ans_45594f0f50d14b75ae434f37d6cec438.jpg

Solve: $\dfrac{4}{5} + \dfrac{2}{3}$

$\dfrac{4}{5}+\dfrac{2}{3}$

$\dfrac{4\times3}{5\times3}+\dfrac{2\times5}{3\times5}$

$\dfrac{12}{50}+\dfrac{10}{15}$

$=\dfrac{12+10}{15}=\dfrac{22}{15}$

Solve: $\dfrac{3}{10} + \dfrac{7}{15}$

$\dfrac{3}{10}+\dfrac{7}{15}$

LCM of

$10$ and

$15$ is

$30$

$=\dfrac{3\times3}{10\times3}+\dfrac{7\times2}{10\times2}$

$\dfrac{9}{30}+\dfrac{14}{30}$

$\dfrac{9+14}{30}$

$\dfrac{23}{30}$

Find the reciprocal of $\displaystyle {3 \over 5}$ , $\displaystyle{9 \over {10}}$ , $\displaystyle{3 \over 9}$ , $\displaystyle{7 \over 4}$ and $\displaystyle{9 \over 2}$ also classify them as proper and improper fractions.

Proper fraction - where the numerator is less than the denominator.

Improper fraction - where the numerator is greater than the denominator.

By using the above definition:

Fraction	Reciprocal	Type of fraction
$\dfrac{3}{5}$	$\dfrac53$	Improper
$\dfrac9{10}$	$\dfrac{10}9$	Improper
$\dfrac39$	$\dfrac93$	Improper
$\dfrac74$	$\dfrac47$	Proper
$\dfrac92$	$\dfrac29$	Proper

Simplify $\dfrac { 3 }{ 4 } +\dfrac { -2 }{ 5 }$ .

Given

$\dfrac{3}{4},\dfrac{-2}{5}$

$\Rightarrow \dfrac{3}{4}+(\dfrac{-2}{5})$

$=\dfrac{3}{4}-\dfrac{2}{5}$

$=\dfrac{3\times 5-2\times 4}{4\times 5}$

$=\dfrac{15-8}{20}$

$=\dfrac{7}{20}$

What can you say about the prime factorisation of the denominator of $27.\overline {142857}$

Let

$x=27.\overline{142857}$

1000000 x=27142857.

$\overline{142857}$

$\Rightarrow$ 1000000 x -x=271428.57.

$\overline{142857}-27.\overline{142857}$

$\Rightarrow 999999x=27142830$

$\Rightarrow x=\frac{27142830}{999999}$

$\Rightarrow x=\frac{27142830}{9\times 111111}$

$\Rightarrow x=\frac{27142830}{3^{2}\times 111111}$

1218292_1440686_ans_f56b33818a4e4502abb1b56979ab62f8.jpg

Mary bought $3\dfrac { 1 }{ 2 }$ m of lace. show used $1\dfrac { 3 }{ 4}$ m of lace for her new dress How much lace in left with her?

Mary bought $3 \dfrac {1}{2} m$ of lace

She used $1 \dfrac {3}{4} m$ of lace for her new dress

left lace = bought $3\dfrac {1}{2} m$ of lace - used $1 \dfrac {3}{4} m$ of lace

$\Rightarrow 3 \dfrac {1}{2}m - 1 \dfrac {3}{4} m = \dfrac {( 2 \times 3) +1}{2}m - \dfrac {( 1\times 4) +3}{4} m = \dfrac { 6 +1}{2}m - \dfrac { 4 +3 }{4} m$

$\Rightarrow \dfrac {7}{2}m - \dfrac {7}{4} m = \dfrac {( 7 \times 4) - ( 7\times 2) }{ 2 \times 4} m = \dfrac {28 -14}{8} m$

$\Rightarrow \dfrac{14}{8}m = \dfrac { 14 \div 2}{ 8 \div 2}m = \dfrac {7}{2} m$

Asha and Samuel have bookshelves of the same size partly filled with books. Asha's shelf is $\dfrac { 5 }{ 6 }$ th full and Samuel's shelf is $\dfrac { 2 }{ 5 }$ th full. Whose bookshelf is more full? By what fraction?

A fraction of books in Asha’s shelf

$=5/6$

A fraction of books in Samuel’s shelf

$=2/5$

We compare the fractions to find whose shelf is more full.

LCM of

$5$ &

$6$ is

$30$

$\dfrac{5}{6}$

$=\dfrac{5\times5}{5\times6}$

$=\dfrac{25}{30}$

$\dfrac{2}{5}$

$=\dfrac{2\times6}{5\times6}$

$=\dfrac{12}{30}$

$\dfrac{25}{30}>\dfrac{12}{30}$

$\dfrac{5}{6}>\dfrac{2}{5}$

So, Asha’s shelf has more books and is more full.

$\dfrac{5}{6}−\dfrac{2}{5}$

$=\dfrac{5\times5}{6\times5}−\dfrac{2\times6}{5\times6}$

$=\dfrac{25}{30}−\dfrac{12}{30}$

$=\dfrac{13}{30}$

Asha’s shelf is more full and by

$=\dfrac{13}{30}$

Add the following fractions.
$\dfrac { 3 }{ 2 } ,\dfrac { 13 }{ 17 }$

Given

$\dfrac{3}{2},\dfrac{13}{17}$

$=\dfrac{3}{2}+\dfrac{13}{17}$

$=\dfrac{3\times 17+13\times 2}{2\times 17}$

$=\dfrac{51+26}{34}$

$=\dfrac{77}{34}$

Add the following fractions.
$\dfrac { -7 }{ 9 } ,\dfrac { -3 }{ 4 }$

Given

$\dfrac{-7}{9},\dfrac{-3}{4}$

$\Rightarrow \dfrac{-7}{9}+\dfrac{-3}{4}$

$\Rightarrow \dfrac{-7\times 4}{9\times 4}-\dfrac{3\times 9}{4\times 9}$

$\Rightarrow \dfrac{-28}{36}-\dfrac{27}{36}$

$\Rightarrow -\dfrac{55}{36}$

Jaidev takes $2\dfrac { 1 }{ 5 }$ minutes to walk across the school ground. Rahul takes $\dfrac { 7 }{ 4 }$ minutes to do the same. who takes less time and by what fraction?

Time taken by Jaidev to walk

$= 2 \dfrac{1}{5}$ minutes

$=\dfrac{11}{5}$ minutes

Time taken by Rahul to walk across the ground

$=\dfrac{7}{4}$ minutes

Now, compare the fractions

$\dfrac{11}{5}$

$=\dfrac{11\times4}{5\times4}$

$=\dfrac{44}{20}$

$\dfrac{7}{4}$

$=\dfrac{7\times5}{4\times5}$

$=\dfrac{35}{20}$

$\dfrac{44}{20}>\dfrac{35}{20}$

$⇒\dfrac{7}{4}<\dfrac{11}{5}$

So, Rahul takes lesser time than Jaidev.

Difference

$=\dfrac{11}{5}−\dfrac{7}{4}$

$=\dfrac{11\times4}{5\times4}−\dfrac{7\times5}{4\times5}$

$=\dfrac{44}{20}−\dfrac{35}{20}$

$=\dfrac{9}{20}$

So, Rahul takes lesser time than Jaidev, by

$\dfrac{9}{20}$ minutes.

$7$ hours $=$ ________ fraction of a day.

1311055_1584623_ans_591d12fe66b943ee85bb6b7688dc8af0.jpg

subract 2 from 6

2 - 6 = -4

Write these fractions appropriately as additions or subtractions:

$\textbf{Step -1: Shaded circle in each rectangle}$

$\text{It is observed that all three rectangles are divided in 6 equal circles.}$

$\text{The first rectangle has all 2 out of 6 circles shaded, whereas the}$

$\text{the second circle has 3 out of 6 circles shaded.}$

$\text{The third circle has 5 out of 6 circles shaded.}$

$\textbf{Step -2: Fractions written as additions.}$

$\text{The three rectangles as a fraction are represented as follows, }$

$\dfrac{2}{6} , \dfrac{3}{6}, \dfrac{5}{6}$

$\text{It is clearly seen that the addition of first and second rectangle }$

$\text{is the fraction represented by the third rectangle.}$

$\text{Hence, }\dfrac{2}{6} + \dfrac{3}{6}=\dfrac{2+3}{6} = \dfrac{5}{6}$

$\textbf{Thus, the fractions are written appropriately as }$

$\mathbf{\dfrac{2}{6} + \dfrac{3}{6}=\dfrac{5}{6}.}$

Simplify : $6+\left \{ \frac{4}{3} (\frac{3}{4}-\frac{1}{3}) \right \}$

1300299_1609024_ans_5809d295fd61486ca84865f97016bdbd.jpg

$3-\dfrac{12}{5}$ .

1297883_1584473_ans_f70a3e0179ce452b949420ac93b2c457.jpg

Find the sum : $\dfrac{3}{4}+\dfrac{5}{6}+\dfrac{11}{8}$

Sum of

$\dfrac{3}{4},\dfrac{5}{6},\dfrac{11}{8}$

$\implies \dfrac{3}{4}+\dfrac{5}{6}+\dfrac{11}{8}=\dfrac{3\times 6}{4\times 6}+\dfrac{5\times 4}{6\times 4}+\dfrac{11\times 3}{8\times 3}$

$= \dfrac{18}{24}+\dfrac{20}{24}+\dfrac{33}{24}$

$=\dfrac{18+20+33}{24}$

$=\dfrac{71}{24}$

Solve:
$\dfrac {8}{15} + \dfrac {3}{15}$

$\dfrac {8}{15} + \dfrac {3}{15} = \dfrac {8 + 3}{15} = \dfrac {11}{15}$ .

Represent $\dfrac {2}{5}$ on a number line.

1670872_1657936_ans_78fdcd01f3934bafb2cc4faba7bce7fe.png

Solve :
$\dfrac { 1 }{ 4 } +\dfrac { 0 }{ 4 }$

$\dfrac{1}{4}+\dfrac{0}{4}$

$\dfrac{1}{4}+0$

$\dfrac{1}{4}$

Express the following as improper fractions:
$7\dfrac {3}{4}$

$7\dfrac {3}{4}$

$= \dfrac {(4\times 7) + 3}{4}$

$= \dfrac {31}{4}$ .

Represent $\dfrac {0}{8}$ and $\dfrac {8}{8}$ on a number line.

$\dfrac{0}{8}=0$ and

$\dfrac{8}{8}=1$

1671752_1657939_ans_680d8c99a907453da3dd0e588b1e8e7b.JPG

Convert each of the following into a mixed fractions:
$\dfrac {145}{9}$

After dividing

$145$ by

$9$ we get

$16$ as quotient as

$1$ as remainder. So,

$\dfrac{145}{9}=16\dfrac{1}{9}$

Convert each of the following into an improper fraction:
$5\dfrac {3}{10}$

1501523_1657954_ans_27ab28c3582e40b7a7d046a54e0bbb0e.jpg

Represent $\dfrac {2}{7}, \dfrac {5}{7}$ and $\dfrac {6}{7}$ on a number line.

1671023_1657938_ans_e6c86fce4fdc4af79e94cee723e4651a.png

Convert each of the following into an improper fraction:
$7\dfrac {1}{4}$

1501518_1657952_ans_b15357d270f1477b97a96c3e3defef79.jpg

Convert each of the following into a mixed fractions:
$\dfrac {226}{15}$

1501515_1657949_ans_5853233915a54f81a18a39a460fd2e9e.jpg

Represent $\dfrac {0}{10}, \dfrac {1}{10}, \dfrac {5}{10}$ and $\dfrac {10}{10}$ on a number line.

1671015_1657937_ans_e0fc5f57577c4de4a4e159a1658d3003.jpg

Convert each of the following into an improper fraction:
$8\dfrac {5}{7}$

1501519_1657953_ans_8187017b9da14203957b563e7f1a11b6.jpg

Convert each of the following into a mixed fractions:
$\dfrac {128}{5}$

1501517_1657951_ans_a2fe8039bd464857b13e07a94c688d96.jpg

Convert each of the following into a mixed fractions:
$\dfrac {28}{9}$

1501510_1657948_ans_d0a9b46fef2b42f4b74fc94adc097d6e.jpg

Solve :
$\dfrac {7}{31} - \dfrac {4}{31} + \dfrac {9}{31}$

1501847_1658029_ans_ebac74dd668440a287f8e145130c6fc7.jpg

Solve:
$\dfrac {3}{22} + \dfrac {7}{22}$

$\dfrac{3}{22} + \dfrac{7}{22}$

$= \dfrac{3 + 7}{22} = \dfrac{10}{22}$

$= \dfrac{5}{11}$

Write these fractions appropriately as addition or subtractions.

$\dfrac {1}{5} + \dfrac {2}{5} = \dfrac {3}{5}$

Solve:
$\dfrac {0}{15} + \dfrac {2}{15} +\dfrac {1}{15}$

1501779_1658028_ans_0152ba47f3bf46d4b5077b255069c9df.jpg

Convert each of the following into an improper fraction:
$12\dfrac {7}{15}$

1501524_1657955_ans_03fd9a11e2dc470baa472a1c3a639d7a.jpg

Solve :
$\dfrac {4}{13} + \dfrac {2}{13} + \dfrac {1}{13}$

$\dfrac{4}{13} + \dfrac{2}{13} + \dfrac{1}{13}$

$=\dfrac{4+2+1}{13}$

$= \dfrac{7}{13}$

Arrange the following fractions in the ascending order:
$\dfrac {3}{5}, \dfrac {1}{5}, \dfrac {4}{5}, \dfrac {2}{5}$

Given,

$\dfrac {3}{5}, \dfrac {1}{5}, \dfrac {4}{5}, \dfrac {2}{5}$

In the ascending order:

$\dfrac {1}{5}, \dfrac {2}{5}, \dfrac {3}{5}, \dfrac {4}{5}$

As denominator is same for all, fraction with greater numerator will be greater.

Arrange the following fractions in the ascending order:
$\dfrac {37}{47}, \dfrac {37}{50}, \dfrac {37}{100}, \dfrac {37}{1000}, \dfrac {37}{85}, \dfrac {37}{41}$

Denominator having less value is more when numerator is same.

Ascending order :

$\dfrac{37}{1000} , \dfrac{37}{100} , \dfrac{37}{85} , \dfrac{37}{50} , \dfrac{37}{47}, \dfrac{37}{41}$

Find the equivalent fraction of $\dfrac {3}{5}$ , having:
numerator $9$

To have numerator

$9$ , we should find out

$9 \div 3=3$

So, we must multiply numerator and denominator by

$3$ to get the equivalent fraction :

$\dfrac{3\times 3}{5 \times 3}=\dfrac{9}{15}$

Solve:
$\dfrac {1}{4} + \dfrac {0}{4}$

1501805_1658026_ans_aeacb43deccf4edaabcdc826bdacca3f.jpg

Replace $\square$ by the correct number.
$\square - \dfrac {1}{5} = \dfrac {1}{2}$

1502106_1658065_ans_475b489c04b94ab2bf4e2d3477b39381.jpg

Find the sum:
$4\dfrac{3}{4}+9\dfrac{2}{5}$

1515345_1675721_ans_17607e4f825e4d36bbaf7f2198a21189.jpg

Find the sum:
$\dfrac{5}{6}+3+\dfrac{3}{4}$

1514908_1675732_ans_0ff8b947b23d4275bd99b36c61ed7eca.jpg

Find the difference of
$\dfrac{21}{25}$ and $\dfrac{18}{20}$

1590739_1675774_ans_25725d0098654098aaddcd556aa444c8.jpg

Replace $\square$ by the correct number :
$\dfrac {1}{2} - \square = \dfrac {1}{6}$ .

1502107_1658066_ans_c19ef87f7f2f47b4b9b5b4af24546267.jpg

Replace $\square$ by the correct number.
$\square - \dfrac {5}{8} = \dfrac {1}{4}$

1501794_1658064_ans_b95ee903abe145379cb687e537c8b178.jpg

Find the difference of
$\dfrac{13}{24}$ and $\dfrac{7}{16}$

1514874_1675747_ans_674b3c7013c040e3907f798b1abd59e6.jpg

Find the sum:
$2\dfrac{3}{5}+4\dfrac{7}{10}+2\dfrac{4}{15}$

1590732_1675744_ans_260a8ef421d84161983ac9419768b828.jpg

Find the sum:
$\dfrac{5}{8}+\dfrac{3}{10}$

1514906_1675718_ans_b554cfd06bb2458284514d3604eb8b67.jpg

Find the difference of
$6$ and $\dfrac{23}{3}$

1515663_1675771_ans_6a3dfe37ae044e15a03df1dd983dbc48.jpg

Simplify:
$\dfrac{2}{3}+\dfrac{1}{6}-\dfrac{2}{9}$

1515667_1675799_ans_434cbc278daf46769d6bd578b6a10836.jpg

Write down the numerator of each of the following rational numbers:
$\dfrac{-17}{-25}$

1590773_1676031_ans_5e3018c4d7634c8195e4bf5e2c53872b.jpg

Write down the numerator of each of the following rational numbers:
$\dfrac{15}{-4}$

For fraction

$\dfrac {15} {-4}$ . numerator is

$15$

Find the difference of
$8-\dfrac{5}{9}$

1514940_1675786_ans_eb989f6e08724fdea1b58fba7395a19b.jpg

Find the difference of
$3\dfrac{3}{10}$ and $\dfrac{7}{15}$

1590741_1675777_ans_33a4a617348b4a6483730d76fdd7a7c9.jpg

Find the difference of
$4\dfrac{3}{10}-1\dfrac{2}{15}$

1514942_1675793_ans_7e427e9398874f13877db11c6041a039.jpg

Simplify:
$7\dfrac{5}{6}-4\dfrac{3}{8}+2\dfrac{7}{12}$

1513281_1675806_ans_9475e261dc924ceeb74b12a3929616e2.jpg

Find the difference of
$9-5\dfrac{2}{3}$

1514941_1675790_ans_c429f5d68d664dfca7c62e061c0e7504.jpg

Write down the numerator of each of the following rational numbers:
$\dfrac{-7}{5}$

Given rational number is

$\dfrac{-7}{5}$

Numerator

$=-7$

Find the difference of
$\dfrac{6}{7}-\dfrac{9}{11}$

1514938_1675782_ans_0eabde0f27e54c6d8fd7afbc6e6009b2.jpg

Write the following rational numbers as integers:
$\dfrac{95}{1}$

$\dfrac{95}{1}$ as a integer van be written as

$95$ .

