Class 7 Maths - Rational Numbers

Does the following pairs represent the same rational number:
$\dfrac {-16}{20}$ and $\dfrac {20}{-25}$

$\cfrac{-16}{20} = \cfrac{-4 \times 4}{5 \times 4} = - \cfrac{4}{5}$

$\cfrac{20}{-25} = \cfrac{4 \times 5}{-5 \times 5} = - \cfrac{4}{5}$

Since both numbers are

$-\cfrac{4}{5}$ , thus they are same.

Does the following pairs represent the same rational number:
$\dfrac {-2}{-3}$ and $\dfrac {2}{3}$

$\cfrac{-2}{-3} = \cfrac{-1 \times 2}{-1 \times 3} = \cfrac{2}{3}$

$\cfrac{2}{3} = \cfrac{1 \times 2}{1 \times 3} = \cfrac{2}{3}$

Since both numbers are

$\cfrac{2}{3}$ , thus they are same.

Which of the following pairs represent the same rational numbers?
$\dfrac {8}{-5}$ and $\dfrac {-24}{15}$

$\cfrac{8}{-5} = \cfrac{1 \times 8}{-1 \times 5} = - \cfrac{8}{5}$

$\cfrac{-24}{15} = \cfrac{-8 \times 3}{5 \times 3} = - \cfrac{8}{5}$

Since both numbers are

$- \cfrac{8}{5}$ , thus they are same.

Multiply:
$-\dfrac{-6}{11}$ by $\dfrac{-55}{36}.$
If the answer is $\dfrac ab$ , then the value of $(a+b)$ is

So comparing with

$\dfrac ab$ mentioned in question, the value of (a+b) becomes 1.
1508815_1671394_ans_b34e5e203b6e487fbd84190698ce0c50.jpg

Multiply:
$-\dfrac{-8}{25}$ by $\dfrac{-5}{16}$

1508816_1671395_ans_7786fed9f8a54a35a1860cade359a836.jpg

Simplify:
$\dfrac{-19}{36}\times 16$

1503665_1671430_ans_33a3af5ed4f94a1c9cc17e600729004f.jpg

Simplify:
$\dfrac{-9}{16}\times \dfrac{-64}{-27}$

1503668_1671433_ans_3ef5031149cf4a54af50c02df594c0f5.jpg

Simplify:
$\dfrac{7}{6}\times \dfrac{-3}{28}$

1503664_1671428_ans_5ddc14130e4b4ead95d670a041e5d83b.jpg

Simplify:
$\dfrac{-13}{9}\times \dfrac{27}{-26}$

1503667_1671432_ans_618686d9c4aa4b80a445e37fbca2a0d9.jpg

Simplify and express the result as a rational number in standard form, and find the value of $a$
$\dfrac{-50}{7}\times \dfrac{14}{3}=\dfrac{-a}{b}$

1503669_1671435_ans_c8abb4b643de4aa391da7e651cd01d3b.jpg

Simplify:
$\dfrac{-5}{9}\times \dfrac{72}{-5}$

$\dfrac{-5}{9}\times \dfrac{72}{-5} = 8$

Simplify:
$\left(\dfrac{13}{5}\times \dfrac{8}{3}\right)-\left(\dfrac{-5}{2}\times \dfrac{11}{3}\right)$

1503748_1671447_ans_358cd30280b5437aad3cb90c9b6cb6a1.jpg

Simplify:
$\left(\dfrac{13}{7}\times \dfrac{11}{26}\right)-\left(\dfrac{-4}{3}\times \dfrac{5}{6}\right)$

1503750_1671448_ans_3e370c06832641ce82bc42d3a8e4dee1.jpg

Simplify:
$\left(\dfrac{1}{2}\times \dfrac{1}{4}\right)+\left(\dfrac{1}{2}\times 6\right)$

1503744_1671442_ans_be5bcb48823c43c5a5306e276f679ec6.jpg

Simplify:
$\left(-5\times \dfrac{2}{15}\right)-\left(6\times \dfrac{2}{9}\right)$

1503745_1671443_ans_3f3969d147f04705a771320c8e412043.jpg

Simplify:
$\left(\dfrac{-4}{3}\times \dfrac{12}{-5}\right)+\left(\dfrac{3}{7}\times \dfrac{21}{15}\right)$

1503747_1671446_ans_e648b8ef24f14734bcc7889138071049.jpg

Simplify:
$\left(\dfrac{8}{5}\times \dfrac{-3}{2}\right)+\left(\dfrac{-3}{10}\times \dfrac{11}{16}\right)$

1503751_1671450_ans_a017812568b6449d9575570cfe869ab7.jpg

Simplify:
$\left(\dfrac{25}{8}\times \dfrac{2}{5}\right)-\left(\dfrac{3}{5}\times \dfrac{-10}{9}\right)$ .

