Fractions - Class 6 Maths - Extra Questions

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

Given ratio is $$ 25 : 100 $$

Now

 $$ 25 : 100 $$

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

$$ = 1 : 4 $$ 

Hence, the simplest form is $$ 1:4$$

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

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.

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

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

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

$$=\dfrac{39}{368}$$

Given that

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

  $$ {{x}^{2}}=\dfrac{16}{9} $$

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

 $$ x=\pm \dfrac{4}{3} $$

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

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

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

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.

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

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

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

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

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

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

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 .

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

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

we have,

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

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

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

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

 $$ \Rightarrow \dfrac{349}{63} $$

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 $$1$$.

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 $$\dfrac{49}{15}$$.

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.

We have,

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

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

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.

Consider the following Equation.

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

 $$ =\dfrac{1}{9} $$

Hence, this is the correct answer.

Consider the following Equation.

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

 $$=1$$

Hence, this is the correct answer.

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.

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.

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.

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

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

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

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.

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

Taking $$LCM$$
$$2 |18,9$$
$$3 |9,9$$
$$3| 3,3$$
   $$|1,1$$

$$ ∴LCM of 9 and 18=2\times3\times3=18 $$

$$ \dfrac{-5}{18} - \dfrac{-2}{9} = \dfrac{-5 \times 1 }{18 \times 1} - \dfrac{\left ( -2 \times 2 \right )}{9 \times 2} $$

$$ = \dfrac{-5 + 4}{18}$$

 $$= \dfrac{-1}{18}$$ 

Taking $$LCM$$
$$2 |22,33$$
$$3 |11,33$$
$$11 |1,11$$
    $$|1,1$$

$$ ∴LCM of 22 and 33=2 \times3 \times 11 = 66 $$

$$ \dfrac{5}{33} - \left ( \dfrac{-9}{22} \right ) = \dfrac{5 \times 2}{33 \times 2} + \dfrac{9 \times 3}{22 \times 3} $$

$$ = \dfrac{10 + 27 }{66} = \dfrac{37}{66}$$

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)

                  $$ =2\times \left[ \left( 12\dfrac{1}{2} \right)+\left( 10\dfrac{2}{3} \right) \right] $$

                  $$ =2\times \left[ \left( 12+\dfrac{1}{2} \right)+\left( 10+\dfrac{2}{3} \right) \right] $$

                  $$ =2\times \left[ \left( 12+10 \right)+\left( \dfrac{1}{2}+\dfrac{2}{3} \right) \right] $$

                  $$ =2\times \left[ 22+\left( \dfrac{3+4}{6} \right) \right] $$

                  $$ =2\times \left[ 22+\dfrac{7}{6} \right] $$

                  $$ =44+\dfrac{7}{3} $$

                  $$ =44+2\dfrac{1}{3} $$

                  $$ =46\dfrac{1}{3}\ cm $$

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 $$15\ m$$.

Simplifying $$\dfrac{5}{6} + \left( {\dfrac{{ - 3}}{{21}}} \right)$$.

$$\dfrac{5}{6} + \left( {\dfrac{{ - 3}}{{21}}} \right)$$

$$ = \dfrac{5}{6} - \dfrac{1}{7}$$

$$ = \dfrac{{5\left( 7 \right) - 1\left( 6 \right)}}{{42}}$$

$$ = \dfrac{{35 - 6}}{{42}}$$

$$ = \dfrac{{29}}{{42}}$$

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 $$2\cfrac{2}{3}$$.

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,

$$x=y-1\,\,\,\,........\,\,\,\,\left( 2 \right)$$

$$x=3-1=2$$

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.

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 $$\dfrac{40}{60}hr\,=\dfrac{2}{3}\,hr.$$

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.

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.