Write down the denominator of the following rational number:
$\dfrac{-15}{-82}$

The given rational number is

$\dfrac{-15}{-82}=\dfrac{Numertor}{Denominator}$

So, the Denominator is

$-82$

Express $\dfrac{2}{5}$ as a rational number with numerator :
$500$

1516297_1676238_ans_7e12c8f61b3b415fae5da0017cf68c3f.jpg

Write the following rational numbers as integers:
$\dfrac{34}{1}$

$\textbf{Step 1: Convert the rational number into integer by making its denominator as 1.}$

$\text{Given number,}$

$\dfrac{34}{1}$

$\text{ as denominator is already we need not to convert into integer}$

$\text{Hence, Integer is 34}$

$\textbf{Hence, Rational number }\boldsymbol{\dfrac{34}{1}\textbf{ can b written as 34.} }$

Express $\dfrac{2}{5}$ as a rational number with numerator :
$-750$

1591454_1676236_ans_e8f3e914562441e292d26458be7f474f.jpg

Write the following rational numbers as integers:
$\dfrac{-73}{1}$

writing as a integer

$\dfrac{-73}{1}=-73$

Write down the numerator of each of the following rational numbers:
$\dfrac{8}{9}$

For fraction

$\dfrac 89$
numerator is

$8$

Write down the denominator of each of the following rational numbers:
$\dfrac{-4}{5}$

1516003_1676044_ans_7f8d9898995342c2878e87de17722fa9.jpg

Write down the denominator of each of the following rational numbers:
$15$

1516015_1676059_ans_8012cc3055b0413a9e9d1b12eb957109.jpg

Write the following integers as rational numbers with denominator $1$ :
$-15$

$-15$ with denominator

$1$ can be written as

$\dfrac {-15}{1}$

Express $\dfrac{-192}{108}$ as a rational number with numerator:
$-48$

1591456_1676262_ans_eaac11a94db043429b2da77302c207dc.jpg

Express $\dfrac{168}{-294}$ as a rational number with denominator:
$-49$

1591464_1676276_ans_00b13aa024294a569181b509357d2088.jpg

Express $\dfrac{-192}{108}$ as a rational number with numerator:
$-16$

1516301_1676246_ans_f75336e61ce245208dbbde675ba35390.jpg

Write $\dfrac{-14}{42}$ in a form so that the numerator is equal to:
$-2$

1516416_1676300_ans_b65aea5899674e6b8de2f47ee5c257bd.jpg

Write $\dfrac{-14}{42}$ in a form so that the numerator is equal to:
$7$

1516448_1676303_ans_6caafa577b65466f8d8a6c1db127158f.jpg

Express $\dfrac{168}{-294}$ as a rational number with denominator:
$1470$

1516415_1676291_ans_f020a4a4f1a24612ad7d3676ec171489.jpg

Express $\dfrac{-192}{108}$ as a rational number with numerator:
$32$

1516316_1676248_ans_8ab52cb0e69048aab1d520909343d9b2.jpg

Write $\dfrac{-14}{42}$ in a form so that the numerator is equal to:
$-70$

1591465_1676306_ans_a26bfa1efa88489daa98e4a21324f24d.jpg

Write $\dfrac{-14}{42}$ in a form so that the numerator is equal to:
$42$

1516449_1676305_ans_43c5ea9276c34576ad99aa0a95103220.jpg

Express $\dfrac{-192}{108}$ as a rational number with numerator:
$64$

1516299_1676244_ans_b776237f60ba421a84f55d94d3c01f98.jpg

Write each of the following rational numbers in the standard form:
$\dfrac{299}{-161}$

1591504_1676361_ans_29f4b93a1cae463eb17a1cc4931d8891.jpg

Write each of the following rational numbers in the standard form:
$\dfrac{68}{-119}$

A rational number

$\dfrac ab$ is said to be in the standard form if

$b$ is positive, and the integers

$a$ and

$b$ have no common divisor other than

$1.$

So,

$\begin{aligned}{}\frac{{68}}{{ - 119}}& = - \frac{{17 \times 4}}{{17 \times 7}}\\ &= - \frac{4}{7}\end{aligned}$

Write each of the following rational numbers in the standard form:
$\dfrac{4}{-16}$

1591497_1676359_ans_1fed5e77a6d04801a294cca7634c76e0.jpg

Write each of the following rational numbers in the standard form:
$\dfrac{-63}{-210}$

A rational number

$\dfrac a b$ is said to be in the standard form if

$b$ is positive, and the integers

$a$ and

$b$ have no common divisor other than

$1.$

1516515_1676363_ans_a96977bdeeab4d4591019db51866cda5.jpg

Write each of the following rational numbers in the standard form:
$\dfrac{-195}{275}$

A rational number a/b is said to be in the standard form if b is positive, and the integers a and b have no common divisor other than 1.
1516519_1676369_ans_d96a4f63a792404b9ffc5f3842386f89.jpg

1516519_1676369_ans_d96a4f63a792404b9ffc5f3842386f89.jpg

Write each of the following rational numbers in the standard form:
$\dfrac{2}{10}$

A rational number

$\dfrac a b$ is said to be in the standard form if

$b$ is positive, and the integers

$a$ and

$b$ have no common divisor other than

$1.$
Given

$\dfrac{2}{10}$

Divide the numerator and denominator by 2, we get

$=\dfrac{1}{5}$

Write each of the following rational numbers in the standard form:
$\dfrac{-15}{-35}$

A rational number

$\dfrac a b$ is said to be in the standard form if

$b$ is positive, and the integers

$a$ and

$b$ have no common divisor other than

$1.$

1516490_1676360_ans_a63d83a437a34e18a9b532833829fb5a.jpg

Fill in the blanks so as to make the statement true:
The standard form of $-1$ is ______

The standard form of

$-1$ is

$\dfrac {-1}{1}$ .

Write each of the following rational numbers in the standard form:
$\dfrac{-8}{36}$

1591496_1676356_ans_5deff82a6ce847c283e12920ea98ff71.jpg

Evaluate each of the following:
$-\dfrac{4}{7} - \dfrac{2}{-3}$

1515131_1676699_ans_58a7f579eb974d12959499cbd6a72aa2.jpg

Evaluate each of the following:
$\dfrac{4}{7} - \dfrac{-5}{-7}$

1516660_1676708_ans_3609b096f3ca438288c558db9fd5de00.jpg

Evaluate each of the following:
$\dfrac{2}{3} - \dfrac{3}{5}$

1515105_1676695_ans_845a620c80514579a3b9a454cbfadcc7.jpg

Simplify:
$\dfrac{-8}{19} + \dfrac{-4}{57}$

1516633_1676662_ans_dfccf4c8983e400293353c469e4ebb18.jpg

Simplify:
$\dfrac{8}{9} + \dfrac{-11}{6}$

1513643_1676651_ans_03d3471524604fa88229b18fb4934cc3.jpg

Add and express the sum as a mixed fraction :
$\dfrac{-31}{6}$ and $\dfrac{-27}{6}$

1591537_1676671_ans_4b9df4c979f243b593f83d0904fde5ed.jpg

Add and express the sum as a mixed fraction :
$\dfrac{24}{7} + \dfrac{-11}{7}$

1591535_1676668_ans_8d1c584e8ca84d66b90cbc8d69fa4896.jpg

Add and express the sum as a mixed fraction :
$\dfrac{-12}{5} + \dfrac{43}{10}$

1517967_1676665_ans_ab9c2d2a40884ddab85163661f8eb51c.jpg

Simplify:
$\dfrac{-5}{16} + \dfrac{7}{24}$

1516259_1676655_ans_5011b33c3b1a45e49faf1def6e091ff8.jpg

Simplify:
$\dfrac{1}{-12} + \dfrac{2}{-15}$

1516262_1676659_ans_ec4df77e93ce404c80cce411663f564b.jpg

What number should be subtracted from $\dfrac{3}{7}$ to get $\dfrac{5}{4}$ ?

1516910_1676756_ans_8f49513ec7b74f228bcd5c9f63556681.jpg

Evaluate each of the following:
$-2 - \dfrac{5}{9}$

1515234_1676714_ans_8e63aad43ed54e5390a00c681151b977.jpg

What should be added to $\displaystyle \left(\frac{1}{2} + \frac{1}{3} + \frac{1}{5}\right)$ to get $3$ ?

1516268_1676764_ans_2410cc7725e04b59939b65f072ccd827.jpg

What should be added to $\left(\dfrac{2}{3} + \dfrac{3}{5}\right)$ to get $\dfrac{-2}{15}$ ?

1518012_1676761_ans_0a25436719e54c80a032213a15bdf66f.jpg

What number should be added to $\dfrac{-5}{11}$ so as to get $\dfrac{26}{33}$ ?

1516637_1676733_ans_ab2e86f9dab54421beacd3df3067b577.jpg

Simplify:
$\displaystyle \frac{-3}{2} + \frac{5}{4} -\frac{7}{4}$

1516274_1676772_ans_a09ce107cbdd4d27addd8ce8401fa2e0.jpg

What should be subtracted from $\left(\dfrac{3}{4} - \dfrac{2}{3} \right)$ to get $\dfrac{-1}{6}$ ?

1515244_1676768_ans_fe4a3b4024f24f95819ef2c19b126b12.jpg

Simplify:
$\displaystyle \frac{5}{4} -\frac{7}{6} -\frac{-2}{3}$

1518029_1676778_ans_05290939b5cf4219bf595d99319272f4.jpg

Simplify:
$\displaystyle \frac{5}{3} - \frac{7}{6} + \frac{(-2)}{3}$

1515029_1676776_ans_9225b3a466384e09b91de585375eeecd.jpg

What number should be added to $\dfrac{-5}{7}$ to get $\dfrac{-2}{3}$ ?

1515236_1676741_ans_2431ff51474a47d58e43c6af1ed5f73c.jpg

Subtract : $\dfrac {-13}{9}$ from $0$ .

Anything subtracted from

$0$ remains the same.

$\therefore 0 - \dfrac {-13}{9} = \dfrac {0 - (-13)}{9} =\dfrac {0 + 13}{9} \Rightarrow \dfrac {13}{9}$ .

Simplify:
$\displaystyle \left(\frac{-9}{4} \times \frac{5}{3}\right) + \left(\frac{13}{2} \times \frac{5}{6}\right)$

1515062_1676874_ans_7e52f38820f64ed5b3a7932fa5b66e46.jpg

Fill in the blanks:
$\dfrac{-4}{13} - \dfrac{-3}{26} = ...$

1516284_1676785_ans_08779f8096354a79b25b43ce0010205c.jpg

Fill in the blanks:
$\dfrac{-7}{9} + ... = 3$

1516300_1676793_ans_3df97cd709d44d70bd67b7ab7b2f0f17.jpg

Fill in the blanks:
$\dfrac{-9}{14}+.... = -1$

1517935_1676789_ans_7fab25eeadf7441dbce15e24bdaf053e.jpg

Subtract :
$\dfrac {-7}{1}$ from $\dfrac {-4}{7}$ .

Since the denominators are different we need to take LCM

The LCM of

$1$ and

$7$ is

$7$

$\therefore$ By multiplying both numerator and denominator with

$7$ we get,

$\dfrac {-7}{1} = \dfrac {-7\times 7}{1\times 7} = \dfrac {-49}{7}$

And by multiplying both numerator and denominator with

$1$ we get,

$\dfrac {-4}{7} = \dfrac {-4\times 1}{7\times 1} = \dfrac {-4}{7}$

Now by subtracting

$\dfrac {-4}{7} - \dfrac {-7}{1} = \dfrac {-4}{7} - \dfrac {-49}{7}$

Since we have the same denominators we can subtract directly,

$\dfrac {-4 - (-49)}{7} = \dfrac {-4 + 49}{7}\Rightarrow \dfrac {45}{7}$ .

Subtract :
$\dfrac {-32}{13}$ from $\dfrac {-6}{5}$ .

Since the denominators are different we need to take LCM
The LCM of

$13$ and

$5$ is

$65$

$\therefore$ By multiplying both numerator and denominator with

$5$ we get,

$\dfrac {-32}{13} = \dfrac {-32 \times 5}{13\times 5} = \dfrac {-160}{65}$

And by multiplying both numerator and denominator with

$13$ we get,

$\dfrac {-6}{5} = \dfrac {-6\times 13}{5\times 13} = \dfrac {-78}{65}$

Now by subtracting

$\dfrac {-6}{5} - \dfrac {-32}{13} = \dfrac {-78}{65} - \dfrac {-160}{65}$
Since we have the same denominators we can subtract directly,

$\dfrac {-78 - (-160)}{65} = \dfrac {-78 + 160}{45} \Rightarrow \dfrac {82}{65}$ .

Subtract :
$\dfrac {-8}{9}$ from $\dfrac {-3}{5}$ .

Since the denominators are different we need to take LCM

The LCM of

$9$ and

$5$ is

$45$

$\therefore$ By multiplying both numerator and denominator with

$5$
we get,

$\dfrac {-8}{9} = \dfrac {-8\times 5}{9\times 5} = \dfrac {-40}{45}$

And by multiplying both numerator and denominator with

$9$ we get,

$\dfrac {-3}{5} = \dfrac {-3\times 9}{5\times 9} = \dfrac {-27}{45}$

Now by subtracting

$\dfrac {-3}{5} - \dfrac {-8}{9} = \dfrac {-27}{45} - \dfrac {-40}{45}$
Since we have the same denominators we can subtract directly,

$\dfrac {-27 - (-40)}{45} = \dfrac {-27 + 40} {45} = \dfrac {13}{45}$ .

Fill in the blanks:
$.... + \dfrac{15}{23} = 4$

1515248_1676796_ans_c8ae24bc57f3404d9d63f98a58309745.jpg

Subtract :
$\dfrac {-9}{7}$ from $\dfrac {-1}{1}$ .

Since the denominators are different we need to take LCM
The LCM of

$7$ and

$1$ is

$7$

$\therefore$ By multiplying both numerator and denominator with

$1$ we get,

$\dfrac {-9}{7} = \dfrac {-9\times 1}{7\times 1} = \dfrac {-9}{7}$

And by multiplying both numerator and denominator with

$7$ we get,

$\dfrac {-1}{1} = \dfrac {-1\times 7}{1\times 7} = \dfrac {-7}{7}$

Now by subtracting

$\dfrac {-1}{1} - \dfrac {-9}{7} = \dfrac {-7}{7} - \dfrac {-9}{7}$

Since we have the same denominators we can subtract directly,

$\dfrac {-7 - (-9)}{7} = \dfrac {-7 + 9}{7}= \dfrac {2}{7}$ .

Find the following ratio in the simplest form:
$24$ to $56$

To convert given ratio

$a:b$ to its simplest form, we divide each term by the

$HCF$ of

$a$ and

$b$

$\therefore 24:56$

$=\dfrac{24}{56}$

$=\dfrac{(24\div 8)}{(56\div 8)}$

$[\because HCF$ of

$24$ and

$56$ is

$8$

$=\dfrac37$

$\therefore$ The simplest form of

$24:56$ is

$3:7$

Using the rearrangement property find the sum:
$\dfrac {4}{3} + \dfrac {3}{5} + \dfrac {-2}{3} + \dfrac {-11}{5}$ .

We shall arrange the numbers of the same denominators

$\left (\dfrac {4}{3} + \dfrac {-2}{3}\right ) + \left (\dfrac {3}{5} + \dfrac {-11}{5}\right )$

Since it has the same denominators we can add them directly

$\left (\dfrac {4 + (-2)}{3}\right ) + \left (\dfrac {3 + (-11)}{5}\right ) = \left (\dfrac {4 - 2}{3}\right ) + \left (\dfrac {3 - 11}{5}\right ) = \dfrac {2}{3} + \dfrac {-8}{5}$

We now take the LCM of

$3, 5$ i.e.

$15$

$\therefore$ By multiplying both numerator and denominator with

$5$

we get,

$\dfrac {2}{3} = \dfrac {2\times 5}{3\times 5} = \dfrac {10}{15}$

And by multiplying both numerator and denominator with

$3$ we get,

$\dfrac {-8}{5} = \dfrac {-8\times 3}{5\times 3} = \dfrac {-24}{15}$

$\dfrac {10}{15} + \dfrac {-24}{15}$ Since it has the same denominators we can add directly

$\dfrac {10 + (-24)}{15} = \dfrac {10 - 24}{15}\Rightarrow \dfrac {-14}{15}$ .

On one day a rickshaw puller earned $Rs. 160$ . Out of his earnings he spent $Rs. 26 \dfrac 35$ on tea and snacks. $Rs. 50 \dfrac {1}{2}$ on food and $Rs. 16 \dfrac 25$ on repairs of the rickshaw. How much did he save on that day?

We know that the given details are

Total earnings

$= Rs. 160$

Earnings spent on tea and snacks

$= Rs. 26 \dfrac 35$

Earning spent on food

$= Rs. 50 \dfrac 12$

Earnings sped on repairs

$= Rs. 16 \dfrac 25$

$\therefore$ Total expenditure

$=$ Earnings spent on tea and snacks

$+$ Earnings spent on food

$+$ Earnings spent on repairs

$= Rs. 26 \dfrac 35 + Rs. 50\dfrac {1}{2} + Rs. 16 \dfrac 25$

$= \dfrac {133} 5 + \dfrac {101}2 + \dfrac {82}5$

By taking LCM for

$5$ and

$2$ is

$10$

$=\dfrac {(133\times 2 + 101\times 5 + 82\times 2)} {10}$

$= \dfrac {(266 + 505 + 164)} {10}$

$= \dfrac {935} {10}$

Total Savings

$=$ Total Earnings

$-$ Total Expenditure

$= Rs. 160 - Rs. \dfrac {935} {10}$

By taking

$10$ as the LCM

$= \dfrac {(160\times 10 - 935\times 1)} {10}$

$= \dfrac {(1600 - 935)} {10}$

$= \dfrac {665} {10} = 66\dfrac {1}{2}$

$\therefore$ Total Savings on that day is

$Rs. 66\dfrac {1}{2}$ .

What number should be added to $\dfrac {-5}{8}$ so as to get $\dfrac {-3}{2}$ ?

Let us consider

$x$

$\dfrac {-5}{8} + x = \dfrac {-3}{2}$

Solving for

$x$ ,

$x = \dfrac {-3}{2} - \dfrac {-5}{8}$

By taking LCM of

$2$ and

$8$ is

$8$

$\dfrac {(-3\times 4) - (-5\times 1)}{8} = \dfrac {-12 - (-5)}{8} = \dfrac {-12 + 5}{8} = \dfrac {-7}{8}$

$x = \dfrac {-7}{8}\therefore \dfrac {-7}{8}$ should be added to

$\dfrac {-5}{8}$ to get

$\dfrac {-3}{2}$ .

What number should be added to $-1$ so as to get $\dfrac {5}{7}$ ?

Let us consider

$x$

$-1 + x = \dfrac {5}{7}$

Solving for

$x$ ,

$x = \dfrac {5}{7} - (-1) = \dfrac {5}{7} + 1$

By taking LCM of

$2$ and

$8$ is

$8$

$\dfrac {5 + (1\times 7)}{7} = \dfrac {5 + 7}{7} = \dfrac {12}{7}$

$x = \dfrac {12}{7}$

$\therefore \dfrac {12}{7}$ should be added to

$-1$ to get

$\dfrac {5}{7}$ .

Find the following ratio in the simplest form:
$84$ paise to Rupees $3$

To convert given ratio

$a:b$ to his simplest form, we divide each term by the

$HCF$ of

$a$ and

$b$

Now,

$84$ paise to Rupees

$3=0.84$ paise to

$300$ paise

$=84:300$

$=\dfrac{84}{300}$

$=\dfrac{(84\div12)}{(300\div 12)}\hspace{1cm}[\because HCF$ of

$84$ and

$300$ is

$12]$

$=\dfrac7{25}$

$\therefore$ The simplest form of

$84$ paise to

$3$ Rupees is

$7:25$

Using the rearrangement property find the sum:
$\dfrac {-6}{7} + \dfrac {-5}{6} + \dfrac {-4}{9} + \dfrac {-15}{7}$ .