If the answer will be $\dfrac ab$ , then the value of $a+b$ will be

So comparing with

$\dfrac ab$ mentioned in question, the value of (a+b) becomes 35.
1503743_1671441_ans_7d23bbfbba40428fa04692fee8b258d3.jpg

Simplify:
$\left(\dfrac{-9}{4}\times \dfrac{5}{3}\right)+\left(\dfrac{13}{2}\times \dfrac{5}{6}\right)$

1503746_1671445_ans_6843bcff4e68492d9edefa37fee705dc.jpg

If $\left(\dfrac{3}{2}\times \dfrac{1}{6}\right)+\left(\dfrac{5}{3}\times \dfrac{7}{2}\right)-\left(\dfrac{13}{8}\times \dfrac{4}{3}\right)=\dfrac{47}{4 \times a}$ , then what is the value of $a?$

1503752_1671451_ans_45db1cc960d54229a6e8de4cc6a355df.jpg

The product of two rational numbers is $\dfrac{-8}{9}$ . If one of the numbers is $\dfrac{-4}{15}$ then find the other.

1509015_1671602_ans_1709de9fd971432f833befa008276bf8.jpg

Fill in the blank:
The product of a rational number and its reciprocal is _____.1

We know that, if

$p$ and

$q$ are two rational numbers such that,

$p\times q=1$

Then,

$q$ is called the reciprocal of

$p$ or vive-verca.

Hence, the product of a rational number and its reciprocal is

$1$ .

Fill in the blank:
The number $0$ is ____ the reciprocal of any number.

We know that, if a rational number

$a$ is the reciprocal of a rational number

$b$

Then, we can say that

$b$ is the reciprocal of

$a$ .

Since,

$0$ does not have any reciprocal.

Therefore,

$0$ is not the reciprocal of any number.

Fill in the blank:
The reciprocal of a negative rational number is _____.

We know that, the product of a number and its reciprocal is

$1$ .

$\therefore$ The product of a

$-$ ve rational number will be

$+$ ve only when the other number is

$-$ ve.

Hence, the reciprocal of a negative rational number is

$\text{negative.}$

Fill in the blank:
Reciprocal of $\dfrac{1}{a}\ ,\ a\neq 0$ is .........

Let, the reciprocal be

$x$

Then,

$\dfrac1a\times x=1$

$\Rightarrow x=a$ [By cross multiplication]

Hence, the reciprocal of

$\dfrac1a$ is

$a$ .

Fill in the blank:
$\dfrac{-4}{5}\times \left(\dfrac{5}{7}+\dfrac{-8}{9}\right)=\left(\dfrac{-4}{5}\times ........\right)+\dfrac{-4}{5}\times \dfrac{-8}{9}$

1508912_1671563_ans_8582974ca6bd429d9ea1b6306a2384f1.jpg

Fill in the blank:
Zero has _____ reciprocal.

We know that, any rational number

$\times 0=0$

$\therefore$ No rational number

$\dfrac pq$ exists such that

$0\times \dfrac pq=1$

$\Rightarrow$ Reciprocal of zero does not exist.

Divide:
$1$ by $\dfrac{1}{2}$

1508992_1671569_ans_07751d56d0c9444d8460057e70e3551d.jpg

Fill in the blank:
The reciprocal of a positive rational number is _____.

We know that, the product of a number and its reciprocal is

$1$ .

$\therefore$ The product of a positive rational number will be positive only when the other number is positive.

Hence, the reciprocal of a positive rational number is

$\text{positive.}$

Simplify the following:
$\sqrt{24}/8 + \sqrt{54}/9$

$\sqrt{24}/8 + \sqrt{54}/9$

Let us simplify the expression,

$\sqrt{24}/8 + \sqrt{54}/9$

$= \sqrt{(4\times 6)}/8 \div \sqrt{(9 \times 6)}/9$

$= 2\sqrt{6}/8 + 3\sqrt{6}/9$

$= \sqrt 6/4 + \sqrt 6/3$

By taking LCM

$= (3\sqrt 6 + 4\sqrt 6)/12$

$= 7\sqrt 6/12$

Compare :
$-1 \dfrac{1}{2}$ and $0$

$-1 \dfrac{1}{2}$ and

$0$
P is on the right of Q
Hence ,

$\dfrac{-3}{2} , 0$ or

$-1 \dfrac{1}{2} < 0$ .

Compare :
$\dfrac{-7}{2}$ and $\dfrac{5}{2}$

$\dfrac{-7}{2}$ and

$\dfrac{5}{2}$
P is on the right of Q
Hence,

$\dfrac{-7}{2} < \dfrac{5}{2}$

Compare :
$-3$ and $2\dfrac{3}{4}$

$-3$ and

$2\dfrac{3}{4}$
P is on the right of Q
Hence ,

$-3 < \dfrac{11}{4}$ or

$-3 < 2\dfrac{3}{4}$ .

Compare :
$\dfrac{1}{4}$ and $0$

$\dfrac{1}{4}$ and

$0$

$\dfrac{1}{4}$ is a positive rational number which is always greater than

$0$ .

Hence ,

$\dfrac{1}{4} > 0$ .

Compare :
$0$ and $\dfrac{3}{4}$

$0$ and

$\dfrac{3}{4}$
P is on the right of Q
Hence,

$0 < \dfrac{3}{4}$

Write the following rational numbers in ascending order:
$\cfrac{1}{3}$ , $\cfrac{-2}{9}$ , $\cfrac{-4}{3}$

$\cfrac{1}{3}$ ,

$\cfrac{-2}{9}$ ,

$\cfrac{-4}{3}$

$\Rightarrow$

$\cfrac{3}{9},\cfrac{-2}{9},\cfrac{-12}{9}$ (Converting into same denominator)
Now

$\cfrac{-12}{9}< \cfrac{-2}{9}< \cfrac{3}{9}$

$\Rightarrow$

$\cfrac{-4}{3}< \cfrac{-2}{9}< \cfrac{1}{3}$

Which of the two rational numbers in each of the following pairs of rational numbers is greater?
$\dfrac{-4}{11}, \dfrac{3}{11}$

1517739_1676413_ans_8b643aa3d1d54a36b30991f00ee4cf64.jpg

Which of the two rational numbers in each of the following pairs of rational numbers is greater?
$\dfrac{-3}{8}, 0$

1591524_1676411_ans_6c66724e994b4363b399c201da4b7c9f.jpg

Which of the two rational numbers in each of the following pairs of rational numbers is smaller?
$\dfrac{-4}{3}, \dfrac{8}{-7}$

1517490_1676422_ans_363185b8b6c649648f522220fbe5ffe1.jpg

Which of the two rational numbers in each of the following pairs of rational numbers is smaller?
$\dfrac{16}{-5}, 3$

1591530_1676421_ans_e3df4218e95c4ebf9f5ac1514e7586fc.jpg

Which of the two rational numbers in each of the following pairs of rational numbers is smaller?
$\dfrac{-12}{5}, -3$

1517770_1676423_ans_3c8f1cf60afd40e48892967de7d09c68.png

Which of the two rational numbers in each of the following pairs of rational numbers is smaller?
$\dfrac{-6}{-13}, \dfrac{7}{13}$

1591528_1676420_ans_5cef1d8abd9e440d90a56ce7e95c9519.jpg

Which of the two rational numbers in each of the following pairs of rational numbers is greater?
$\dfrac{5}{-8}, \dfrac{-7}{12}$

1591526_1676418_ans_2ec647a7c8d248a1b930c15f178a5ad0.jpg

Add the following rational numbers:
$\dfrac{-8}{11}$ and $\dfrac{-4}{11}$

1515239_1676635_ans_4bfbfe21d9024999953c106e5c0cefeb.jpg

Add the following rational numbers:
$\dfrac{-5}{7}$ and $\dfrac{3}{7}$

1518011_1676626_ans_fbc7897c7d4e44ee9593aa785d07d67b.jpg

Add the following rational numbers:
$\dfrac{6}{13}$ and $\dfrac{-9}{13}$

1516281_1676637_ans_d40963d4207b494dac6979cf36a2dd61.jpg

Add the following rational numbers:
$\dfrac{-7}{27}$ and $\dfrac{11}{18}$

1516254_1676646_ans_17fb4119dcd742f1930525454cb733ad.jpg

Subtract the first rational number from the second in each of the following:
$\dfrac{3}{8}, \dfrac{5}{8}$

1517398_1676682_ans_5b5ae867a7cc4dbc97411541b54ff5e1.jpg

Fill in the blanks:
$\dfrac {-16}{7} + ... = ... + \dfrac {-16}{7} = \dfrac {-16}{7}$ .

Addition of

$0$ to any fraction results in the same answer
So hereby we are adding

$0$ to the above equation and also by using commutative property
i.e.

$a + b = b + a$

$\therefore \dfrac {-16}{7} + 0 = 0 + \dfrac {-16}{7} = \dfrac {-16}{7}$ .

Which of the two rational numbers is greater in each of the following pairs?
$\cfrac79$ or $\cfrac{-5}9$

Since both denominators are same therefore compare the numerators only.

So, we have

$7 > -5$

$\therefore \cfrac79>\cfrac{-5}9$

Simplify:
$\dfrac{6}{5}\times\dfrac{3}{7}-\dfrac{1}{5}\times\dfrac{3}{7}$

$\dfrac{6}{5}\times\dfrac{3}{7}-\dfrac{1}{5}\times\dfrac{3}{7}$

$=\dfrac{6}{5}\times\dfrac{3}{7}-\dfrac{1}{5}\times\dfrac{3}{7}$

$=\dfrac{18}{35}-\dfrac{3}{35}$

$=\dfrac{18-3}{35}$

$=\dfrac{15}{35}$

$=\dfrac{3}{7}$

Solve
$\dfrac{5}{13}-\dfrac{-8}{26}$

$\dfrac{5}{13}-\dfrac{-8}{26}$

$=\dfrac{5}{13}-\left(\dfrac{-8}{26}\right)$

$=\dfrac{(5\times 2)-(-8\times 1)}{26}$

$=\dfrac{10+8}{26}$

$=\dfrac{18}{26}=\dfrac{9}{13}$

Multiply:
$-24$ and $\dfrac{5}{16}$

$-24$ and

$\dfrac{5}{16}$
It can be written as

$= \dfrac{-24 \times 5}{16}$
By further calculation

$= \dfrac{-3 \times 5}{2}$
So we get

$= \dfrac{-15}{2}$

Fill in the following blanks:
The number ___ and _ are their own reciprocals.