We arrange the numbers as such

$\dfrac {-6}{7} + \dfrac {-15}{7} + \dfrac {-5}{6} + \dfrac {-4}{9}$

$\dfrac {-6}{7} + \dfrac {-15}{7}$ Since it has the same denominators we can add directly

$\dfrac {-6 + (-15)}{7} = \dfrac {-6 - 15}{7}\Rightarrow \dfrac {-21}{7}$ we can further divide by

$3$ i.e.

$\dfrac {-21}{7} = \dfrac {-3}{1}$

LCM of

$6$ and

$9$ is

$18$

By multiplying both numerator and denominator with

$3$ we get,

$\dfrac {-5}{6} = \dfrac {-5\times 3}{6\times 3} = \dfrac {-15}{18}$

And by multiplying both numerator and denominator with

$2$ we get,

$\dfrac {-4}{9} = \dfrac {-4\times 2}{9\times 2} = \dfrac {-8}{18}$

$\dfrac {-15}{18} + \dfrac {-8}{18}$ Since it has the same denominators we can add directly

$\dfrac {-15 + (-8)}{18} = \dfrac {-15 - 8}{18}\Rightarrow \dfrac {-23}{18}$

$\therefore \dfrac {-3}{1} + \dfrac {-23}{18}$ We take the LCM of

$1$ and

$18$ i.e.

$18$

By multiplying both numerator and denominator with

$18$ we get,

$\dfrac {-3}{1} = \dfrac {-3\times 18}{1\times 18} = \dfrac {-54}{18}$

And by multiplying both numerator and denominator with

$1$ we get,

$\dfrac {-23}{18} = \dfrac {-23 \times 1}{18\times 1} = \dfrac {-23}{18}$

$\dfrac {-54}{18} + \dfrac {-23}{18}$ Since it has the same denominators we can add directly

$\dfrac {-54 + (-23)}{18} = \dfrac {-54 - 23}{18} \Rightarrow \dfrac {-77}{18}$ .

Using the rearrangement property find the sum:
$\dfrac {-8}{3} + \dfrac {-1}{4} + \dfrac {-11}{6} + \dfrac {3}{8}$ .

We arrange the numbers as such

$\dfrac {-8}{3} + \dfrac {-11}{6} + \dfrac {-1}{4} + \dfrac {3}{8}$

We take the LCM of

$3, 6$ and

$4, 8$

$\therefore$ LCM of

$3$ and

$6$ is

$6$

By multiplying both numerator and denominator with

$2$ we get,

$\dfrac {-8}{3} = \dfrac {-8\times 2}{3\times 2} = \dfrac {-16}{6}$

And by multiplying both numerator and denominator with

$1$ we get,

$\dfrac {-11}{6} = \dfrac {-11\times 1}{6\times 1} = \dfrac {-11}{6}$

$\dfrac {-16}{6} + \dfrac {-11}{6}$ Since it has the same denominators we can add directly

$\dfrac {-16 + (-11)}{6} = \dfrac {-16 - 11}{6}\Rightarrow \dfrac {-27}{6}$

LCM of

$4$ and

$8$ is

$8$

By multiplying both numerator and denominator with

$2$ we get,

$\dfrac {-1}{4} = \dfrac {-1\times 2}{4\times 2} = \dfrac {-2}{8}$

And by multiplying both numerator and denominator with

$1$ we get,

$\dfrac {3}{8} = \dfrac {3\times 1}{8\times 1} = \dfrac {3}{8}$

$\dfrac {-2}{8} + \dfrac {3}{8}$ Since it has the same denominators we can add directly

$\dfrac {-2 + 8}{8}\Rightarrow \dfrac {6}{8}$

$\therefore \dfrac {-27}{6} + \dfrac {6}{8}$ We take the LCM of

$6$ and

$8$ i.e.

$24$

By multiplying both numerator and denominator with

$4$ we get,

$\dfrac {-27}{6} = \dfrac {-27\times 4}{6\times 4} = \dfrac {-108}{24}$

And by multiplying both numerator and denominator with

$3$ we get,

$\dfrac {1}{8} = \dfrac {1\times 3}{8\times 3} = \dfrac {3}{24}$

$\dfrac {-2}{8} + \dfrac {3}{8}$ Since it has the same denominators we can add directly

$\dfrac {-108 + 3}{24}\Rightarrow \dfrac {-105}{24}$

$\dfrac {-105}{24}$ Can be further divided

$\dfrac {-105 \div 3}{24\div 3} \Rightarrow \dfrac {-35}{8}$ .

Using the rearrangement property find the sum:
$\dfrac {-13}{20} + \dfrac {11}{14} + \dfrac {-5}{7} + \dfrac {7}{10}$ .

We arrange the numbers as such

$\dfrac {-13}{20} + \dfrac {11}{14} + \dfrac {-5}{7} + \dfrac {7}{10}$

We take the LCM of

$20, 10$ and

$14, 7$

$\therefore$ LCM of

$20$ and

$10$ is

$20$

By multiplying both numerator and denominator with

$1$ we get,

$\dfrac {-13}{20} = \dfrac {-13\times 1}{20\times 1} = \dfrac {-13}{20}$

And by multiplying both numerator and denominator with

$2$ we get,

$\dfrac {7}{10} = \dfrac {7\times 2}{10\times 2} = \dfrac {14}{20}$

$\dfrac {-13}{20} + \dfrac {14}{20}$ Since it has the same denominators we can add directly

$\dfrac {-13 + 14}{20}\Rightarrow \dfrac {1}{20}$

LCM of

$14$ and

$7$ is

$14$

By multiplying both numerator and denominator with

$1$ we get,

$\dfrac {11}{14} = \dfrac {11\times 1}{14\times 1} = \dfrac {11}{14}$

And by multiplying both numerator and denominator with

$2$ we get,

$\dfrac {-5}{7} = \dfrac {-5\times 2}{7\times 2} = \dfrac {-10}{14}$

$\dfrac {11}{14} + \dfrac {-10}{14}$ Since it has the same denominators we can add directly

$\dfrac {11 + (-10)}{14} = \dfrac {11 - 10}{14}\Rightarrow \dfrac {1}{14}$

$\therefore \dfrac {1}{20} + \dfrac {1}{14}$ We take the LCM of

$20$ and

$14$ i.e.

$140$

By multiplying both numerator and denominator with

$7$ we get,

$\dfrac {1}{20} = \dfrac {1\times 7}{20\times 7} = \dfrac {7}{140}$

And by multiplying both numerator and denominator with

$10$ we get,

$\dfrac {1}{14} = \dfrac {1\times 10}{14\times 10} = \dfrac {10}{140}$

$\dfrac {7}{140} + \dfrac {10}{140}$ Since it has the same denominators we can add directly

$\dfrac {7 + 10}{140} \Rightarrow \dfrac {17}{140}$ .

What number should be subtracted from $\dfrac {-2}{3}$ to get $\dfrac {-1}{6}$ ?

Let us consider

$x$

$\dfrac {-2}{3} - x = \dfrac {-1}{6}$

Solving for

$x$ ,

$x = \dfrac {-2}{3} - \dfrac {-1}{6}$

By taking LCM of

$3$ and

$6$ is

$6$

$\dfrac {(-2\times 2) - (-1\times 1)}{6} = \dfrac {-4 - (-1)}{6} = \dfrac {-4 + 1}{6} = \dfrac {-3}{6}$ we further can divide to lowest terms,

$\dfrac {-3\div 3}{6\div 3} = \dfrac {-1}{2}$

$x = \dfrac {-1}{2} \therefore \dfrac {-1}{2}$ should be subtracted from

$\dfrac {-2}{3}$ to get

$\dfrac {-1}{6}$ .

Find the each of the following ratios in the simplest form:
$2.4\ km$ to $900\ m$

To convert given ratio

$a:b$ to his simplest form, we divide each term by the

$HCF$ of

$a$ and

$b$

Now,

$2.4\ km$ to

$900\ m=2400\ m:900\ m$

$=2400:900$

$=\dfrac{2400}{900}$

$=\dfrac{2400\div 300}{900\div 300}\hspace{1cm}[\because HCF$ of

$2400$ and

$900$ is

$300]$

$=\dfrac83$

$\therefore$ The simplest form of

$2400:900$ is

$8:3$

Find the following ratio in the simplest form:
$4\ kg$ to $750\ g$

To convert given ratio

$a:b$ to his simplest form, we divide each term by the

$HCF$

$a$ and

$b$

We have,

$4\ kg$ to

$750\ g=4000\ g$ to

$750\ g$ (

$1 \ kg = 1000\ g$ )

$=4000:750$

$=\dfrac{4000}{750}$

$=\dfrac{(4000\div 250)}{(750\div 250)}\hspace{1cm}[\because HCF$ of

$4000$ and

$750$ is

$250]$

$=\dfrac{16}3$

$\therefore$ The simplest form of

$4000:750$ is

$16:3$

Find the following ratio in the simplest form:
$1.8\ kg$ to $6\ kg$

To convert given ratio

$a:b$ to his simplest form, we divide each term by the

$HCF$ of

$a$ and

$b$

We have,

$1.8\ kg$ to

$6\ kg=1.8:6$

$=\dfrac{1.8}6$

$=\dfrac{1.8\times10}{6\times10}\hspace{1cm}[$ Multiplying numerator and denominator by

$10]$

$=\dfrac{18}{60}$

$=\dfrac{18\div 6}{60\div 10}\hspace{1cm}[\because HCF$ of

$18$ and

$60$ is

$6]$

$=\dfrac3{10}$

$\therefore$ The simplest form of

$1.8:6$ is

$3:10$

Express the following ratio in simplest form:
$85:561$

To convert given ratio

$a:b$ to his simplest form, we divide each term by the

$HCF$ of

$a$ and

$b$ .

$\therefore \ 85:561$

$=\dfrac{85}{561}$

$=\dfrac{85\div 17}{561\div 17}\hspace{1cm}[\because HCF$ of

$85=17\times5$ and

$561=17\times 3\times 11$ is

$17]$

$=\dfrac5{33}$

Hence, the simplest form of

$85:561$ is

$5:33$ .

Express the following ratio in simplest form:
$36:90$

To convert given ratio

$a:b$ to his simplest form, we divide each term by the

$HCF$ of

$a$ and

$b$

Now,

$36:90$

$=\dfrac{36}{90}$

$=\dfrac{36\div 18}{90\div18}\hspace{1cm}[\because HCF$ of

$36$ and

$90$ is

$18]$

$=\dfrac25$

$=2:5$

Hence, the simplest form of

$36:90$ is

$2:5$

Express the following ratio in simplest form:
$480:384$

To convert given ratio

$a:b$ to its simplest form, we divide each term by the

$HCF$ of

$a$ and

$b$ .

We have

$480:384$

$=\dfrac{480}{384}$

$=\dfrac{480\div 96}{384\div 96}\hspace{1cm}[\because \ HCF$ of

$480$ and

$384$ is

$96]$

$=\dfrac54$

Hence, the simplest form of

$480:384$ is

$5:4$ .

Express each of the following ratio in the simplest form:
$777:1147$

To convert given ratio

$a:b$ to his simplest form, we divide each term by the

$HCF$ of

$a$ and

$b$

$\therefore \ 777:1147$

$=\dfrac{777}{1147}$

$=\dfrac{777\div 37}{1147\div 37}\hspace{1cm} [\because HCF$ of

$777$ and

$1147$ is

$37]$

$=\dfrac{21}{31}$

Hence, the simplest form of

$777:1147$ is

$21:31$

Express the following ratio in the simplest form:
$186:403$

To convert given ratio

$a:b$ to his simplest form, we divide each term by the

$HCF$ of

$a$ and

$b$

$\therefore \ 186:403$

$=\dfrac{186}{403}$

$=\dfrac{186\div 31}{403\div 31}\hspace{1cm}[\because HCF$ of

$186$ and

$403$ is

$31]$

$=\dfrac6{13}$

Hence, the simplest form of

$186:403$ is

$6:13$

Express the following ratio in simplest form:
$324:144$

To convert given ratio

$a:b$ to his simplest form, we divide each term by the

$HCF$ of

$a$ and

$b$

$\therefore \ 324:144$

$=\dfrac{324}{144}$

$=\dfrac{324\div 36}{144\div 36}\hspace{1cm}[\because HCF$ of

$324$ and

$144$ is

$36]$

$=\dfrac94$

Hence, the simplest form of

$324:144$ is

$9:4$

Find the following ratio in the simplest form:
$48$ Minutes to $1$ Hour

To convert given ratio

$a:b$ to his simplest form, we divide each term by the

$HCF$ of

$a$ and

$b$

$\therefore 48$ Minutes to

$1$ hour

$=48$ min

$:60$ min

$=48:60$

$=\dfrac{48}{60}$

$=\dfrac{48\div 12}{60\div 12}\hspace{1cm}$

$[\because HCF$ of

$48$ and

$60$ is

$12]$

$=\dfrac45$

$\therefore$ The simplest form of

$48:60$ is

$4:5$

Write the following ratio in the simplest form:
$3$ weeks: $30$ days

To convert given ratio

$a:b$ to his simplest form, we divide each term by the

$HCF$ of

$a$ and

$b$

$\therefore 3$ weeks :

$30$ days

$\\=21$ days :

$30$ days

$=21:30$

$=\dfrac{21}{30}$

$=\dfrac{21\div 3}{30\div 3}$

$\hspace{1cm} [\because HCF$ of

$21$ and

$30$ is

$3]$

$=\dfrac7{10}$

$\therefore$ Simplest form of

$21:30$ is

$7:10$

Write the following ratio in the simplest form:
$48\ min:2\ hrs\ 40\ min$

To convert given ratio

$a:b$ to his simplest form, we divide each term by the

$HCF$ of

$a$ and

$b$

$\therefore48$ min :

$2$ hrs

$40$ min

$=48$ min :

$160$ min

$=\dfrac{48}{160}$

$=\dfrac{48\div 16}{160\div 16}$

$[\because HCF$ of

$48$ and

$160$ is

$16]$

$=\dfrac3{10}$

$\therefore$ Simplest form of

$48$ min

$:2$ hrs

$40$ min is

$3:10$

Write the following ratio in the simplest form:
$1\ L\ 35\ mL: 270\ mL$

To convert given ratio

$a:b$ to his simplest form, we divide each term by the

$HCF$ of

$a$ and

$b$

$\therefore1\ L\ 35\ mL: 270\ mL=1035\ mL: 270\ mL$

$=1035:270$

$=\dfrac{1035}{270}$

$=\dfrac{1035\div 45}{270\div 45}$

$[\because HCF$ of

$1035$ and

$270$ is

$45]$

$=\dfrac{23}6$

$\therefore$ Simplest form of

$1035:270$ is

$23:6$

Find the difference:
$\dfrac58-\dfrac7{12}$

From the above prime factorization of

$8$ and

$12,$

$LCM(8,12)=2\times 2\times 2\times 3$

$=24$

Now,

$\dfrac{5}{8} - \dfrac{7}{{12}} = \dfrac{{5 \times 3}}{{8 \times 3}} - \dfrac{{7 \times 2}}{{12 \times 2}}$

$= \dfrac{{15}}{{24}} - \dfrac{{14}}{{24}}$

$= \dfrac{{15 - 14}}{{24}}$

$= \dfrac{{1}}{{24}}$

Hence,

$\dfrac58-\dfrac7{12}=\dfrac1{24}$

1630929_1777300_ans_4087f762d92c417dac206dc169649364.JPG

Write the following ratio in the simplest form:
$4\ kg : 2\ kg\ 500\ g$

To convert given ratio

$a:b$ to his simplest form, we divide each term by the

$HCF$ of

$a$ and

$b$

$\therefore 4\ kg : 2\ kg\ 500\ g$

$=4000\ g : 2500\ g$

$=\dfrac{4000}{2500}$

$=\dfrac{4000\div 500}{2500 \div 500}$

$[\because HCF$ of

$4000$ and

$2500$ is

$500]$

$=\dfrac85$

$\therefore$ Simplest form of

$4\ kg:2\ kg \ 500\ g$ is

$8:5$

Write the following ratio in the simplest form:
Rupees $6.30$ : Rupees $16.80$

To convert given ratio

$a:b$ to his simplest form, we divide each term by the

$HCF$ of

$a$ and

$b$

$\therefore$ Rupees

$6.30$ : Rupees

$16.80$

$=\dfrac{6.30}{16.80}$

$=\dfrac{63}{168}\hspace{1cm} [$ Multiplying numerator and denominator by

$10]$

$=\dfrac{63\div 21}{168\div21}\hspace{1cm}[\because HCF$ of

$63$ and

$168$ is

$21]$

$=\dfrac38$

$\therefore$ Simplest form of Rupees

$6.30$ : Rupees

$16.80$ is

$3:8$

Write the following ratio in the simplest form:
$3\ m\ 5\ cm : 35\ cm$

To convert given ratio

$a:b$ to his simplest form, we divide each term by the

$HCF$ of

$a$ and

$b$

$\therefore 3\ m\ 5\ cm:35\ cm$

$=305\ cm:35\ cm$

$=305:35$

$=\dfrac{305}{35}$

$=\dfrac{305\div 5}{35\div 5}$

$[\because HCF$ of

$305$ and

$35$ is

$5]$

$=\dfrac{61}7$

$\therefore$ Simplest form of

$305:30$ is

$61:7$

Find the sum
$3\dfrac{4}{5}+2\dfrac{3}{10}+1\dfrac{1}{15}$

First, convert each mixed fraction into an improper fraction.
We get,

$3\dfrac{4}{5}=\dfrac{18}{5}$

$2\dfrac{3}{10}=\dfrac{23}{10}$

$1\dfrac{1}{15}=\dfrac{16}{15}$

Then, to find out

$\dfrac{19}{5}+\dfrac{23}{10}+\dfrac{16}{15}$ , we need to calculate the LCM of the denominators.

LCM of

$5,10\, \, and \, \, 15=30$

Now,let us change each of the given fraction into an equivalent fraction having 30 as the denominator

$\dfrac{19}{5}\times\dfrac{6}{6}=\dfrac{114}{30}$

$\dfrac{23}{10}\times\dfrac{3}{3}=\dfrac{69}{30}$

$\dfrac{16}{15}\times\dfrac{2}{2}=\dfrac{32}{30}$

Now,

$\dfrac{114}{30}+\dfrac{69}{30}+\dfrac{32}{30}=$

$\dfrac{(114+69+32)}{30}$

$=\dfrac{215}{30}$

$=\dfrac{43}{6}$

$=7\dfrac{1}{6}$

Find the difference
$3\dfrac{1}{5}-\dfrac{7}{10}$

Convert mixed fraction into improper fraction,

$=3\dfrac{1}{5}=\dfrac{16}{5}$

$=\dfrac{16}{5}-\dfrac{7}{10}$
For subtraction of two unlike fractions,first change them to the like fractions.
LCM of

$5,10 =10$
Now, let us change each of the given fraction into an equivalent fraction having

$10$ as the denominator.

$=\dfrac{16}{5}\times \dfrac{2}{2}=\dfrac{32}{10}$

$=\dfrac{7}{10}\times \dfrac{1}{1}=\dfrac{7}{10}$
Then,

$=\dfrac{32}{10}-\dfrac{7}{10}$

$\dfrac{32-7}{10}$

$[divide\ by\ 5]$

$=\dfrac{5}{2}$

$=2\dfrac{1}{2}$

Divide the sum of $(65/12)$ and $(8/3)$ by their difference.

First we have to find the sum

$(65/12)+(8/3)$

$LCM$ of

$12$ and

$3$ is

$12$
Then,

$(65\times 1)/(12\times 1)=(65/12)$

and

$(8\times 4)/(3\times 4)=(32/12)$

Now,

$=(65+32)/12$

$=(97/12)$

Now find the difference of

$(65/12)-(8/3)$

$LCM$ of

$12$ and

$3$ is

$12$
Then,

$(65\times 1)/(12\times 1)=(65/12)$

$(8\times 4)/(3\times 4)=(32/12)$
Now,

$=(65-32)/12$

$=(33/12)$

Now divide

$(97/12)\div (33/12)$

$=(97\times 12)\times (12/33)$

$=(97\times 12)/(12\times 33)$

$=(97\times 1)/(1\times 33)$

$=(97/33)$

Find the sum
$\cfrac{5}{9} + \cfrac{3}{9}$

For adding two like fractions, the numerators are added and the denominator remains the same.

$\dfrac{5}{9}+\dfrac{3}{9}$

$=\dfrac{5+3}{9}$

$=\dfrac{8}{9}$

Find the difference
[2(5/9)]-[1(7/15)]

First we convert mixed fraction into improper fraction, and then we find the difference.

$[2(5/9)]=(23/9)$
and

$[1(7/15)]=(22/15)$
We get,

$=(23/9)-(22/15)$
LCM of

$9,15=45$
Now.let us change each of the given fraction into an equivalent fraction having

$45$ as the denominator.

$[(23/9)x(5/5)]=(115/45)$

and

$[(22/15)x(3/3)]=(66/45)$

Then,

$=(115/45)-(66/45)$

$=(115-66)/45$

$=(49/45)$

$=[1(4/45)]$

Find the sum
(8/9)+(7/12)

For addition of two unlike fractions,first change them to the like fractions.
LCM of 9,12=36
Now,let us change each of the given fraction into an equivalent fraction having 36 as the denominator.

$\dfrac{8}{9}\times \dfrac{4}{4}=\dfrac{32}{36}$

$\dfrac{7}{12}\times \dfrac{3}{3}=\dfrac{21}{36}$

Now.add the like fractions,

$=\dfrac{32}{36}$

$+\dfrac{21}{36}$

$=\dfrac{32+21}{36}$

$=\dfrac{53}{36}$

$=[1\dfrac{17}{36}]$

Find the difference
$\left( {\dfrac{5}{7}} \right) - \left( {\dfrac{2}{7}} \right)$

The subtraction of fractions can be performed in a manner similar to that of addition.

For subtracting two like fractions, the numerators are subtracted and the denominator remains the same.

$\begin{aligned}{}\left( {\frac{5}{7}} \right) - \left( {\frac{2}{7}} \right)& = \frac{{5 - 2}}{7}\\ &= \frac{3}{7}\end{aligned}$

Find the difference
(5/6)-(3/4)

For subtraction of two unlike fractions,first change them to the like fractions.

LCM of 6,4 =12

Now, let us change each of the given fraction into an equivalent fraction having 12 as the denominator.

=[(5/6)x(2/2)]=(10/12)

=[(3/4)x(3/3)]=(9/12)

Now,

=(10/12)-(9/12)

=[(10-9)/12]

=(1/12)

Find the difference
$7-\left[4\left(\cfrac{2}{3}\right)\right]$

$7-\left[4\left(\cfrac{2}{3}\right)\right]$

We have to covert mixed fraction in improper fraction(a fraction in which numerator is greater than denominator).

$\therefore 4\left(\cfrac{2}{3}\right)=\cfrac{4\times 3+2}{3}=\cfrac{14}{3}$

$\Rightarrow 7-\cfrac{14}{3}$

$7$ can also be written as

$\cfrac{7}{1}$

$\Rightarrow \cfrac71-\cfrac{14}{3}$

To subtract rational numbers their denominator should be same, so we can simply subtract their numerator and denominator will remain common.

$LCM$ of

$1$ and

$3$ is

$3$

So, let's make numbers with same denominator equal to their

$LCM$

$\cfrac{7}1=\cfrac{(7\times (3))}{(1\times (3))}=\cfrac{21}{3}$

So,

$\cfrac71-\cfrac{14}{3}=\cfrac{21}{3}-\cfrac{14}{3}$

$=\cfrac{21-14}{3}$

$=\cfrac{7}{3}$

$=\cfrac{6+1}{3}$

$=2+\cfrac13$

$=2\left(\cfrac13\right)$

Find the difference
$3\dfrac{3}{10}-1\dfrac{7}{15}$

Convert mixed fraction into improper fraction,and then find the difference.

$=3\dfrac{3}{10}=\dfrac{33}{10}$

$=1\dfrac{7}{15}=\dfrac{22}{15}$
we get,

$=(33/10)-(22/15)$
LCM of

$10,15=30$
Now,let us change each of the given fraction into an equivalent fraction having 10 as the denominator.

$=\dfrac{33}{10 } \times \dfrac{3}{3}=\dfrac{99}{30}$

$=\dfrac{22}{15}\times \dfrac{2}{2}=\dfrac{44}{30}$
Then,

$\dfrac{99}{30}-\dfrac{44}{30}$

$=\dfrac{99-44}{30}$

$=\dfrac{55}{30}$

$=\dfrac{11}{6}$

$=1\dfrac{5}{6}$

Simplify: $8-[4(1/2)]-[2(1/4)]$

Convert mixed fraction into improper fraction, and then find the difference.

$=[4(1/2)]=(9/2)$

$=[2(1/4)]=(9/4)$

LCM of

$1,2,4=4$
Now, let us change each of the given fraction into an equivalent fraction having 4 as the denominator.

$=(9/2)\times (2/2)=(18/4)$
and

$=(9/4)\times (1/1)=(9/4)$
and

$=(8/1)\times (4/4)=(32/4)$

Then,

$=(32/4)-[(18/4)-(9/4)]$

$=[(32-18-9)/4]$

$=[(32-27)/4]$

$=(5/4)$

$=[1(1/4)]$

What number must be added to each term of the ratio $9:16$ to make the ratio $2:3$

Let the number be

$x$
Then the two term becomes

$=(9+x):(16+x)=2:3$

$=(9+x)/(16+x)=(2/3)$
By cross multiplication

$=3\times (9+x)=2\times (16+x)$

$=27+3x=32+2x$
Transporting

$27$ to RHS and

$2x$ to LHS

$=3x-2x=32-27=x=5$
Hence the number be added to each of the ratio

$9:16$ to make the ratio

$2:3$ is

$5$

Simplify: (2/3)+(5/6)-(1/9)

LCM of

$3,6,9=18$
Now let us change each of the given fraction into an equivalent fraction having

$18$ as the denominator.

$(2/3)x(6/6)=(12/18)$
and

$(5/6)x(3/3)=(15/18)$
and

$(1/9)x(2/2)=(2/18)$

Now,

$=(12/18)+(15/18)-(2/18)$

$=(12+15-2)/18$

$=(27-2)/18$

$=(25/18)$

$=[1(7/18)]$

Aneeta bought $3\dfrac{3}{4}$ kg apples and $4\dfrac{1}{2}$ kg guava. What is the total weight of fruits purchased by her?

The total weight of fruits bought by Aneeta =

$\left( 3\dfrac{3}{4}+4\dfrac{1}{2}\right)$

First convert each mixed fraction into improper fraction

$\therefore3\dfrac{3}{4}=\dfrac{(3\times4)+3}{4}=\dfrac{15}{4}$

and

$4\dfrac{1}{2}=\dfrac{(4\times2)+1}{2}=\dfrac{9}{2}$

$\therefore\text{Total weight of fruits}=\dfrac{15}{4}+\dfrac{9}{2}$

Now, LCM of 4 & 2 =4

$\therefore$ We change each of the given fraction into an equivalent fraction having 4 as the denominator

$\dfrac{15}{4}\times\dfrac{1}{1}=\dfrac{15}{4}$
and

$\dfrac{9}{2}\times\dfrac{2}{2}=\dfrac{18}{4}$

$\therefore\text{Total weight of fruits}=\dfrac{15}{4}+\dfrac{9}{2}$

$=\dfrac{15}{4}+\dfrac{18}{4}$

$=\dfrac{15+18}{4}$

$=\dfrac{33}{4}$

$=8\dfrac{1}{4}$

$\therefore\text{The total weight of fruits purchased by Aneeta}= 8\dfrac{1}{4}\text{ kg}$

What number must be subtracted from each term of the ratio $17:33$ so that the ratio becomes $7:15$ ?

Let the number be

$x$
Then the two term becomes

$=(17-x):(33-x)=7:15$

$=(17-x)/(33-x)=(7/15)$
By cross multiplication,

$=15\times (17-x)=7\times (33-x)$

$=255-15x=231-7x$
Transporting

$-15x$ to RHS and

$231$ to LHS

$=255-231=15x-7x$

$8x=24$

$x=3$
Hence, the number be subtracted from each term of the ratio

$17:33$ to make the ratio

$7:15$ is

$3$

A number representing a part of a _____ is called a fraction.

Example:

$\dfrac{1}{4}, \dfrac{3}{4}, \dfrac{3}{6}, etc.$

$\therefore \textbf{A number representing a part of a whole is called a fraction.}$

$13 \dfrac{5}{18}$ Is a _____ fraction.

A fraction is formed by combining a whole number and a fraction is called mixed fraction.

$\textbf{Hence, } 13 \dfrac{15}{18}$

$\textbf{is a mixed fraction}$ .

$\dfrac{7}{19}$ Is a _____ fraction.

A fraction whose numerator is less than the denominator is called a proper fraction.

$\therefore \dfrac{7}{19} \textbf{is a proper fraction.}$

$\dfrac{18}{8}$ Is a _____ fraction.

$\dfrac{18}{8}$ Is a

$improper$ fraction.

A fraction with denominator greater than the numerator is called a _____ fraction.

Example:

$\dfrac{2}{5} , \dfrac{3}{8} , etc.$

$\therefore \textbf{A fraction with a denominator greater than the numerator is called Proper fraction.}$

$\dfrac{67}{14} - \dfrac{24}{14}$ =_____.

$\dfrac{67}{14} - \dfrac{14}{14} = \dfrac {43}{14}$
Exaplanation ,

$\Rightarrow \dfrac { 67}{14} - \dfrac {24}{14} = \dfrac { 67 -24}{14} = \dfrac {43}{14}$

$\dfrac{17}{9} + \dfrac{41}{9}$ =_____.

$\dfrac{17}{9} + \dfrac{41}{9} = \dfrac {58}{9}$
Explanation

$\Rightarrow \dfrac {17}{8} + \dfrac {41}{9} = \dfrac { 17 + 41}{8} = \dfrac { 58}{9}$

$\dfrac{11}{16}.....\dfrac{14}{15}$

Make it as like a fraction , multiply in numerator and denominator with the same digit.

$\dfrac { 11}{16} \times \dfrac {15}{15} = \dfrac { 165}{240}$

$\dfrac {14}{15} \times \dfrac {16}{16} = \dfrac {224}{240}$

$\dfrac { 224}{240} > \dfrac {165}{240}$ So

$\dfrac {11}{16} < \dfrac {14}{15}$

The fraction $\frac{6}{15}$ in simplest form is _____.

Simplifying the fraction into its simplest form by dividing numerator and denominator by the HCF of numerator and denominator equal 3, we get

$\dfrac{6}{15} = \dfrac{6÷3}{15÷3}$

$= \dfrac{2}{5}$

$\therefore \textbf{The fraction 6/15 in simplest form is 2/5.}$

The fraction $\dfrac{17}{34}$ in simplest form is _____.

Simplifying the fraction into its simplest form by dividing numerator and denominator by the HCF of numerator and denominator equal 17, we get

$\dfrac{17}{34} = \dfrac{17÷17}{34÷17}$

$= \dfrac{1}{2}$

$\therefore \textbf{The fraction 17/34 in simplest form is 1/2}$

$\dfrac{12}{75}.....\dfrac{32}{200}$

Make it as like a fraction , multiply in numerator and denominator with the same digit.

$\dfrac {12}{75} \times \dfrac {8}{8} = \dfrac {96}{600}$

$\dfrac {32}{200} \times \dfrac {3}{3} = \dfrac {96}{600}$

$\dfrac {96}{600} = \dfrac {96}{600}$ so

$\dfrac {12}{75} = \dfrac {32}{200}$

Express $6 \dfrac{2}{3}$ as an improper fraction.

we have,

$6 \dfrac {2}{3} = \dfrac { ( 6 \times 3) +2 }{3} = \dfrac {18 +2}{3} = \dfrac {20}{3}$
This is improper fraction.

Katrina rode her bicycle $6\dfrac{1}{2} km$ in the morning and $8\dfrac{3}{4} km$ in the evening. Find the distance traveled by her altogether on that day.

Given

Distance traveled by Katrina in the morning

$= 6\dfrac 12 = \dfrac {( 6 \times 2) +1 }{2} = \dfrac { 12+1}{2}=\dfrac {13}{2}\ km$

Distance traveled by Katrina in the evening

$= 8 \dfrac {3}{4} = \dfrac {( 8 \times 4) +3 }{4} = \dfrac { 32 +3}{4} = \dfrac {35}{4} \ km$

$\therefore$ Total distance traveled by her

$=\dfrac {13}{2} + \dfrac {35}{4}\\ = \dfrac { 13}{2}\times \dfrac 22+ \dfrac {35}{4}\times\dfrac 11 \hspace{1.5cm}[\because \text {LCM of }2 \text { and } 4=4]\\ =\dfrac {26}4+\dfrac {35}4\\ =\dfrac{26+35}4\\=\dfrac{61}4=15\dfrac 14 \ km$

Add $1 \dfrac{1}{4}$ and $6 \dfrac{1}{2}$ .

We have:.

$1\dfrac {1}{4} = \dfrac {( 1\times 4) +1 }{ 4}=\dfrac {5}{4}$

And

$6 \dfrac {1}{2} =\dfrac {( 6 \times 2) + 1}{2} =\dfrac {13}{2}$

$\therefore\ 1 \dfrac{1}{4} + 6 \dfrac{1}{2}= \dfrac {5}{4}+\dfrac {13}{2}\\ =\dfrac {5}{4}\times\dfrac11+\dfrac {13}{2}\times\dfrac22\\=\dfrac {5}{4}+\dfrac {26}4\\=\dfrac {31}4=7\dfrac 34$

Add the fractions $\dfrac{3}{8}$ and $\dfrac{2}{3}$ .

$LCM$ of

$8$ and

$3=24$

$\therefore \dfrac {3}{8} + \dfrac {2}{3}\\ = \dfrac38\times \dfrac33 +\dfrac23\times\dfrac23\\=\dfrac9{24}+\dfrac{16}{24}\\= \dfrac {9+16}{24} \\= \dfrac {25}{24}$

(i) What number should be added to $\dfrac{5}{12}$ to get $2 \dfrac{2}{8}$ ?
(ii) What number should be subtracted from $5$ to get $1 \dfrac{5}{13}$ ?

(i) Let the number to be added

$= x$

$\therefore \dfrac{5}{12} + x = 2 \dfrac{3}{8}$

$\therefore \, Required \,number\, \left(x\right) = 2 \dfrac{3}{8} - \dfrac{5}{12}$

$x = \dfrac{19}{8} - \dfrac{5}{12}$

$= \dfrac{57 - 10 }{24} = \dfrac{47}{24} = 1\dfrac{23}{24}$
(ii) Let the number to be subtracted

$= x$

$\therefore 5 - x = 1\dfrac{1}{13}$
Required number

$\left(x\right) = 5 - 1\dfrac{5}{13}$

$x = \dfrac{5}{1} - \dfrac{18}{13}$

$= \dfrac{65 - 18}{13} = \dfrac{47}{13} = 3 \dfrac{8}{13}$

$\dfrac{18}{15}.....1.3$

Since convert

$1.3$ into fraction from , we get

$\dfrac {13}{10}$

$\dfrac {18}{15}...... \dfrac {13}{10}$
Make it as like a fraction , multiply in numerator and denominator with the same digit.

$\dfrac {18}{15} \times \dfrac {10}{10} = \dfrac {180}{150}$ and

$\dfrac {13}{10} \times \dfrac {15}{15} = \dfrac {195}{150}$

$\dfrac { 180}{150} < \dfrac {195}{150}$ So,

$\dfrac {18}{15} < 1.3$

Subtract $1 \dfrac{1}{4}$ from $6 \dfrac{1}{2}$ .

We have:

$1 \dfrac {1}{4} =\dfrac{(1\times 4)+1}4=\dfrac 54$

And

$6 \dfrac {1}{2}=\dfrac{(6\times 2)+1}2=\dfrac {13}2$

$LCM$ of

$2$ and

$4=4$

$\therefore 6\dfrac 12-1\dfrac 14=\dfrac {13}2-\dfrac54\\=\dfrac {13}2\times\dfrac 22-\dfrac54\times\dfrac 11\\=\dfrac {26}4-\dfrac 54\\=\dfrac {26-5}4\\=\dfrac {21}4=5\dfrac 14$

$\dfrac{8}{15}.....\dfrac{95}{14}$

Make it as like a fraction , multiply in numerator and denominator with the same digit.

$\dfrac {8}{15} \times \dfrac {14}{14} = \dfrac {112}{210}$

$$ \dfrac {95}{14} \times \dfrac {15}{15} = \dfrac {1425}{210}

$\dfrac {1425}{210} > \dfrac {112}{210}$ So,

$\dfrac {8}{15} < \dfrac {95}{14}$

Subtract $\dfrac{1}{6}$ from $\dfrac{1}{2}.$

$LCM$ of

$2$ and

$6=6$

$\therefore\dfrac {1}{2} - \dfrac {1}{6}\\ = \dfrac12\times\dfrac33-\dfrac16\times\dfrac11\\=\dfrac36-\dfrac16\\=\dfrac{3-1}6\\=\dfrac 26=\dfrac13$

On an average $\frac{ 1}{10}$ the food eaten is turned into organism's own body and is available for the next level of the consumer in a food chain. What fraction of the food eaten is not available for the next level?

let the total food for the eaten is

$1$ and out of these

$\dfrac {1}{10}$ of the food eaten is turned into organism own body.
Now available for the next level of the consumer in a food chain is

$\Rightarrow 1 - \dfrac {1}{10} = \dfrac { 10-1}{10} = \dfrac {9}{10}$

Sunil purchased $12 \dfrac {1}{2}$ litres of juice on Monday and $14 \dfrac {3}{4}$ litres of juice on Tuesday . How many litres of juice did he purchase together in two days?