The numbers

$1$ and

$-1$ are their own reciprocals.

Simplify:
$1+\dfrac{-4}{5}$

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

Simplify:
$0+\dfrac{-3}{5}$

$0 + \dfrac{-3}{5} = \dfrac{0}{1} + \dfrac{(-3)}{5} = \dfrac{5\times 0 + (-3)}{5} = \dfrac{0-3}{5} = \dfrac{-3}{5}$

Simplify:
$\dfrac{-16}{21}\times \dfrac{14}{5}$

1503657_1671427_ans_dc93717479184284a2e4a37ad0166581.jpg

Multiply:
$-\dfrac{8}{-9}$ by $\dfrac{-7}{-16}$

1503677_1671403_ans_60a2efca64c44c83bb6c6d30f559fd3b.jpg

Multiply:
$-\dfrac{6}{7}$ by $\dfrac{-49}{36}$

1503679_1671398_ans_b0ad3c9c880845a48686b19f3b694fa9.jpg

Multiply:
$-\dfrac{-8}{9}$ by $\dfrac{3}{64}$

1503655_1671404_ans_2086aad40d7144dbabc1e42dafeb29df.jpg

solve
$2\frac{1}{3} + 1\frac{1}{4} + 2\frac{5}{6} + 3\frac{7}{{12}}$

On solving fraction, we get

$=\dfrac{7}{3}+\dfrac{5}{4}+\dfrac{17}{6}+\dfrac{43}{12}$

$=\dfrac{120}{12}$

$=10$

Find the rational number that should be subtracted from $\dfrac{-5}{7}$ to get $-1$ ?

1351279_1251235_ans_35aff5eeb8d34780892b1d02042450f7.PNG

Subtract :
$0$ from $\dfrac{-3}{8}$

$0$ from

$\dfrac{-3}{8}$
It can be written as

$= \dfrac{-3}{8} - 0$

$= \dfrac{-3}{8}$

Which of the two rational numbers in each of the following pairs of rational numbers is greater?
$\dfrac{5}{2}, 0$

1517738_1676412_ans_4e7d7a3d69514cc785e24adc74d6eb3a.jpg

Which of the two rational numbers in each of the following pairs of rational numbers is greater?
$\dfrac{5}{9}, \dfrac{-3}{-8}$

1591525_1676417_ans_d9c9cddbc5ff42418a0a5684af06a161.jpg

Which of the two rational numbers in each of the following pairs of rational numbers is greater?
$\dfrac{-5}{8}, \dfrac{3}{-4}$

1517742_1676416_ans_b97bc558a03c449a941c4f000bec143c.jpg

Which of the two rational numbers in each of the following pairs of rational numbers is greater?
$\dfrac{-7}{12}, \dfrac{5}{-12}$

1517740_1676414_ans_7258755080eb430bad74d1bd4add9359.jpg

Which of the two rational numbers in each of the following pairs of rational numbers is greater?
$\dfrac{4}{-9}, \dfrac{-3}{-7}$

1517741_1676415_ans_586fca57a0154a2cbe996e9df83eef0f.jpg

Fill in the blanks:
$-9 + \dfrac {-21}{8} = (...) + (-9)$ .

As we all know commutative property states,

$a + b = b + a$
We can simply solve by using the property

$\therefore -9 + \dfrac {-21}{8} = \left (\dfrac {-21}{8}\right ) + (-9)$ .

Give four rational numbers equivalent to $\dfrac {4}{9}$

The given fraction is

$\dfrac{4}{9}$

Now,

$\cfrac{4}{9}=\cfrac{4}{9}\times\cfrac{2}{2}=\cfrac{8}{18}$

$\cfrac{4}{9}=\cfrac{4}{9}\times\cfrac{3}{3}=\cfrac{12}{27}$

$\cfrac{4}{9}= \cfrac{4}{9}\times\cfrac{4}{4}=\cfrac{16}{36}$

$\cfrac{4}{9}=\cfrac{4}{9}\times\cfrac{5}{5}=\cfrac{20}{45}$

Hence, four equivalent rational numbers are

$\cfrac{8}{18},\cfrac{12}{27},\cfrac{16}{36}$ and

$\cfrac{20}{45}$ .

Arrange $\displaystyle \frac{5}{8}, \frac{3}{16}, -\frac{1}{4}$ and $\displaystyle \frac{17}{32}$ in descending order of their magnitudes.Identify the largest rational number.

Consider the given numbers

$\dfrac {5}{8},\dfrac {3}{16},-\dfrac {1}{4}$ and

$\dfrac {17}{32}$ .