Juice purchased by Sunil on Monday

$= 12 \dfrac {1}{2}$ litres

$= \dfrac {25}{2}$ litrres

Juice purchased by Sunil on Tuesday

$= 14 \dfrac {3}{4}$ litres

$= \dfrac {59}{4}$ litres

$\therefore$ Total juice purchased

$= \dfrac {25}{2} + \dfrac {59}{4}$

$= \dfrac { ( 25 \times 2) +( 59 \times 1) }{ 4 } \\ = \dfrac { 50 +59}{4} \\$

$= \dfrac {109}{4}$ litres

Nasir travelled $3 \frac {1}{2}$ km in a bus and then walked $1 \frac {1}{8} km$ to reach a town . How much did he travel to reach the town ?

Nasir travel $3 \dfrac {1}{2} km$ in bus = $\dfrac {7}{2} km$

Walked $1 \dfrac {1}{8} km$ to reach a town = $\dfrac {9}{8} km$

$\therefore$ Total distance covered by nasir

$\Rightarrow ( \dfrac {7}{2} + \dfrac {9}{8}) km = \dfrac {( 7 \times 8) +( 9 \times 2) }{2 \times 8} km = \dfrac { 56 + 18 }{16} km$

$\Rightarrow \dfrac {74}{16} km = \dfrac { 74 \div 2 }{ 16 \div 2}km =\dfrac {37}{8} km = 4 \dfrac {5}{8}km$

The fish caught by neetu was of weight $3 \frac {3}{4} Kg$ and the fish caught by Narendra was of weight $2 \frac {1}{2} kg$
How much more did Neetu's fish weigh than that of Narendra ?

the weight of fish caught by neetu $3 \dfrac {3}{4} kh = \dfrac {15}{4} kg$

The weight of fish caught by narendra $2 \dfrac {1}{2} kg = \dfrac {5}{2} kg$

$\therefore$ Weight of Neetu more than narendra

$\Rightarrow ( \dfrac { 15}{4} - \dfrac {5}{2}) kg = \dfrac { ( 15 \times 2) -( 5 \times 4 ) }{ 4 \times 2} kg = \dfrac { 30 -20 }{8} kg$

$\Rightarrow \dfrac {10}{8} kg = \dfrac {10 \div 2 }{ 8 \div 2} kg = \dfrac {5}{4} kg$

Write a pair of fractions whose sum is $\frac {7}{11}$ and difference is $\frac {2}{11}$ .

Let the pair fraction $F_1$ and $F_2$

$\therefore$ According to the question

$F_1 + F_2 = \dfrac {7}{11}......(i)$

$F_1 - F_2 = \dfrac {2}{11}......(ii)$

Adding the equation (i) and (ii) we get that

$2F_1 = \dfrac {7}{11} + \dfrac {2}{11} = \dfrac { 7 + 2 }{11} = \dfrac {9}{11}$

$\Rightarrow F_1 = \dfrac {9}{11} \times \dfrac {1}{2} = \dfrac {9}{22}$

Put the value of $F_1$ in equation (i) we get that

$\Rightarrow \dfrac {9}{11} +F_2 = \dfrac {7}{11}$

$\Rightarrow F_2 = \dfrac {7}{11} - \dfrac {9}{22} = \dfrac {7}{11} \times \dfrac {2}{2} - \dfrac {9}{22} = \dfrac {14}{22} - \dfrac {9}{22} = \dfrac { 14 -9 }{22}$

$\Rightarrow F_2 = \dfrac {5}{22}$

A bamboo of length $2 \dfrac{3}{4}$ metre broke into two pieces. One piece was $\dfrac{7}{8}$ metre long. How long is the other pieces ?

Let of original pieces of bamboo

$= 2\dfrac{3}{4} = \dfrac{11}{4}$ metre
Length of one piece of bamboo

$= \dfrac{7}{8}$ metre
Length of other pieces

$= \dfrac{7}{8} \,metre - \dfrac{7}{8}\, metre$

$= \left(\dfrac{11}{4} - \dfrac{7}{8}\right)$ metre

$= \left(\dfrac{11 \times 2 - 7 \times 1}{8}\right)$ m

$\left[LCM = \left(4,8\right) = 8 \right]$

$= \dfrac{22 - 7}{8} = \dfrac{15}{8}$ metre or

$1\dfrac{7}{8}$ metre

When Sunita weighted herself on monday. She found that she had gained $1 \dfrac {1}{4}$ kg. earlier her weight was $46 \dfrac {3}{8} kg$ What was her weight on monday ?

Anita had gained

$1 \dfrac {1}{4} kg = \dfrac {5}{4} kg$

Earlier her weight was

$46 \dfrac {3}{8} kg = \dfrac { 371}{7} kg$

$\therefore$ Total wight gain by anita =

$\dfrac {5}{4} kg + \dfrac { 371}{8} kg$

$\Rightarrow \dfrac {( 5\times 8) + ( 371 \times 4 ) }{ 4 \times 8} kg = \dfrac { 40 + 1484}{32} kg$

$\Rightarrow \dfrac {1524}{32} kg = \dfrac {1524 \div 4 }{ 32 \div 4} kg = \dfrac {381}{8} kg$

What is wrong in the following additions ?

(a) Equal denominators added

(b) Numerator and denominator are added.

Nazima gave $2 \frac {3}{4}$ litres out of $5 \frac {1}{2} litres$ of juice she purchased to her friends.how many litres of juice is left with her ?

Nazima have $5 \dfrac {1}{2} litres = \dfrac {11}{2} litres$

he gave $2 \dfrac {3}{4} litres$ to her friend $= \dfrac {11}{4} litres$

$\therefore$ juice is left

$\Rightarrow \dfrac {11}{2} lit - \dfrac {11}{4} lit = \dfrac {( 11 \times 4 ) -( 11 \times 2) }{ 2 \times 4} lit = \dfrac { 44 -22}{8} litres$

$\Rightarrow \dfrac {22}{8} litres = \dfrac { 22 \div 2}{8 \div 2}litres = \dfrac {11}{4} litres$

Neelam's father needs $1 \dfrac {3}{4}m$ of cloth for the skirt of neelam's new dress and $\dfrac {1}{2} m$ for the scarf. How much cloth must he but in all?

Neelam's father needs $1 \dfrac {3}{4} m ( \dfrac {7}{4} m )$ of cloth for the skirt of

Neelam's new dress and $\dfrac {1}{2}$ m for the scarf

Total cloths buy by her father

$\Rightarrow ( \dfrac {7}{4} + \dfrac {1}{1})m = \dfrac {( 7 \times 20 + ( 1 \times 4) }{ 4 \times 2}m = \dfrac {14+4}{8} m$

$\Rightarrow \dfrac {18}{8}m = \dfrac { 18 \div 2}{ 8 \div 2}m = \dfrac {9}{4} m$ or $2 \dfrac {1}{4} m$

Asha and Samuel have bookshelves of the same size partly filled with books. Asha's shelf is $\dfrac{5}{6}$ full and Samuel's shelf is $\dfrac{3}{5}$ full. Whose bookshelf is more full and by what fraction?

Asha's shelf is

$\dfrac{5}{6}$ th fukk of book Samuel's shelf is

$\dfrac{3}{5}$ th full of book.

$\left[LCM \left(6,5\right) = 30\right]$
Like fractions of

$\dfrac{5}{6} \, and\,\dfrac{3}{5}$ is :

$\Rightarrow \dfrac{5}{6} \times \dfrac{5}{5} = \dfrac{25}{30}$

$\Rightarrow \dfrac{3}{5} \times \dfrac{6}{6} = \dfrac{18}{30}$

$\therefore \dfrac{25}{30} > \dfrac{18}{30}$

$\therefore$ Asha's bookshelf is more full by fraction

$\dfrac{25}{30} - \dfrac{18}{30}$ .
i.e. by fraction

$\dfrac{25 - 18}{30} = \dfrac{7}{30}$

Will be quotient
$7\frac{1}{6} +3\frac{2}{3}$ be a fraction greater
than 1.5 or less than 1.5? Explain.

1903310_1790209_ans_3363b9684c1442ecae9c316ccffa2f3c.jpg

Reemu read $\dfrac{1}{5}th$ pages of a book. If she reads further $40$ pages, she would have read $\dfrac{7}{10}$ th pages of the book. How many pages are left to be read?

First we will assume the pages in the book be $x.$

$\dfrac23$ of $x + 40 = \dfrac7{10}$ of $x$

$\Rightarrow \dfrac{2x}3 + 40 = \dfrac{7x}{10}$

$\Rightarrow \dfrac{7x}{10} - \dfrac{2x}{3} = 40$

$\Rightarrow \dfrac{21x-20x}{30} = 40$

$\Rightarrow \dfrac{x}{30} = 40$

$\Rightarrow x = 1200$

So, there are $1200$ pages in the book from which $1200 \times \dfrac7{10} = 840$ pages are read.

So this means there are $1200-840=360$ pages left to be read.

Using suitable rearrangement and find the sum:
$-5 + \dfrac{7}{10} + \dfrac{3}{7} + (-3) + \dfrac{5}{14} + \dfrac{-4}{5}$

Given

$-5 + \dfrac{7}{10} + \dfrac{3}{7} + (-3) + \dfrac{5}{14} + \dfrac{-4}{5}$

After rearrangment, We get

$=\left\{-5 + (-3)\right\} +\left\{ (-\dfrac{4}{5}) + \dfrac{7}{10} \right\}+\left\{ \dfrac{3}{7} + \dfrac{5}{14}\right\}\\=\left\{-5 -3\right\} +\left\{ \dfrac{7}{10} -\dfrac{8}{10}\right\}+\left\{ \dfrac{6}{14} + \dfrac{5}{14}\right\}\\= -8 + \dfrac{7 - 8}{10} + \dfrac{6 + 5}{14}\\= -8 - \dfrac{1}{10} + \dfrac{11}{14}\\= \dfrac{-560 - 7 +55}{70}\\= \dfrac{-512}{70}\\= \dfrac{-256}{35}$

In the morning, a milkman filled $5\frac{1}{5}$ L of milk in his can. He sold to Renu, Kamla and Renuka $\frac{3}{4 }L$ each; to Shadma he sold $\frac{7}{8}$ L; and to Jassi he gave $1\frac{1}{2}L$ . How much milk is left in the can?

2001992_1790293_ans_6868486fffd648b28cddf1d01d9487f9.png

Four friends had a competition to see how far they could hop on one foot. the table given shows the distance covered by each.
Name Seema Nancy Megha Soni
Distance covered (km) $\dfrac{1}{25}$ $\dfrac{1}{32}$ $\dfrac{1}{40}$ $\dfrac{1}{20}$
(a) How farther did Soni hop than Nancy?
(b) What is the total distance covered by Seema and Megha?
(c) Who walked farther, Nancy or Megha?

(a) Distance covered by Soni

$= \dfrac{1}{20} \,km$

Distance covered by Nancy

$= \dfrac{1}{32}\,km$

$\therefore$ Difference between Soni's and Nancy's hop

$= \dfrac{1}{20} - \dfrac{1}{32} = \dfrac{8 - 5}{160} = \dfrac{3}{160}\ km$

(b) Distance covered by Seema

$= \dfrac{1}{25}\,km$

Distance covered by Megha

$= \dfrac{1}{40}$

Total distance covered by Seema and Megha

$= \dfrac{1}{25} + \dfrac{1}{40} = \dfrac{8 + 5}{200} = \dfrac{13}{200}\ km$

$= \dfrac{1}{32}\,km=\dfrac5{160}\ km$

distance covered by Megha

$= \dfrac{1}{40}\,km=\dfrac4{16}\ km$

Comparing these numbers, we get

$\dfrac5{160}>\dfrac4{16}$

$\therefore \dfrac{1}{32}> \dfrac{1}{40}$

Hence, Nancy walked further.

The diagram shows the wingspans of different species of birds. Use the diagram to answer the question given below:
(a) How much longer is the wingspan of an Albatross than the wingspan of a Sea gull ?
(b) How much longer is the wingspan of a Golden eagle than the wingspan of a Blue jay ?

Given

The length of wingspan of Blue jay

$= \dfrac{41}{100}\,m$

The length of wingspan of Golden eagle

$= 2\dfrac{1}{2} \ m= \dfrac{5}{2} \,m$

The length of wingspan of Albatross

$= 3\dfrac{3}{5}\ m=\dfrac{18}5\ m$

The length of wingspan of Sea gull

$= 1\dfrac{7}{10}\ m=\dfrac{17}{10}\ m$

(a) Difference between the lengths of wingspan of Albatross and Sea gull

$=\left(\dfrac{18}{5} - \dfrac{17}{10}\right)\ \ m\\=\dfrac{36 - 17}{10} \ \ m\\= \dfrac{19}{10} \ m\\= 1\dfrac{9}{10}\ m$

(ii) Difference between the lengths of wingspan of Golden eagle and Blue jay

$= \left(\dfrac{5}{2} - \dfrac{41}{100}\right)\ m\\=\dfrac{250 - 41}{100} \ m\\= \dfrac{209}{100} \ m\\= 2\dfrac{9}{100}\ m$

From the following fractions, separate:
(i) Proper fractions
(ii) Improper fractions:
2 / 9, 4 / 3, 7 / 15, 11 / 20, 20 / 11, 18 / 23 and 27 / 35

(i) A fraction whose numerator is less than denominator is known as proper fractions

The proper fractions are $2 / 9,\ 7 / 15,\ 11 / 20, \ 18 / 23\ and\ 27 / 35$

(ii) A fraction whose numerator is greater than denominator is known as improper fractions

The improper fractions are $4 / 3\ and\ \ 20 / 11$

The table given below shows the distances, in kilometres, between four villages of a state. To find the distance between two villages, locate the square where the roe for one village and the column for the other village intersect.
(a) Compare the distance between Himgaon and Rawalpur to Sonapur and Ramgarh ?
(b) If you drove from Himgaon to Sonapur and then from Sonapur to Rawalpur, how far would you drive ?

(a) Distance between Himgaon and Rawalpur =

$98\dfrac{3}{4} \,km = \dfrac{395}{4}\ km$

Distance between Sonapur and Ramgarh

$= 40\dfrac{2}{3} \,km = \dfrac{122}{3} \,km$

On subtracting,

$\dfrac{395}{4} - \dfrac{122}{3} = \dfrac{1185}{12} - \dfrac{488}{12} = \dfrac{697}{12} =58.08 \,km$

$\therefore$ distance between Sonapur and Ramgarh is more than between Himgton and Rawalpur

(b) Distance between Himgton to Sonpur

$= 100\dfrac{5}{6}\ km=\dfrac{605}6\ km$

Distance between Sonapur to Rawalpur

$= 16\dfrac{1}{2}=\dfrac{33}2\ km$

$\therefore$ Total distance

$= \dfrac{605}{6} + \dfrac{33}{2} = \dfrac{605 + 99}{6} = \dfrac{704}{6} \,km=117.34\ km$

Simplify-
$\dfrac{3}{7} + \dfrac{-2}{21} \times \dfrac{-5}{6}$

$\dfrac{3}{7} + \dfrac{-2}{21} \times \dfrac{-5}{6}$

$= \dfrac{3}{7} + \dfrac{5}{63}$

$= \dfrac{27 + 5}{63}$

$= \dfrac{32}{63}$

Evaluate the following:

(i) $\dfrac{4}{3} + \dfrac{7}{8}$ (ii) $8 \dfrac{1}{2} - 3 \dfrac{5}{8}$

(ii) $\dfrac{5}{12}+ \dfrac{1}{18}- \dfrac{2}{9}$

$(i) \ \dfrac{4}{3} + \dfrac{7}{8}$

$= \dfrac{32+21}{24}$ (L.C.M. of

$3, 8 =24$ )

$= \dfrac{53}{24} = 2 \dfrac{5}{24}$

$(ii) \ 8\dfrac{1}{2} - 3\dfrac{5}{8}$

$= \dfrac{17}{2}- \dfrac{29}{8}$

$= \dfrac{68-29}{8}$ (L.C.M. of

$2, 8 =8$ )

$\dfrac{39}{8} = 4 \dfrac{7}{8}$

$(iii) \ \dfrac{5}{12}+ \dfrac{1}{18} - \dfrac{2}{9}$

$= \dfrac{15+2-8}{36}$ (L.C.M. of

$12, 8, 9 =36$ )

$\dfrac{17-8}{36} = \dfrac{9}{36} = \dfrac{9 \div 9}{36 \div 9} = \dfrac{1}{4}$

Add the following fractions:
(i) $1\dfrac { 3 }{ 4}$ and $\dfrac{3 }{ 8}$
(ii) $\dfrac{2}{ 5}$ , $2\dfrac { 3 }{ 15}$ and $\dfrac{7} { 10}$
(iii) $1\dfrac { 7 }{ 8}$ , $1\dfrac { 1 }{ 2}$ and $1\dfrac { 3 }{ 4}$
(iv) $3\dfrac { 3 }{ 4}$ , $2\dfrac { 1 }{ 6}$ , and $1\dfrac { 5 }{ 8}$
(v) $2\dfrac { 8 }{ 11}$ , $\dfrac{11}{ 18}$ and $3\dfrac { 5 }{ 6}$

(i) $1\frac { 3 }{ 4 }$ and 3 / 8

The given fractions can be added as follows

7 / 4 + 3 / 8 = (7 × 2) / (4 × 2) + 3 / 8

= 14 / 8 + 3 / 8

= 17 / 8

= $2\frac { 1 }{ 8 }$

Hence,
is the addition of given fractions

(ii) 2 / 5,
and 7 / 10

The given fractions can be added as follows

2 / 5 + 33 / 15 + 7 / 10 = (2 × 6) / (5 × 6) + (33 × 2) / (15 × 2) + (7 × 3) / (10 × 3)

= 12 / 30 + 66 / 30 + 21 / 30

= 99 / 30

= 33 / 10

Hence,
is the addition of given fractions

(iii)
,
and

The given fractions can be added as follows

15 / 8 + 3 / 2 + 7 / 4 = (15 × 1) / (8 × 1) + (3 × 4) / (2 × 4) + (7 × 2) / (4 × 2)

= 15 / 8 + 12 / 8 + 14 / 8

= 41 / 8

Hence,
is the addition of the given fractions

(iv) $3\frac { 3 }{ 4 } \quad$ , $2\frac { 1 }{ 6 } \quad$ and $1\frac { 5 }{ 8 } \quad$

The given fractions can be added as follows

15 / 4 + 13 / 6 + 13 / 8 = (15 × 6) / (4 × 6) + (13 × 4) / (6 × 4) + (13 × 3) / (8 × 3)

= 90 / 24 + 52 / 24 + 39 / 24

= 181 / 24

= $7\frac { 13 }{ 24 } \quad$

Hence,
$7\frac { 13 }{ 24 } \quad$ is the addition of given fractions

(v)

$2\frac { 8 }{ 9 } \quad$ , 11 / 18 and $3\frac { 5 }{ 6 } \quad$

The given fractions can be added as follows

26 / 9 + 11 / 18 + 23 / 6 = (26 × 2) / (9 × 2) + 11 / 18 + (23 × 3) / (6 × 3)

= 52 / 18 + 11 / 18 + 69 / 18

= 132 / 18

= 22 / 3

= $7\frac { 1 }{ 3 } \quad$

Hence,

$7\frac { 1 }{ 3 } \quad$ is the addition of given fractions

A skirt that is $35\dfrac{7}{8}\,cm$ long has a hem of $3\dfrac{1}{8} \,cm.$ How long will the skirt be if the hem is let down ?

Given

Length of skirt

$= 35\dfrac{7}{8}\,cm = \dfrac{297}{8} \,cm$
Length of hem

$= 3\dfrac{1}{8}\ cm = \dfrac{25}{8}\ cm$

$\therefore$ Total length of skirt

$= \dfrac{287 }{8}+ \dfrac{25}{8}\\$

$=\dfrac{287 + 25}{8}\\$

$= \dfrac{312}{8} \\= 39\,cm$

Hence, The skirt will be

$39 \ cm$ long if the hem is let down

Classify each fraction given below as decimal or vulgar fraction, proper or improper fraction and mixed faction
11/10

11/10 is a decimal and improper fraction.

Simplify the following:

(i) $7 \dfrac{3}{4} - 3 \dfrac{5}{6} + \dfrac{7}{8}$

(ii) $6 \dfrac{1}{8} - 2 \dfrac{1}{12} - 5 \dfrac{1}{10} + 3 \dfrac{7}{25}$

$(i) \ 7\dfrac{3}{4} - 3 \dfrac{5}{6} + \dfrac{7}{8} = \dfrac{31}{4} - \dfrac{23}{6}+ \dfrac{7}{8}$

(L.C.M. of

$4, 6$ and

$8=24$ )

$=\dfrac{31 \times 6 - 23 \times 4 + 7 \times 3}{24}$

$= \dfrac{186-92+21}{24}$

$= \dfrac{207 -92}{24} = \dfrac{115}{24} = 4 \dfrac{19}{24}$

$(ii) \ 6\dfrac{1}{8} - 2 \dfrac{1}{12} - 5\dfrac{1}{10} + 3 \dfrac{7}{25}$

$= \dfrac{49}{8}-\dfrac{25}{12}- \dfrac{51}{10}+ \dfrac{82}{25}$

(L.C.M. of

$8, 12, 10$ and

$25$ is

$600$ )

$=\dfrac{49 \times 75 - 25 \times 50 -51 \times 60+ 82 \times 24}{600}$

$= \dfrac{3675-1250-3060+1968}{600}= \dfrac{1333}{600}$

$2\dfrac{133}{600}$

Convert given fractions into fractions with equal numerators:
6/13, 15/23 and 12/17

$6/13, 15/23$ and

$12/17$
Here the LCM of

$6, 15$ and

$12$ is

$60$

$6/13 = \left(6 \times 10 \right)/\left(13 \times 10 \right) = 60/130$

$15/23 = \left(15 \times 4 \right)/\left(23 \times 4 \right) = 60/92$

$12/17 = \left(12 \times 5 \right)/\left(17 \times 5 \right) = 60/85$
Therefore,

$60/130, 60/92$ and

$60/85$ are the required fractions.

Convert given fractions into fractions with equal numerators:
15/19, 25/28, 9/11 and 45/47

$15/19, 25/28, 9/11$ and

$45/47$
Here the LCM of

$15, 25, 9$ and

$45$ is

$225$

$15/19 = \left(15 \times 15 \right)/\left(19 \times 15 \right) = 225/285$

$25/28 = \left(25 \times 9 \right)/\left(11 \times 25 \right) = 225/252$

$9/11 = \left(9 \times 25 \right)/\left(11 \times 25 \right) = 225/275$

$145/47 = \left(45 \times 5 \right)/\left(47 \times 5 \right) = 225/235$
Therefore,

$225/285, 225/252, 225/275$ and

$225/235$ are the required fractions.

Classify each fraction given below as decimal or vulgar fraction, proper or improper fraction and mixed faction
$3 \dfrac{2}{9}$

$3 \dfrac{2}{9}$ is a mixed fraction.

Classify each fraction given below as decimal or vulgar fraction, proper or improper fraction and mixed faction
18/7

$18/7$ is a vulgar and improper fraction.

Convert given fractions into fractions with equal denominators:
5/6 and 7/9

$5/6$ and

$7/9$
Here the LCM of

$6$ and

$9$ is

$18$

$5/6 = \left(5 \times 3 \right)/\left(6 \times 3 \right) = 15/18$

$7/9 = \left(7 \times 2 \right)/\left(9 \times 2 \right) = 14/18$
Therefore,

$15/18$ and

$14/18$ are the required fractions.

Put the given fractions in ascending order by making denominators equal:
$1/3, 2/5, 3/4$ and $1/6$

$1/3, 2/5, 3/4$ and

$1/6$
Here the LCM of

$3, 5, 4$ and

$6$ is

$60$

$1/3 = \left(1 \times 20 \right)/\left(3 \times 20 \right) = 20/60$

$2/5 = \left(2 \times 12 \right)/\left(5 \times 12 \right) = 24/60$

$3/4 = \left(3 \times 15 \right)/\left(4 \times 15 \right) = 45/60$

$1/6 = \left(1 \times 10 \right)/\left(6 \times 10 \right) = 10/60$
So we get

$10/60 < 20/60 < 24/60 < 45/60$
It can be written as

$1/6 < 1/3 < 2/5 < 3/4$
Therefore,

$1/6, 1/3, 2/5$ and

$3/4$ are in ascending order.

Convert given fractions into fractions with equal denominators:
$4/5, 17/20, 23/20$ and $11/16$

$4/5, 17/20, 23/20$ and

$11/16$
Here the LCM of

$5, 20, 40$ and

$16$ is

$80$

$4/5 = \left(4 \times 16 \right)/\left(5 \times 16 \right) = 64/80$

$17/20 = \left(17 \times 4 \right)/\left(20 \times 4 \right) = 68/80$

$23/40 = \left(23 \times 2 \right)/\left(40 \times 2 \right) = 46/80$

$11/16 = \left(11 \times 5 \right)/\left(16 \times 5 \right) = 55/80$
Therefore,

$64/80, 68/80, 46/80$ and

$55/80$ are the required fractions.

Convert given fractions into fractions with equal numerators:
8/9 and 12/17

$8/9$ and

$12/17$
Here the LCM of

$8$ and

$12$ is

$24$

$8/9 = \left(8 \times 3 \right)/\left(9 \times 3 \right) = 24/27$

$12/17 = \left(12 \times 2 \right)/\left(17 \times 2 \right) = 24/34$
Therefore,

$24/27$ and

$24/34$ are the required fractions.

Convert given fractions into fractions with equal denominators:
2/3, 5/6 and 7/12

$2/3, 5/6$ and

$7/12$
Here the LCM of

$3, 6$ and

$12$ is

$12$

$2/3 = \left(2 \times 4 \right)/\left(3 \times 4 \right) = 8/12$

$5/6 = \left(5 \times 2 \right)/\left(6 \times 2 \right) = 10/12$

$7/12 = 7/12$
Therefore,

$8/12, 10/12$ and

$7/12$ are the required fractions.

Classify each fraction given below as decimal or vulgar fraction, proper or improper fraction and mixed faction
13/20

$13/20$ is a decimal and proper fraction.

The heights of two vertical poles, above the earth's surface, are $14 \dfrac{1}{2}$ m and $22 \dfrac{1}{3}$ m respectively. How much higher is the second pole as compared with the height of the first pole?

It is given that
Height of first pole above the earth's surface

$=14 \dfrac{1}{2}$ m
Height of second pole above the earth's surface

$=22 \dfrac{1}{3}$ m
So height of second pole when compared to first pole =

$22 \dfrac{1}{3}$ m

$-14 \dfrac{1}{2}$ m
We can write it as

$= 67/3 - 29/2$
LCM of 3 and 2 is 6

$= \left(134 - 87 \right)/6$
By further calculation

$\dfrac{47}{6}=7\dfrac{5}{6}m$

Arrange the given fractions in descending order by making numerators equal :
$3/7, 4/9, 5/7$ and $8/11$

$3/7, 4/9, 5/7$ and

$8/11$
Here the LCM of

$3, 4, 5$ and

$8$ is

$120$

$3/7 = \left(3 \times 40 \right)/\left(7 \times 40 \right) = 120/280$

$4/9 = \left(4 \times 30 \right)/\left(9 \times 30 \right) = 120/270$

$5/7 = \left(5 \times 24\right)/\left(7 \times 24 \right) = 120/168$

$8/11 = \left(8 \times 15 \right)/\left(11 \times 15 \right) = 120/165$
So we get

$120/165 > 120/168 > 120/270 > 120/280$
It can be written as

$8/11 > 5/7 > 4/9 > 3/5$
Therefore,

$8/11, 5/7, 4/9$ and

$3/7$ are in descending order.

A student bought $4 \dfrac{1}{3}$ m of yellow ribbon, $6 \dfrac{1}{6}$ m of red ribbon and $3 \dfrac{2}{9}$ m of blue ribbon decorating a room. How many meters of ribbon did he buy?

It is given that
Length of yellow ribbon

$= 4 \dfrac{1}{3} m = 13/3 m$
Length of red ribbon

$= 6 \dfrac{1}{6} m = 37/6 m$
Length of blue ribbon

$= 3 \dfrac{2}{9} m = 29/9 m$
So the total length

$= 13/3 + 37/6 + 29/9$
LCM of

$3, 6$ and

$9$ is

$18$

$= \left(78 + 111 + 58 \right)/18$
So we get

$= 247/18$

$= 13 \dfrac{13}{18} m$

Arrange the given fractions in descending order by making numerators equal:
$1/10, 6/11, 8/11$ and $3/5$

$1/10, 6/11, 8/11$ and

$3/5$
Here the LCM of

$1, 6, 8$ and

$3$ is

$24$

$1/10 = \left(1 \times 24 \right)/\left(10 \times 24 \right) = 24/240$

$6/11 = \left(6 \times 4 \right)/\left(11 \times 4 \right) = 24/44$

$8/11 = \left(8 \times 3\right)/\left(11 \times 3 \right) = 24/33$

$3/5 = \left(3 \times 8 \right)/\left(5 \times 8 \right) = 24/40$
So we get

$24/33 > 24/40 > 24/44 > 24/240$
It can be written as

$8/11 > 3/5 > 6/11 > 1/10$
Therefore,

$8/11, 3/5, 6/11$ and

$1/10$ are in descending order.

Put the given fractions in ascending order by making denominators equal:
$5/6, 7/8, 11/12$ and $3/10$

$5/6, 7/8, 11/12$ and

$3/10$
Here the LCM of

$6, 8, 12$ and

$10$ is

$240$

$5/6 = \left(5 \times 40 \right)/\left(6 \times 40 \right) = 200/240$

$7/8 = \left(7 \times 30 \right)/\left(8 \times 30 \right) = 210/240$

$11/12 = \left(11 \times 20\right)/\left(12 \times 20 \right) = 220/240$

$3/10 = \left(3 \times 24 \right)/\left(10 \times 24 \right) = 72/240$
So we get

$72/240 < 200/240 < 210/240 < 220/240$
It can be written as

$3/10 < 5/6 < 7/8 < 11/12$
Therefore,

$3/10, 5/6, 7/8$ and

$11/12$ are in ascending order.

From a sack of potatoes weighing 120 kg, a merchant sells portions weighing $6 kg, 5 \dfrac{1}{4} kg, 9 \dfrac{1}{2} kg$ and $9 \dfrac{3}{4}$ kg respectively.
(i) How many kg did he sell?
(ii) How many kg are still left in the sack?

Weight of potato

$= 120 kg$
(i) potatoes sold by merchant

$= 6 kg + 5 \dfrac{1}{4} kg + 9 \dfrac{1}{2} kg + 9 \dfrac{3}{4} kg$
It can be written as

$= \left(6 + 21/4 + 19/2 + 39/4 \right)$ kg
LCM of 1, 4 and 2 is 4

$= \left(24 + 21 + 38 + 39 \right)/4$
By further calculation

$= 122/4$

$= 61/2 kg$

$= 30 \dfrac{1}{2} kg$
(ii) Potatoes left in the sack

$= 120 kg - 30 \dfrac{1}{2} kg$
It can be written as

$= \left(120/1 - 61/2 \right)$ kg
By further calculation

$= \left(240 - 61 \right)/2$ kg
So we get

$= 179/2 kg$

$= 89 \dfrac{1}{2} kg$

Reduce to a single fraction:
$2/3 - 1/6$

$2/3 - 1/6$
Here the LCM of

$3$ and

$6$ is

$6$

$= \left (2 \times 2 \right )/ \left (3 \times 2 \right ) - 1/6$
By further calculation

$= 4/6 - 1/6$
So we get
=

$\left (4 - 1 \right )/6$

$= 3/6$

$= 1/2$

Arrange the given fractions in descending order by making numerators equal:
$5/6, 4/15, 8/9$ and $1/3$

$5/6, 4/15, 8/9$ and

$1/3$
Here the LCM of

$5, 4, 8$ and

$1$ is

$40$

$5/6 = \left(5 \times 8 \right)/\left(6 \times 8 \right) = 40/48$

$4/15 = \left(4 \times 10 \right)/\left(15 \times 10 \right) = 40/150$

$8/9 = \left(8 \times 5\right)/\left(9 \times 5 \right) = 40/45$

$1/3 = \left(1 \times 40 \right)/\left(3 \times 40 \right) = 40/120$
So we get

$40/45 > 40/48 > 40/120 > 40/150$
It can be written as

$8/9 > 5/6 > 1/3 > 4/15$
Therefore,

$8/9, 5/6, 1/3$ and

$4/15$ are in descending order.

Put the given fractions in ascending order by making denominators equal:
$5/7, 3/8, 9/14$ and $20/21$

$5/7, 3/8, 9/14$ and

$20/21$
Here the LCM of

$7, 8, 14$ and

$21$ is

$168$

$5/7 = \left(5 \times 24 \right)/\left(7 \times 24 \right) = 120/168$

$3/8 = \left(3 \times 21 \right)/\left(8 \times 21 \right) = 63/168$

$9/14 = \left(9 \times 12\right)/\left(14 \times 12 \right) = 108/168$

$20/21 = \left(20 \times 8 \right)/\left(21 \times 8 \right) = 160/168$
So we get

$63/168 < 108/168 < 120/168 < 160/168$
It can be written as

$3/8 < 9/14 < 5/7 < 20/21$
Therefore,

$3/8, 9/14, 5/7$ and

$20/21$ are in ascending order.

Reduce to a single fraction:
$3/5 - 1/10$

$3/5 - 1/10$
Here the LCM of

$5$ and

$10$ is

$10$

$= \left (3 \times 2 \right )/ \left (5 \times 2 \right ) - 1/10$
By further calculation

$= 6/10 - 1/10$
So we get

$= \left(6 - 1 \right)/10$

$= 5/10$

$= 1/2$

Subtract :
$\dfrac{5}{8} From \dfrac{-3}{8}$

1974156_1837779_ans_9e435bee8dcb4ddcadabeb81a8a728a4.jpeg

A rectangular field is $16 \dfrac{1}{2} m$ long and $12 \dfrac{2}{5} m$ wide. Find the perimeter of the field.

Dimensions of rectangular field are
Length

$= 16 \dfrac{1}{2} m$
Breadth

$= 12 \dfrac{2}{5} m$
So the perimeter =

$2 \left( l + b\right)$
Substituting the values
=

$2 \left( 16 \dfrac{1}{2} + 12 \dfrac{2}{5}\right)$
It can be written as
=

$2 \left( 33/2 + 62/5\right)$
LCM of 2 and 5 is 10

$= 2 \times \left[\left(33 \times 5 \right)/\left(2 \times 5 \right) + \left(62 \times 2 \right)/\left(5 \times 2 \right)\right]$
By further calculation

$= 2 \times \left[\left(165 + 124 \right)/10\right]$

$= 2 \times 289/10$

$= 289/5 m$

$= 57 \dfrac{4}{5} m$

Draw number lines and locate the points on them:
a) $\frac{1}{2}, \frac{1}{4}, \frac{3}{4},\frac{4}{4}$
b) $\frac{2}{5}, \frac{3}{5}, \frac{8}{5},\frac{4}{5}$
c) $\frac{2}{5}, \frac{3}{5}, \frac{8}{5},\frac{4}{5}$

1878012_1866520_ans_8fbead6c090b4874837eeaacc2597847.png

Which is greater, 3/5 or 7/10 and by how much?

By cross multiplication

$3 \times 10 = 30$ and

$7 \times 5 = 35$
Here

$30$ is smaller than

$35$
So

$3/5 < 7/10$
We know that difference between

$7/10$ and

$3/5$

$= 7/10 - 3/5$
LCM of

$10$ and

$5$ is

$10$

$= \left(7 \times 1 \right)/\left(10 \times 1 \right) - \left(3 \times 2 \right)/\left(5 \times 2 \right)$
So we get

$= \left(7 - 6 \right)/4$

$= 1/10$
Therefore,

$7/10$ is greater than

$3/5$ by

$1/10.$

Subtract:
2/9 from 4/5

$2/9$ and

$4/5$
It can be written as

$= 4/5 - 2/9$
LCM of

$5$ and

$9$ is

$45$

$= \left(4 \times 9 \right)/\left(5 \times 9 \right) - \left(2 \times 5 \right)/\left(9 \times 5 \right)$
By further calculation

$= 36/45 - 10/45$

$= \left(36 - 10\right)/45$

$= 26/45$

A bought $3 \dfrac{3}{4}$ kg of wheat and $2 \dfrac{1}{2}$ kg of rice. Find the total weight wheat and rice bought.

It is given that
Weight of wheat

$= 3 \dfrac{3}{4} kg = 15/4 kg$
Weight of rice

$= 2 \dfrac{1}{2} kg = 5/2 kg$
So the total weight of wheat and rice

$= 15/4 + 5/2$
Here the LCM of 4 and 2 is 4

$= \left(15 \times 1 \right)/\left(4 \times 1 \right) + \left(5 \times 2 \right)/\left(2 \times 2 \right)$
So we get

$= \left(15 + 10 \right)/4$

$= 25/4 kg$

$= 6 \dfrac{1}{4} kg$

Solve:
$\frac{5}{7} + \frac{1}{3}$

$= \dfrac{(5 \times 3) + (1 \times 7)}{21}$ (Taking 21 as L.C.M.)