The LCM of the denominators

$8,16,4,32$ is

$32$ . Therefore, the given numbers will be:

$1]\dfrac { 5\times 4 }{ 8\times 4 } =\dfrac { 20 }{ 32 }$

$2]\dfrac { 3\times 2 }{ 16\times 2 } =\dfrac { 6 }{ 32 }$

$3]-\dfrac { 1\times 8 }{ 4\times 8 } =-\dfrac { 8 }{ 32 }$

$4]\dfrac {17}{32}$

Thus, the numbers in descending order are shown below:

$\dfrac { 20 }{ 32 } ,\dfrac { 17 }{ 32 } ,\dfrac { 6 }{ 32 } ,-\dfrac { 8 }{ 32 } \\ \Rightarrow \dfrac { 5 }{ 8 } ,\dfrac { 17 }{ 32 } ,\dfrac { 3 }{ 16 } ,-\dfrac { 1 }{ 4 }$

Thus the largest rational number is

$\dfrac{5}{8}$ .

Multiply $\dfrac{6}{13}$ by the reciprocal of $\dfrac{-7}{16}$ .

Reciprocal of

$\dfrac{-7}{16}$ is

$\dfrac{16}{-7}$

Now according to the question,

$\dfrac6{13}\times\dfrac{16}{-7} = \dfrac{-96}{91}$

Does the following pair represents the same rational number:
$\dfrac {-7}{21}$ and $\dfrac {3}{9}$

The given rational numbers are

$\dfrac {-7}{21}$ and

$\dfrac {3}{9}$

Now,

$\cfrac{-7}{21} = \cfrac{-7 \div 7}{21 \div 7} = - \cfrac{1}{3}$

and,

$\cfrac{3}{9} = \cfrac{3 \div 3}{9 \div 3} = \cfrac{1}{3}$

Since, the numbers have different sign, they are not same.

The sum of two rational numbers is $\dfrac{-4}{5}$ .If one of them is $\dfrac{-11}{20}$ , find the other

$\because\,\dfrac{-11}{20}+$ a rational number

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

$\therefore$ the required rational number

$=\dfrac{-4}{5}-\left(\dfrac{-11}{20}\right)$

$=\dfrac{-4}{5}+\dfrac{11}{20}$

$=\dfrac{-4\times 4}{5\times 4}+\dfrac{11}{20}$

$=\dfrac{-16}{20}+\dfrac{11}{20}$

$=\dfrac{-16+11}{20}=\dfrac{-5}{20}=\dfrac{-1}{4}$

Thus, the other rational number

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

Rewrite the following rational numbers in the simplest form:
$\cfrac{-8}{6}$

$\cfrac{-8}{6}=\cfrac{-8\div 2}{6\div 2}=\cfrac{-4}{3}$ (H.C.F of

$8$ and

$6$ is

$2$ )

Rewrite the following rational numbers in the simplest form:
$\cfrac{-44}{72}$

$\cfrac{-44}{72}=\cfrac{-44\div 4}{72\div 4}=\cfrac{-11}{18}$ (H.C.F of

$44$ and

$72$ is

$4$ )

Rewrite the following rational number in the simplest form:
$\cfrac{25}{45}$

Given:

$\cfrac{25}{45}$

Dividing the numerator and denominator by

$5$ we get,

$=\cfrac{25\div 5}{45\div 5}=\cfrac{5}{9}$ (

$\because$ H.C.F of

$25$ and

$45$ is

$5$ )

Simplify:
$\dfrac{-13}{8}+\dfrac{5}{36}$

$\dfrac{-13}{8} + \dfrac{5}{36} = \dfrac{-13\times 9+5\times 2}{72} = \dfrac{-117+10}{72}=\dfrac{-107}{72}$

Add the following rational numbers:
$\dfrac{6}{13}$ and $\dfrac{-9}{13}$ .

$\dfrac{6}{13}$ +

$\dfrac{-9}{13} = \dfrac{6 + (-9)}{13} = \dfrac{6-9}{13} = \dfrac{-3}{13}$

Simplify:
$\dfrac{-16}{9}+\dfrac{-5}{12}=\dfrac{-39.5 \times a}{36}$

What is the value of $'a'?$

$\dfrac{-16}{9} + \dfrac{-5}{12} = \dfrac{-16\times 4 + (-5)\times 3}{36} = \dfrac{-64-15}{36} = \dfrac{-79}{36}=\dfrac{-39.5 \times 2} {36}$

So, value of

$a=2$

Add the following rational numbers:
$\dfrac{5}{-9}$ and $\dfrac{7}{3}$

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

$\dfrac{7}{3} = \dfrac{-5}{9} + \dfrac{7}{3} = \dfrac{(-5)+7\times 3}{9} = \dfrac{-5+21}{9} = \dfrac{-5+21}{9} = \dfrac{16 }{9}$

Simplify:
$\dfrac{7}{9}+\dfrac{3}{-4}$

$\dfrac{7}{9} + \dfrac{3}{-4} = \dfrac{7}{9} + \dfrac{(-3)}{4} = \dfrac{(7\times 4)+(-3)\times 9}{36} = \dfrac{28-27}{36} = \dfrac{1}{36}$