$= \dfrac{15 + 7}{21} = \dfrac{22}{21}$

Reduce to a single fraction:
$2/3 - 3/5 + 3 - 1/5$

$2/3 - 3/5 + 3 - 1/5$
Here the LCM of

$3$ and

$5$ is

$15$

$= \left (2 \times 5 \right )/ \left (3 \times 5 \right ) - \left (3 \times 3 \right )/ \left (5 \times 3 \right ) + \left (3 \times 15 \right )/ 15 - \left (1 \times 3 \right )/ \left (5 \times 3 \right )$
By further calculation

$= 10/15 - 9/15 + 45/15 - 3/15$
So we get
=

$\left (10 - 9 + 45 - 3 \right )/15$

$= 43/15$

$= 2 \dfrac{13}{15}$

Reduce to a single fraction:
$2/3 - 1/5 + 1/10$

$2/3 - 1/5 + 1/10$
Here the LCM of

$3, 5$ and

$10$ is

$30$

$= \left (2 \times 10 \right )/ \left (3 \times 10 \right ) - \left (1 \times 6 \right )/ \left (5 \times 6 \right ) + \left (1 \times 3 \right )/ \left (10 \times 3 \right )$
By further calculation

$= 20/30 - 6/30 + 3/30$
So we get
=

$\left (20 - 6 + 3 \right )/30$

$= 17/30$

Reduce to a single fraction:
1/4 + 5/6 - 1/12

$1/4 + 5/6 - 1/12$
Here the LCM of

$4, 6$ and

$12$ is

$12$

$= \left (1 \times 3 \right )/ \left (4 \times 3 \right ) - \left (5 \times 2 \right )/ \left (6 \times 2 \right ) - 1/12$
By further calculation

$= 3/12 + 10/12 - 1/12$
So we get
=

$\left (3 + 10 - 1 \right )/12$

$= 12/12$

$= 1$

Fill in the missing fraction
(a) $\frac{7}{10} - \Box = \frac{3}{10}$
(b) $\Box - \frac{3}{21} = \frac{5}{21}$
(c) $\Box - \frac{3}{6} = \frac{3}{6}$
(d) $\Box + \frac{5}{27} = \frac{12}{27}$

(a)

$\frac{7}{10} - \Box = \frac{3}{10}$

$\Box = \frac{7}{10} - \frac{3}{10} = \frac{7 - 3}{10} = \frac{4}{10} = \frac{2}{5}$
(b)

$\Box - \frac{3}{21} = \frac{5}{21}$

$\Box = \frac{5}{21} + \frac{3}{21} = \frac{5 + 3}{21} = \frac{8}{21}$
(c)

$\Box - \frac{3}{6} = \frac{3}{6}$

$\Box = \frac{3}{6} + \frac{3}{6} = \frac{3 + 3}{6} = \frac{6}{6} = 1$
(d)

$\Box + \frac{5}{27} = \frac{12}{27}$

$\Box = \frac{12}{27} - \frac{5}{27} = \frac{12 - 5}{27} = \frac{7}{27}$

Solve:
$4 \frac{2}{3} + 3 \frac{1}{4}$

$\frac{(3 \times 4) + 2}{3} + \frac{(3 \times 4) + 1}{4}$

$= \frac{14}{3} + \frac{13}{4}$

$= \frac{(14 \times 4) + (13 \times 3)}{12} = \frac{56 + 39}{12} = \frac{95}{12}$

Solve:
$\frac{4}{3} - \frac{1}{2}$

$\frac{(4 \times 2) - (1 \times 3)}{6} = \frac{8 - 3}{6} = \frac{5}{6}$

Complete the addition - subtraction box.

$\frac{2}{3} + \frac{4}{3} = \frac{2 + 4}{3} = \frac{6}{3} - 2$

$\frac{1}{3} + \frac{2}{3} = \frac{1 + 2}{3} = \frac{3}{3} - 1$

$\frac{4}{3} - \frac{2}{3} = \frac{4 - 2}{3} = \frac{2}{3}$

$\frac{1}{3} + \frac{2}{3} = \frac{2}{3} = 1$
1879498_1866710_ans_b4f4d5804ec44ceeb8ae4f46d24a2693.png

Solve:
$\frac{16}{5} - \frac{7}{5}$

$\frac{16}{5} - \frac{7}{5} = \frac{16 -7}{5} = \frac{9}{5}$

Fill in the blanks:
$\Box - \frac{1}{5} = \frac{1}{2}$

$\frac{7}{10} - \frac{1}{5} = \frac{1}{2}$

Fill in the blanks:
$\frac{1}{2} - \Box = \frac{1}{6}$

$\frac{1}{2} - \frac{1}{3} = \frac{1}{6}$

Solve
(a) $\frac{1}{18} + \frac{1}{18}$
(b) $\frac{8}{15} + \frac{3}{15}$
(c) $\frac{7}{7} - \frac{5}{7}$
(d) $\frac{1}{22}+ \frac{21}{22}$
(e) $\frac{12}{15} - \frac{7}{15}$
(f) $\frac{5}{8} + \frac{3}{8}$
(g) $1 - \frac{2}{3} \Big( 1 = \frac{3}{3}\Big)$
(h) $\frac{1}{4} + \frac{0}{4}$

(a)

$\frac{1}{18} + \frac{1}{18} = \frac{1 + 1}{18} = \frac{2}{18} = \frac{1}{9}$ (b)

$\frac{8}{15} + \frac{3}{15} = \frac{8 + 3}{15} = \frac{11}{5}$
(c)

$\frac{7}{7} - \frac{5}{7} = \frac{7 - 5}{7} = \frac{2}{7}$ (d)

$\frac{1}{22}+ \frac{21}{22} = \frac{1 + 21}{22} = \frac{22}{22} = 1$
(e)

$\frac{12}{15} - \frac{7}{15} = \frac{12 - 7}{15} = \frac{5}{15} = \frac{1}{3}$ (f)

$\frac{5}{8} + \frac{3}{8} = \frac{5 + 3}{8} = \frac{8}{8} = 1$
(g)

$1 - \frac{2}{3} \Big( 1 = \frac{3}{3}\Big) = 1 - \frac{2}{3} = \frac{3}{3} - \frac{2}{3} = \frac{3 - 2}{3} = \frac{1}{3}$
(h)

$\frac{1}{4} + \frac{0}{4} = \frac{1}{4} + 0 = \frac{1}{4}$

Fill in the blanks:
$\Box - \frac{5}{8} = \frac{1}{4}$

$\Box - \frac{5}{8} = \frac{1}{4}$

$\Box = \frac{1}{4}+\frac{5}{8}$

$\Box = \frac{1}{4}+\frac{5}{8} =\frac{7}{8}$

$\frac{7}{8} - \frac{5}{8} = \frac{1}{4}$

Complete the addition - subtraction box.

$\frac{1}{2} + \frac{1}{3} = \frac{(1 \times 3) + (1 \times 2)}{6} = \frac{3 + 2}{6} = \frac{5}{6}$

$\frac{1}{2} - \frac{1}{3} = \frac{(1 \times 4) + (1 \times 3)}{12} = \frac{4 + 3}{12} = \frac{7}{12}$

$\frac{1}{2} - \frac{1}{3} = \frac{(1 \times 3) - (1 \times 2)}{6} = \frac{3 - 2}{6} = \frac{1}{6}$

$\frac{1}{3} - \frac{1}{4} = \frac{(1 \times 4) - (1 \times 3)}{12} = \frac{4 - 3}{12} = \frac{1}{12}$

$\frac{1}{6} + \frac{1}{12} = \frac{(1 \times 2) + 1}{12} = \frac{2 + 1}{12} = \frac{3}{12} = \frac{1}{4}$
1879501_1866713_ans_12d74ec095714a37b489ffc897af2ec6.png

Simply the following :
(i) $\left( \frac { 25 }{ 9 } +\frac { 12 }{ 3 } \right) +\frac { 3 }{ 5 }$

$\left( \dfrac { 25\times3+12\times9 }{ 9\times3 } \right) +\dfrac { 3 }{ 5 }$

$\dfrac { 75+108 }{ 27 } +\dfrac { 3 }{ 5 } =\dfrac { 183 }{ 27 } +\dfrac { 3 }{ 5 } =\dfrac { 61 }{ 9 } +\dfrac { 3 }{ 5 }$

$=\dfrac { 61\times5+3\times9 }{ 9\times5 } =\dfrac { 305+27 }{ 45 }$

$=\dfrac{332}{45}$

Represent the following mixed infinite decimal periodic fractions as common fractions:
$\displaystyle\dfrac{2^{-2} + 2^{0}}{\left(\dfrac{1}{2}\right)^{-2} - 5 \cdot (-2)^{-2} + \left(\dfrac{2}{3}\right)^{-2}}$

$\displaystyle\dfrac{2^{-2} + 2^{\circ}}{\left(\dfrac{1}{2}\right)^{-2} - 5 \cdot (-2)^{-2} + \left(\dfrac{2}{3}\right)^{-2}}$

$=\dfrac{\left(\dfrac{1}{2}\right)^2+1}{\left(4\right)^2-5.\left(\dfrac{1}{-2}\right)^2+\left(\dfrac{3}{2}\right)^2}$

$=\dfrac{\dfrac{5}{4}} {16- \dfrac{5}{4} +\dfrac{9}{4}}$

$=\dfrac{\dfrac{5}{4}}{16+\dfrac{4}{4}}$

$=\dfrac{\dfrac{5}{4} }{\dfrac{15}{1}}$

$=\dfrac{5}{4}\times\dfrac{1}{15}$

$=\dfrac{1}{12}$

Represent the following mixed infinite decimal periodic fractions as common fractions:
$\displaystyle\dfrac{(0. 6)^{0} - (0.1)^{-1}}{\left(\dfrac{3}{2^3}\right)^{-1} \cdot \left(\dfrac{3}{2}\right)^3 + \left(-\dfrac{1}{3}\right)^{-1}}$

$\displaystyle\dfrac{(0. 6)^{0} - (0.1)^{-1}}{\left(\dfrac{3}{2^3}\right)^{-1} \cdot \left(\dfrac{3}{2}\right)^3 + \left(-\dfrac{1}{3}\right)^{-1}}$

$= \dfrac{1-10}{\dfrac{8}{3} .\dfrac{27}{8}-3}$ .....

$(x^0=1)$

$=\dfrac{-9}{9-3}$

$=\dfrac{-9}{6}$

$=\dfrac{-3}{2}$

Solve the following:
$\displaystyle\frac{x - 1}{x^{3/4} + x^{1/2}}\, \cdot \frac{x^{1/2} + x^{1/4}}{x^{1/2}+ 1}\, \cdot x^{1/4} + 1$ for $x = 16$

Here the value of

$x=16$ so , put the value of

$x$ in the given equation we

$\dfrac{16-1}{16^{\dfrac{3}{4}}+16^{\dfrac{1}{2}}}$ .

$\dfrac{16^{\dfrac{1}{2}}+ 16^{\dfrac{1}{4}}}{16^{\dfrac{1}{2}}+1}$ .

$16^{\dfrac{1}{4}}+1$

$\dfrac{16-1}{(2^4)^{\dfrac{3}{4}}+(2^4)^{\dfrac{1}{2}}}$ .

$\dfrac{(2^4)^{\dfrac{1}{2}}+(2^4)^{\dfrac{1}{4}}}{(2^4)^{\dfrac{1}{2}}+1}$ .

$(2^4)^{\dfrac{1}{4}}+1$

$\dfrac{16-1}{2^3+2^2}$ .

$\dfrac{2^2+2}{2^2+1}$ .

$2+1$

$\dfrac{15}{12}$ .

$\dfrac{6}{5}.2+1$

$\therefore$ The answer is

$3+1=4$

Solve the following:
$\displaystyle\dfrac{a^3 - a - 2b - \dfrac{b^2}{a}}{\left ( 1 - \sqrt{\dfrac{1}{a} + \dfrac{b}{a^2}} \right )(a + \sqrt{a + b})}: \left ( \dfrac{a^3 + a^2 + ab + a^2b}{a^2 - b^2} + \dfrac{b}{a - b} \right )$ for $a =23, b = 22$

Put the value of

$a$ and

$b$ in the given expression.

We have ,

$\dfrac{23^3-23-(2\times 22)-\dfrac{22^2}{23}}{(1-\sqrt{\dfrac{1}{23}+\dfrac{22}{23^2}})(a+\sqrt{22+23})} :$

$(\dfrac{23^3+23^2+23\times22+23^2\times22}{23^2-22^2}+\dfrac{22}{23-22})$

Now simplify the expression,

$\dfrac{\dfrac{(23^4-23^2-(2\times 23\times22)-22^2) (23-\sqrt{23+22})}{23}}{(\dfrac{23-\sqrt{23^2+22}}{23})(23+\sqrt{23+22})(23-\sqrt{23+22})}:$

$\dfrac{23^3+23^2+(23\times22)+(23^2\times22)+22(23+22)}{23^2-22^2}$

$\dfrac{\dfrac{(277816)(23-\sqrt{45}}{23}}{\dfrac{(23-\sqrt{45)}(529-45)}{23}}:$

$\dfrac{25370}{45}$

$\dfrac{(277816)(23+\sqrt{45})}{23}\times\dfrac{23}{(23-\sqrt{45})(484)}:\dfrac{25370}{45}$

$\dfrac{277816}{484}$ :

$\dfrac{25370}{45}$

$\dfrac{277816}{484}\times \dfrac{45}{25370}=1.0181316$ hence this the final answer.

Represent the following mixed infinite decimal periodic fractions as common fractions:
$\displaystyle\frac{3\dfrac{1}{3} \sqrt{\dfrac{9}{80}} - \dfrac{5}{4} \sqrt{\dfrac{4}{5}} + 5 \sqrt{\dfrac{1}{5}} + \sqrt{20} - 10\sqrt{0.2}} {3\dfrac{1}{2} \sqrt{32} - \sqrt{4\dfrac{1}{2}} + 2 \sqrt{\dfrac{1}{8}} + 6\sqrt{\dfrac{2}{9}} - 140\sqrt{0.02}} \, \sqrt{\dfrac{2}{3}}$

$\displaystyle\frac{3\dfrac{1}{3} \sqrt{\dfrac{9}{80}} - \dfrac{5}{4} \sqrt{\dfrac{4}{5}} + 5 \sqrt{\dfrac{1}{5}} + \sqrt{20} - 10\sqrt{0.2}} {3\dfrac{1}{2} \sqrt{32} - \sqrt{4\dfrac{1}{2}} + 2 \sqrt{\dfrac{1}{8}} + 6\sqrt{\dfrac{2}{9}} - 140\sqrt{0.02}} \, \sqrt{\dfrac{2}{3}}$

$=\dfrac{\dfrac{10}{3}\dfrac{3}{4\sqrt{5}}-\dfrac{5}{4}.\dfrac{2}{\sqrt{5}}+\dfrac{5}{\sqrt{5}}+2\sqrt{5}-\dfrac{10}{\sqrt{5}}} {\dfrac{7}{2}.4\sqrt{2} -\dfrac{3}{\sqrt{2}}+\dfrac{1}{\sqrt{2}} .\dfrac{6\sqrt{2}}{3} -\dfrac{140}{5\sqrt{2}}}$

$\sqrt{\dfrac{2}{3}}$

$= \dfrac{\sqrt{5} -\sqrt{5} +\sqrt{5} +2\sqrt{5}-2\sqrt{5}} {\dfrac{28}{\sqrt{2}}-\dfrac{3}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}+\dfrac{4}{\sqrt{2}} -\dfrac{28}{\sqrt{2}}}$

$\sqrt{\dfrac{2}{3}}$

$=\dfrac{\sqrt{5}}{\sqrt{2}}$

$\sqrt{\dfrac{2}{3}}$

$=\sqrt{\dfrac{5}{3}}$

Represent the following mixed infinite decimal periodic fractions as common fractions:
Find $x$ if $\displaystyle\frac{0.1 (6) + 0.(3)}{0.(3) + 1.1 (6)} \, x = 10$
Calculate (5 31)

Change the mixed infinite decimal into a common fraction first

$0.1(6)$ in common fraction is

$\dfrac{15}{90}$

$0.(3)$ in common fraction is

$\dfrac{1}{3}$

$1.1(6)$ in common fraction is

$\dfrac{105}{90}$

Now put the value in the given expression, we have

$\dfrac{\dfrac{15}{90}+\dfrac{1}{3}}{\dfrac{1}{3}+\dfrac{105}{90}}x$

After solving the expression, we get

$0.3333333x$

This can also be written as

$\dfrac{1}{3}x$

which is equal to

$10$

$\therefore \dfrac{1}{3}x=10$

$\Rightarrow x=30$

Represent the following mixed infinite decimal periodic fractions as common fractions:
$\displaystyle\, \left ( \frac{1}{(a^{1/2} + b^{1/2})^{-2}} \right ) - \left ( \frac{\sqrt{a} - \sqrt{b}}{a^{3/2} - b^{3/2}} \right )^{-1} (ab)^{-1/2}$

$[(\cfrac{1}{a^{\frac{1}{2}}+b^{\frac{1}{2}})^{-2}})-(\cfrac{\sqrt a-\sqrt b}{a^{\frac{3}{2}}-b^{\frac{3}{2}}})^{-1}](ab)^{\frac{-1}{2}}$

$=[\sqrt a+\sqrt b)^{2}-\cfrac{(\sqrt a)^{3}-(\sqrt b)^{3}}{\sqrt a-\sqrt b}]\sqrt {ab}$

$=[a+b+2\sqrt {ab}-\cfrac{(\sqrt a-\sqrt b)(a+b+\sqrt {ab})}{(\sqrt a-\sqrt b)}]\sqrt {ab}$

$=(a+b+2\sqrt{ab}-a-b-\sqrt{ab})]\sqrt{ab}$

$=\sqrt{ab}\times \sqrt{ab}$

$=ab$

Subtract:
$\dfrac{4}{5}$ from $\dfrac{1}{2}$

$\dfrac {4}{5}-\dfrac {1}{2}$

As denominator is not same

$\dfrac {4(2)-1(5)}{5\times 2}=\dfrac {8-5}{10}=\dfrac {3}{10}$

$\dfrac{1}{2}+\dfrac{3}{4}+\dfrac{5}{3}$ .

1304526_1581651_ans_265ddc8c96384c6a832deb01d187e99a.jpg

Represent the following mixed infinite decimal periodic fractions as common fractions:
$\displaystyle\, 2(a + b)^{-1} (ab)^{1/2}\left ( 1 + \frac{1}{4}\left ( \sqrt{\frac{a}{b}} - \sqrt{\frac{b}{a}} \right )^2 \right )^{1/2}$

$2(a+b)^{-1}(ab)^{\frac{1}{2}}(1+\cfrac{1}{4}(\sqrt{\cfrac{a}{b}}-\sqrt{\cfrac{b}{a}})^{2})^{\cfrac{1}{2}}$

$=2(a+b)^{-1}\sqrt{ab}\cfrac{(4ab+(a-b)^{2})^{\frac{1}{2}}}{(4ab)^{\frac{1}{2}}}$

$=2(a+b)^{-1}\sqrt{ab}\cfrac{((a+b)^{2})^{\frac{1}{2}}}{2\sqrt {ab}}$

$=(a+b)^{-1}(a+b)=1$

Can a mixed fraction have the form $2 \dfrac{5}{2}$ ?

$2\cfrac{5}{2}$ , 5 in the numerator is greater than the denominator 2. So, it is not the form of mixed fraction.

$4\cfrac{1}{2}$ is the actual mixed fraction.

Compare the fractions.
$\dfrac{1}{3}\square \dfrac{7}{8}$ .