Add the following rational numbers:
$\dfrac {7}{-18}$ and $\dfrac {8}{27}$

Let us add the fractions

$\dfrac {7}{-18} + \dfrac {8}{27}$
Since the denominator is negative we need to change to positive by multiplying with

$-1$

$\dfrac {7}{-18} = \dfrac {7\times -1}{-18\times -1} = \dfrac {-7}{18}$
Since it has the difference denominator we need to take LCM
LCM of the denominators

$18$ and

$27$ is

$54$
So by multiplying the numerator and denominator by

$3$ we get,

$\dfrac {-7}{18} = \dfrac {-7\times 3}{18\times 3} = \dfrac {-21}{54}$
And by multiplying the numerator and denominator by

$2$ we get,

$\dfrac {8}{27} = \dfrac {8\times 2}{27\times 2} = \dfrac {16}{54}$
Now we can add the fractions since it has the same denominators

$\dfrac {-21}{54} + \dfrac {16}{54} = \dfrac {-21 + 16}{54} \Rightarrow \dfrac {-5}{54}$ .

Rewrite the following rational numbers in the simplest form:
$\cfrac{-8}{10}$

$\cfrac{-8}{10}=\cfrac{-8\div 2}{10\div 2}=\cfrac{-4}{5}$ (H.C.F of

$8$ and

$10$ is

$2$ )

Simplify:
$\dfrac{1}{-12}+\dfrac{2}{15}$

Given,

$\dfrac1{-12}+\dfrac2{15}$

$=-\dfrac1{12}+\dfrac2{15}$

$=-\dfrac{1\times5}{12\times5}+\dfrac{2\times4}{15\times4}\qquad[\because LCM$ of

$12$ and

$15$ is

$60]$

$=-\dfrac5{60}+\dfrac8{60}$

$=\dfrac{-5+8}{60}$

$=\dfrac3{60}=\dfrac1{20}$

Hence,

$\dfrac1{-12}+\dfrac2{15}=\dfrac1{20}$

Add the following rational numbers:
$\dfrac{3}{4}$ and $\dfrac{-5}{8}$

$\dfrac{3}{4}$ +

$\dfrac{-5}{8} = \dfrac{3}{4} + \left(\dfrac{-5}{8}\right) = \dfrac{(3\times 2)+(-5)}{8} = \dfrac{6-5}{8} = \dfrac{1}{8}$

Multiply:
$-\dfrac{-5}{17}$ by $\dfrac{51}{-60}$

1508814_1671392_ans_88ec4cbbfbcc4ef08d10af2eac561602.jpg

Multiply:
$\dfrac{-2}{9}$ by $\dfrac{5}{11}$

1508807_1671373_ans_b356e73ef3074ab0980148e684652bfa.jpg

Multiply:
$\dfrac{-3}{17}$ by $\dfrac{-5}{-4}$

1508809_1671377_ans_2f4285176ca94e0da5db42cf4f91b1a0.jpg

Multiply:
$\dfrac{7}{9}$ by $\dfrac{36}{-11}$

1508810_1671381_ans_33806da0a8d24505b2b0b0c6180c6173.jpg

Multiply:
$\dfrac{5}{7}$ by $\dfrac{-3}{4}$

The multiplication is done as follows:

$\dfrac57\times\dfrac{-3}4$

$=\dfrac{5\times-3}{7\times 4}$

$=\dfrac{-15}{28}$

Multiply:
$\dfrac{-11}{13}$ by $\dfrac{-21}{7}$

1508811_1671383_ans_af8ebb7f1cb34049bc8789fcf881ef43.jpg

Multiply:
$-\dfrac{3}{5}$ by $-\dfrac{4}{7}$

1508812_1671385_ans_c3d6019833074a08955e7dbda843b18b.jpg

Multiply:
$-\dfrac{15}{11}$ by $7$

1508813_1671387_ans_60f045461e914239b304eff68615af97.jpg

By what number should we multiply $\dfrac{-1}{6}$ so that the product may be $\dfrac{-23}{9}$ ?

Let the number be

$x$ .

Accordng to the question,

$\dfrac{-1}6\times x=\dfrac{-23}9$

$\Rightarrow \dfrac{-1\times x}6=\dfrac{-23}9$

$\Rightarrow \dfrac{-x}6=\dfrac{-23}9$

$\Rightarrow -x\times 9=-23\times 6$ [ By cross multiplication ]

$\Rightarrow x=\dfrac{-23\times6}{-9}$

$\Rightarrow x=\dfrac{138}{9}$

$\Rightarrow x=\dfrac{46}{3}$ [ Dividing numerator and denominator by

$3$ ]

Hence,

$\dfrac{46}3$ is the required number to be multilied.

Add the following rational numbers:
$0$ and $\dfrac {-2}{5}$ .

Let us add the fractions

$0 + \dfrac {-2}{5}$
By adding

$0$ to anything gives us the same answer

$\therefore 0 + \dfrac {-2}{5}\Rightarrow \dfrac {-2}{5}$ .

Add the following rational numbers:
$\dfrac {-7}{3}$ and $\dfrac {1}{3}$ .

Let us add the fractions

$\dfrac {-7}{3} + \dfrac {1}{3}$

Since it has the same denominator we can add directly,

$\dfrac {-7 + 1}{3} = \dfrac {-6}{3}$ Further we can divide with the common divisor i.e. by

$3$ ,

We get,

$\dfrac {-6}{3} \Rightarrow -2$ .

Add the following rational numbers:
$\dfrac {5}{6}$ and $\dfrac {-1}{6}$ .

Let us add the fractions

$\dfrac {5}{6} + \dfrac {-1}{6}$

Since it has the same denominator we can add directly,

$\dfrac {5 + (-1)}{6} = \dfrac {5 - 1}{6} = \dfrac {4}{6}$ further we can divide with the common divisor i.e. by

$2$ ,

We get,

$\dfrac {4}{6}\Rightarrow \dfrac {2}{3}$ .

Add the following rational numbers:
$\dfrac {-11}{8}$ and $\dfrac {5}{8}$ .

Let us add the fractions

$\dfrac {-11}{8} + \dfrac {5}{8}$
Since it has the same denominator we can add directly,

$\dfrac {-11 + 5}{8} = \dfrac {-6}{8}$ further we can divide with the common divisor i.e. by

$2$ ,
We get,

$\dfrac {-6}{8} \Rightarrow \dfrac {-3}{4}$ .

Add the following rational numbers:
$\dfrac {-6}{11}$ and $\dfrac {-4}{11}$ .

Let us add the fractions

$\dfrac {-6}{11} + \dfrac {-4}{11}$

Since it has the same denominators we can add directly,

$\dfrac {-6 + (-4)}{11}$

The multiplication of

$(+) (-)$ is

$(-). \therefore$ We get,

$\dfrac {-6 - 4}{11}\Rightarrow \dfrac {-10}{11}$ .

Add the following rational numbers:
$\dfrac {-2}{5}$ and $\dfrac {4}{5}$ .

Let us add the fractions

$\dfrac {-2}{5} + \dfrac {4}{5}$
Since it has the same denominator we can add directly,

$\dfrac {-2 + 4}{5}\Rightarrow \dfrac {2}{5}$ .

Write $\dfrac{5}{6}$ as a fraction with numerator $60$ .

If we multiply

$6$ by

$10$ it becomes

$60$

$\therefore \dfrac {5}{6} = \dfrac {5}{6} \times \dfrac {10}{10} = \dfrac { 5 \times 10}{6 \times 10} = \dfrac {50}{60}$

A reciprocal of a fraction is obtained by
inverting it upside down.

True

Fill in the boxes with the correct symbol 0 , > , < or =.
$\dfrac{3}{7} \square \dfrac{-5}{6}$

Since,

$\dfrac{-5}{6}$ is a negative rational number.

Hence,

$\dfrac{3}{7}>\dfrac{-5}{6}$

1 is the only number which is its own reciprocal.

True

Write the following number in the form $\dfrac{p}{q}$ , where p and q are integers.
Reciprocal of $1$

Reciprocal of

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

Write the following as rational number in their standard forms:
$240÷(-840)$

$240÷(-840)$

$\Rightarrow 240÷(-840)$

$\Rightarrow \dfrac{240}{-840}$

HCF is

$120.$
Now,

$=\dfrac{120÷120}{-840÷120}$

$=\dfrac{2}{-7}$

$=\dfrac{-2}{7}$

Given that $\dfrac{p}{q}$ and $\dfrac{r}{s}$ are two rational numbers with different denominators and both of them are in standard form. To compare these rational numbers we say that:
$\dfrac{\square}{\square}<\dfrac{\square}{\square},$ if $p\times s<r\times q$

$p \times s<r\times q$

By transferring sides...

$\Rightarrow \dfrac{p}{q}<\dfrac{r}{s}$

Which is greater in the following?
$-3\dfrac{5}{7},3\dfrac{1}{9}$

$-3\dfrac{5}{7},3\dfrac{1}{9}$
Here, make the denominator same of both rational numbers.

$\Rightarrow -3\dfrac{5}{7}=\dfrac{(-3\times7)+5}{7}=\dfrac{-21+5}{7}=\dfrac{-26}{7}$

$\Rightarrow 3\dfrac{1}{9}=\dfrac{(3\times9)+1}{9}=\dfrac{28}{9}$
Here, make the denominator same of both rational numbers.

$\Rightarrow \dfrac{-26\times9}{7\times9}=-\dfrac{234}{63}$

$\Rightarrow \dfrac{28\times7}{9\times7}=\dfrac{196}{63}$

$\therefore -3\dfrac{5}{7}<3\dfrac{1}{9}$
Therefore, the greater number is

$3\dfrac{1}{9}.$

Write the following as a rational number in their standard forms:

$115÷207$

$115÷207$
By using prime factorization, we get

$\Rightarrow \dfrac{115}{207}$

$\Rightarrow 115=5\times23$

$\Rightarrow 207=3\times23\times3$
HCF is

$23.$
Now,

$\Rightarrow \dfrac{115÷23}{207÷23}$

$=\dfrac{5}{9}$

Find the reciprocal of the following:
$\dfrac{3}{13}\div \dfrac{-4}{64}$

$\dfrac{3}{13}÷\dfrac{-4}{65}$

$=\dfrac{3}{13}\times\dfrac{65}{-4}$

$=\dfrac{65\times3}{13\times(-4)}$

$=\dfrac{15}{-4}$
Therefore, the reciprocal is

$\dfrac{-4}{15}.$

Taking $x=\dfrac{-4}{9},y=\dfrac{5}{12}$ and $z=\dfrac{7}{18},$ find:
the sum of reciprocals of $x$ and $y$ .