$\frac{1}{3},\frac{7}{8}$

taking LCM

$\Rightarrow \frac{1}{3}\times \frac{8}{8} , \frac{7}{8}\times \frac{3}{3}$

$\frac{8}{24},\frac{21}{24} \Rightarrow \frac{8}{24},\frac{21}{24}$

Now , we can compare & find out that

$\frac{1}{3}< \frac{7}{8}$

1179290_1304895_ans_e71e75fe58004a30af5809531e09048e.jpg

Compare the fractions.
$\dfrac{1}{6}\square \dfrac{3}{4}$ .

$\frac{1}{6},\frac{3}{4}$

taking LCM

$\Rightarrow$

$\frac{1}{6},\frac{3}{4}$

$\frac{2,9}{12}$ LCM

$= 2\times 2\times 3 =12$

clearly,

$\frac{2}{12}< \frac{9}{12}$ So

$\frac{1}{6}< \frac{3}{4}$

1179291_1304905_ans_281a24b09a404244a932c1959dda2fc3.jpg

Find the difference of $\frac{7}{12}$ and $-\frac{19}{30}$

1323323_1588494_ans_634665f5c4a24418918eefc5c7da0602.jpg

Simplify:
$\displaystyle \left(\frac{13}{9} \times \frac{-15}{2}\right) + \left(\frac{7}{3} \times \frac{8}{5}\right) + \left(\frac{3}{5} \times \frac{1}{2}\right)$

2039852_1676882_ans_e83dbe3ed52346b583596fbeb8ed67b8.JPG

Simplify:
$\displaystyle \frac{-2}{5} - \frac{-3}{10} - \frac{-4}{7}$

2039847_1676781_ans_acfb56be542b4eabb9a022a7da59ab9e.JPG

Simplify:
$\displaystyle \left(\frac{3}{11} \times \frac{5}{6}\right) - \left(\frac{9}{12} \times \frac{4}{3}\right) + \left(\frac{5}{13} \times \frac{6}{15}\right)$

2039853_1676887_ans_d9bbce7dcaf74609b70f6e9da9c65f3c.JPG

$\dfrac{6}{3}-\left(\dfrac{7}{15}\right)$ .

1311711_1586982_ans_d1389500d2ea42a59cf9c57051f0d2b2.jpg

Name	Seema	Nancy	Megha	Soni
Distance covered (km)	$\dfrac{1}{25}$	$\dfrac{1}{32}$	$\dfrac{1}{40}$	$\dfrac{1}{20}$

Fractions - Class 6 Maths - Extra Questions

Represent −56\dfrac{-5}{6} on the number line.

Convert the fraction in its simplest form.108:100108 : 100 is 2725\displaystyle \frac{27}{25}if true then enter 11 and if false then enter 00.

The simplest form of the ratio 25:10025:100 is ____

Reduce to standard form 36−24 \displaystyle \frac{36}{-24}

Satypal walks 23 \displaystyle \frac{2}{3} km from a place P towards east and then there 157 \displaystyle 1\frac{5}{7} km towards west Where will he be now from P ?

Verify the following(54+−12)+−32=54+(−12+−32)\left( \dfrac{5}{4} + \dfrac{-1}{2} \right) + \dfrac{-3}{2} = \dfrac{5}{4} + \left( \dfrac{-1}{2} + \dfrac{-3}{2} \right)

Represent 53\dfrac {5}{3} and −53-\dfrac {5}{3} on the number line

Simplify 25+37+(−65)+(−137)\dfrac{2}{5} + \dfrac{3}{7} + \left(\dfrac{-6}{5} \right ) + \left( \dfrac{-13}{7}\right )

Represent the given number on the number line: 97\dfrac{9}{7}

If xb+c−a=yc+a−b=xa+b−c\displaystyle\frac{x}{b+c-a}=\displaystyle\frac{y}{c+a-b}=\frac{x}{a+b-c}, then rove that x+y+za+b+c=x(y+z)+y(z+x)+z(x+y)2(ax+by+cz)\displaystyle\frac{x+y+z}{a+b+c}=\displaystyle\frac{x(y+z)+y(z+x)+z(x+y)}{2(ax+by+cz)}.

Find the factor which will rationalise:3√3−√2\sqrt [ 3 ]{ 3 } -\sqrt { 2 }

Find the factor which will rationalise:3√3−1\sqrt [ 3 ]{ 3 } -1

Find two proper fractions in their lowest terms having 1212 and 88 for their denominators and such that their difference is 124\dfrac{1}{24}.

Find the factor which will rationalise:6√5+3√2\sqrt [ 6 ]{ 5 } +\sqrt [ 3 ]{ 2 }

Find two fractions having 77 and 99 for their denominators, and such that their sum is 110631\dfrac{10}{63}.

Convert the following mixed fraction into improper fraction.1029\displaystyle 10\frac{2}{9}.

Convert the following mixed fraction into improper fraction.127\displaystyle 1\frac{2}{7}.

Convert the following mixed fraction into improper fraction.879\displaystyle 8\frac{7}{9}.

Convert the following improper fraction into mixed fraction.94\displaystyle\frac{9}{4}.

Convert the following improper fraction into mixed fraction.274\displaystyle\frac{27}{4}.

Solve the following:58+16\cfrac { 5 }{ 8 } +\cfrac { 1 }{ 6 } = ab\dfrac ab . Find a + b

Fill in the blank, with appropriate answer by doing addition and subtraction.

Which rational number should be subtracted from −56\dfrac{-5}{6} to get 49\dfrac{4}{9} ?

Using suitable rearrangement and find the sum, 47+(−49)+37+(−139){4 \over 7} + \left( {{{ - 4} \over 9}} \right) + {3 \over 7} + \left( {{{ - 13} \over 9}} \right)

Solve(12)3+(13)3−(56)3{\left( {{1 \over 2}} \right)^3} + {\left( {{1 \over 3}} \right)^3} - {\left( {{5 \over 6}} \right)^3}

If 95\dfrac{9}{5} + 116\dfrac{11}{6} is m30\dfrac{m}{30} then m equals to:

Add the following rational numbers:12+45\dfrac{1}{2}+\dfrac{4}{5}

Add35and23{3 \over 5}{\rm{ and }}{2 \over 3}

What should be subtracted from −2-2 to get 38\frac{3}{8} ?

Add2and−112{\rm{2 and - 1}}{1 \over 2}

If [−25+15]−23=−m15\left[ \frac{-2}{5} +\frac{1}{5} \right] - \frac{2}{3}=\dfrac{-m}{15}.Find mm

Find5314−201196{{53} \over {14}} - {{201} \over {196}}

Rationalise the denominator of 5√3−√5\dfrac{5}{\sqrt{3}-\sqrt{5}}

What should be added to −2-2 to get 38\frac{3}{8} ?

Evaluate:(i)37+−49−−117−79(i) \dfrac{3}{7} + \dfrac{-4}{9} - \dfrac{-11}{7} - \dfrac{7}{9} (ii)23+−45−13−25(ii) \dfrac{2}{3} + \dfrac{-4}{5} - \dfrac{1}{3} - \dfrac{2}{5} (iii)47−−89−−137+179(iii) \dfrac{4}{7} -\dfrac{-8}{9} - \dfrac{-13}{7} + \dfrac{17}{9}

Convert it to improper fraction 3343\frac{3}{4}

solve (2−35)\left(2 - \dfrac{3}{ 5}\right)

Show the fraction 25,35,45\dfrac{2}{5},\dfrac{3}{5},\dfrac{4}{5} and 55\dfrac{5}{5} on a number line.

Add: 512+38\dfrac{5}{12} +\dfrac{3}{8}

Write the additive inverse of each of 28\dfrac{2}{8} and −59\dfrac{-5}{9}

Express 177\dfrac {17}{7} as mixed fraction.

Find (i) 5−314=5-3\dfrac {1}{4}=...............

Convert it to mixed fraction 345\frac{34}{5}

Solve :21+7821+\dfrac {7}{8}

What is : 320−58\dfrac{3}{20}-\dfrac{5}{8} ?

7−312=7-3\dfrac{1}{2}=_______

Solve :36+12536+\dfrac {12}{5}

Solve :9+329+\dfrac {3}{2}

Solve: 35+27\dfrac{3}{5} + \dfrac{2}{7}

Add −25\dfrac{-2}{5} and its multiplicative inverse.

Solve 8x6+338=12548\dfrac{x}{6} + 3\dfrac{3}{8} = 12\dfrac{5}{4}

Solve :15+5715+\dfrac {5}{7}

If 1617+217=m17\dfrac {16}{17}+\dfrac {2}{17}=\dfrac{m}{17}.Find mm

Add the following47\dfrac{4}{7} and 57\dfrac{5}{7}

Solve2−352-\dfrac{3}{5}

Simplify: 178×3151\dfrac{7}{8}\times 3\dfrac{1}{5}

Write the following rational number in decimal form 2323⋅52\dfrac {23}{2^{3}\cdot 5^{2}}

The sum of two rational numbers is 421\dfrac { 4 }{ 21}. If one of them is 57\dfrac { 5 }{ 7 }, find the other.

Express each of the following as a fraction in simplest form:6623%66\dfrac { 2 }{ 3 }\%

Simplify 2316\dfrac {23}{16}

Simplify 12+52+310\dfrac{1}{2} + \dfrac{5}{2} + \dfrac{3}{{10}}

Solve 212−1452\dfrac{1}{2} - 1\dfrac{4}{5}

Write the additive inverse of 439\dfrac{{43}}{9}\,

Soham poured some water in two fish bowls. He poured 4584\dfrac{5}{8} liters of water in the first dish bowl and 2562\dfrac{5}{6} liters in the second bowl. How much water Soham pour in the first bowls altogether?

Work out :498−2454\dfrac{9}{8} - 2\dfrac{4}{5}

Express each of the following as a fraction in simplest form:3312%33\dfrac { 1 }{ 2 }\%

Subtract: 312−23\dfrac { 31 }{ 2 } -\dfrac { 2 }{ 3 }

How do represent fraction on number line34\dfrac{3}{4}

Add the following fractions.(a) 712\dfrac{7}{12} and 38\dfrac{3}{8} (b) 916\dfrac{9}{16} and 712\dfrac{7}{12}.

Add.613+2136\dfrac{1}{3}+2\dfrac{1}{3}

If 57+911=11811a\dfrac{5}{7}+\dfrac{9}{11}=\dfrac{118}{11a}. Find aa

What number should be added in −78\dfrac{-7}{8} to get 49\dfrac{4}{9}

Evaluate 696+4192\dfrac{6}{{96}} + \dfrac{4}{{192}}

In a camp, there are 420420 boys and 180180 girls. Find the ratio of boys to girls:

Add 696+492\dfrac{6}{{96}} + \dfrac{4}{{92}}

What fraction is 270g270g of 3kg3 kg.

Subtract 25−5\dfrac {2}{5}-5

Simplify:856−338+172:\,8\dfrac{5}{6} - 3\dfrac{3}{8} + 1\dfrac{7}{2}.

Simplify: 223+312 2\dfrac{2}{3} +3\dfrac{1}{2}

Represent $\dfrac{-5}{6}$ on the number line.

Convert the fraction in its simplest form.
$108 : 100$ is $\displaystyle \frac{27}{25}$
if true then enter $1$ and if false then enter $0$ .

The simplest form of the ratio $25:100$ is ____

Reduce to standard form $\displaystyle \frac{36}{-24}$

Satypal walks $\displaystyle \frac{2}{3}$ km from a place P towards east and then there $\displaystyle 1\frac{5}{7}$ km towards west Where will he be now from P ?

Verify the following
$\left( \dfrac{5}{4} + \dfrac{-1}{2} \right) + \dfrac{-3}{2} = \dfrac{5}{4} + \left( \dfrac{-1}{2} + \dfrac{-3}{2} \right)$

Represent $\dfrac {5}{3}$ and $-\dfrac {5}{3}$ on the number line

Simplify $\dfrac{2}{5} + \dfrac{3}{7} + \left(\dfrac{-6}{5} \right ) + \left( \dfrac{-13}{7}\right )$

$\dfrac{9}{7}$

If $\displaystyle\frac{x}{b+c-a}=\displaystyle\frac{y}{c+a-b}=\frac{x}{a+b-c}$ , then rove that $\displaystyle\frac{x+y+z}{a+b+c}=\displaystyle\frac{x(y+z)+y(z+x)+z(x+y)}{2(ax+by+cz)}$ .

Find the factor which will rationalise:
$\sqrt [ 3 ]{ 3 } -\sqrt { 2 }$

Find the factor which will rationalise:
$\sqrt [ 3 ]{ 3 } -1$

Find two proper fractions in their lowest terms having $12$ and $8$ for their denominators and such that their difference is $\dfrac{1}{24}$ .

Find the factor which will rationalise:
$\sqrt [ 6 ]{ 5 } +\sqrt [ 3 ]{ 2 }$

Find two fractions having $7$ and $9$ for their denominators, and such that their sum is $1\dfrac{10}{63}$ .

Convert the following mixed fraction into improper fraction.
$\displaystyle 10\frac{2}{9}$ .

Convert the following mixed fraction into improper fraction.
$\displaystyle 1\frac{2}{7}$ .

Convert the following mixed fraction into improper fraction.
$\displaystyle 8\frac{7}{9}$ .

Convert the following improper fraction into mixed fraction.
$\displaystyle\frac{9}{4}$ .

Convert the following improper fraction into mixed fraction.
$\displaystyle\frac{27}{4}$ .

Solve the following:
$\cfrac { 5 }{ 8 } +\cfrac { 1 }{ 6 }$ = $\dfrac ab$ . Find a + b

Which rational number should be subtracted from $\dfrac{-5}{6}$ to get $\dfrac{4}{9}$ ?

Using suitable rearrangement and find the sum,
${4 \over 7} + \left( {{{ - 4} \over 9}} \right) + {3 \over 7} + \left( {{{ - 13} \over 9}} \right)$

Solve
${\left( {{1 \over 2}} \right)^3} + {\left( {{1 \over 3}} \right)^3} - {\left( {{5 \over 6}} \right)^3}$

If $\dfrac{9}{5}$ + $\dfrac{11}{6}$ is $\dfrac{m}{30}$ then m equals to:

Add the following rational numbers:
$\dfrac{1}{2}+\dfrac{4}{5}$

Add
${3 \over 5}{\rm{ and }}{2 \over 3}$

What should be subtracted from $-2$ to get $\frac{3}{8}$ ?

Add
${\rm{2 and - 1}}{1 \over 2}$

If $\left[ \frac{-2}{5} +\frac{1}{5} \right] - \frac{2}{3}=\dfrac{-m}{15}$ .Find $m$

Find
${{53} \over {14}} - {{201} \over {196}}$

Rationalise the denominator of $\dfrac{5}{\sqrt{3}-\sqrt{5}}$

What should be added to $-2$ to get $\frac{3}{8}$ ?

Evaluate:
$(i) \dfrac{3}{7} + \dfrac{-4}{9} - \dfrac{-11}{7} - \dfrac{7}{9}$

$(ii) \dfrac{2}{3} + \dfrac{-4}{5} - \dfrac{1}{3} - \dfrac{2}{5}$

$(iii) \dfrac{4}{7} -\dfrac{-8}{9} - \dfrac{-13}{7} + \dfrac{17}{9}$

Convert it to improper fraction $3\frac{3}{4}$

solve $\left(2 - \dfrac{3}{ 5}\right)$

Show the fraction $\dfrac{2}{5},\dfrac{3}{5},\dfrac{4}{5}$ and $\dfrac{5}{5}$ on a number line.

Add: $\dfrac{5}{12} +\dfrac{3}{8}$

Write the additive inverse of each of $\dfrac{2}{8}$ and $\dfrac{-5}{9}$

Express $\dfrac {17}{7}$ as mixed fraction.

Find (i) $5-3\dfrac {1}{4}=$ ...............

Convert it to mixed fraction $\frac{34}{5}$

$21+\dfrac {7}{8}$

What is : $\dfrac{3}{20}-\dfrac{5}{8}$ ?

$7-3\dfrac{1}{2}=$ _______

$36+\dfrac {12}{5}$

Solve : $9+\dfrac {3}{2}$

Solve: $\dfrac{3}{5} + \dfrac{2}{7}$

Add $\dfrac{-2}{5}$ and its multiplicative inverse.

Solve $8\dfrac{x}{6} + 3\dfrac{3}{8} = 12\dfrac{5}{4}$

$15+\dfrac {5}{7}$

If $\dfrac {16}{17}+\dfrac {2}{17}=\dfrac{m}{17}$ .Find $m$

Add the following
$\dfrac{4}{7}$ and $\dfrac{5}{7}$

Solve
$2-\dfrac{3}{5}$

Simplify: $1\dfrac{7}{8}\times 3\dfrac{1}{5}$

Write the following rational number in decimal form
$\dfrac {23}{2^{3}\cdot 5^{2}}$

The sum of two rational numbers is $\dfrac { 4 }{ 21}$ . If one of them is $\dfrac { 5 }{ 7 }$ , find the other.

Express each of the following as a fraction in simplest form:
$66\dfrac { 2 }{ 3 }\%$

Simplify $\dfrac {23}{16}$

Simplify $\dfrac{1}{2} + \dfrac{5}{2} + \dfrac{3}{{10}}$

Solve $2\dfrac{1}{2} - 1\dfrac{4}{5}$

Write the additive inverse of $\dfrac{{43}}{9}\,$

Soham poured some water in two fish bowls. He poured $4\dfrac{5}{8}$ liters of water in the first dish bowl and $2\dfrac{5}{6}$ liters in the second bowl. How much water Soham pour in the first bowls altogether?

Work out : $4\dfrac{9}{8} - 2\dfrac{4}{5}$

Express each of the following as a fraction in simplest form:
$33\dfrac { 1 }{ 2 }\%$

Subtract: $\dfrac { 31 }{ 2 } -\dfrac { 2 }{ 3 }$

How do represent fraction on number line
$\dfrac{3}{4}$

Add the following fractions.
(a) $\dfrac{7}{12}$ and $\dfrac{3}{8}$ (b) $\dfrac{9}{16}$ and $\dfrac{7}{12}$ .

Add.
$6\dfrac{1}{3}+2\dfrac{1}{3}$

If $\dfrac{5}{7}+\dfrac{9}{11}=\dfrac{118}{11a}$ . Find $a$

What number should be added in $\dfrac{-7}{8}$ to get $\dfrac{4}{9}$

Evaluate $\dfrac{6}{{96}} + \dfrac{4}{{192}}$

In a camp, there are $420$ boys and $180$ girls. Find the ratio of boys to girls:

Add $\dfrac{6}{{96}} + \dfrac{4}{{92}}$

What fraction is $270g$ of $3 kg$ .

Subtract $\dfrac {2}{5}-5$

Simplify $:\,8\dfrac{5}{6} - 3\dfrac{3}{8} + 1\dfrac{7}{2}$ .

Simplify: $2\dfrac{2}{3} +3\dfrac{1}{2}$

Mother bought two chickens. One weight $1\dfrac{3}{4}\ kg$ and the other $1\dfrac{1}{2}\ kg$ . How much did they both weigh together?

How much less is $28$ km than $42.6$ km?