Given, $x=\dfrac{-4}{9},y=\dfrac{5}{12}$ and $z=\dfrac{7}{18},$

Reciprocals of $x$ and $y$ are $\dfrac{1}{x}$ and $\dfrac{1}{y}$

Sum of reciprocals $=\dfrac{1}{x}+\dfrac{1}{y}$

$\Rightarrow \dfrac{1}{x}+\dfrac{1}{y}$

$= \dfrac{-9}{4}+\dfrac{12}{5}$

$=\dfrac{-45+48}{20}$

$=\dfrac{3}{20}$

Find the value of the following:
$4\dfrac {5}{8} \div (-4/9)$ .

$4\dfrac {5}{8} \div (-4/9)$
This can be written as,

$= 37/8\div (-4/9)$

$= 37/ 8\times 9/-4$
We get,

$= 333/ -32$

$= \left \{333\times (-1) \right \} / \left \{-32 \times (-1)\right \}$

$= -333 / 32$

$= -10\dfrac {13}{32}$ .

Taking
$x=\dfrac{-4}{9},y=\dfrac{5}{12}$ and $z=\dfrac{7}{18},$ find:
$(xy)\times z$

Given,$$

x=\dfrac{-4}{9},y=\dfrac{5}{12}

$and$ z=\dfrac{7}{18},$$

To find:

$(x÷y)\times z=$

$=\left(\dfrac{-4}{9}÷\dfrac{5}{12}\right)\times\dfrac{7}{18}$

$=\dfrac{-4\times12\times7}{9\times5\times18}$

$=\dfrac{-56}{135}$

Taking
$x=\dfrac{-4}{9},y=\dfrac{5}{12}$ and $z=\dfrac{7}{18},$ find:
$x-(y+z)$

To find:

$x-(y+z)$

$\Rightarrow x-(y+z)$

$=\dfrac{-4}{9}-\left( \dfrac{5}{12}÷\dfrac{7}{18}\right)$

$=\dfrac{-4}{9}-\left( \dfrac{15+14}{36}\right)$

$=\dfrac{-16-29}{36}$

$=\dfrac{-45}{36}$

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

Taking
$x=\dfrac{-4}{9},y=\dfrac{5}{12}$ and $z=\dfrac{7}{18},$ find:

the reciprocal of $x+y.$

Here,

$x+y$

$\Rightarrow x+y=$

$\dfrac{-4}{9}+\dfrac{5}{12}$

$=\dfrac{-16+15}{36}$

$=\dfrac{-1}{36}$

Therefore, reciprocal of x+y

$=\dfrac{1}{\frac{-1}{36}}=-36$

What should be divided by $\dfrac{1}{2}$ to obtain the greatest negative integer?

As we know the greatest negative integer is

$-1$ .

Let

$y$ be the number.

according to the question,

$\Rightarrow \dfrac{1}{2}÷y=-1$

$\Rightarrow \dfrac{1}{2}\times\dfrac{1}{y}=-1$

$\Rightarrow y=\dfrac{-1}{2}$

Therefore,

$\dfrac{-1}{2}$ is the required number.

Taking
$x=\dfrac{-4}{9},y=\dfrac{5}{12}$ and $z=\dfrac{7}{18},$ find:
$x÷(y÷z)$

Given,

$x=\dfrac{-4}{9},y=\dfrac{5}{12}$ and

$z=\dfrac{7}{18},$

To find:

$x÷(y÷z)$

$\Rightarrow x÷(y÷z)$

$=\dfrac{-4}{9}÷\left(\dfrac{5}{12}÷\dfrac{7}{18}\right)$

$=\dfrac{-4}{9}÷\dfrac{5\times18}{12\times7}$

$=\dfrac{-4\times12\times7}{9\times5\times18}$

$=\dfrac{-336}{810}$

$=\dfrac{-56}{135}$

Taking
$x=\dfrac{-4}{9},y=\dfrac{5}{12}$ and $z=\dfrac{7}{18},$ find:
$x+(y+z)$

Given,

$x=\dfrac{-4}{9},y=\dfrac{5}{12}$ and

$z=\dfrac{7}{18},$

To find:

$x+(y+z)$

$\Rightarrow x+(y+z)$

$=\dfrac{-4}{9}+\left( \dfrac{5}{12}+\dfrac{7}{18}\right)$

$=\dfrac{-4}{9}+\dfrac{15+14}{36}$

$=\dfrac{-4\times4+29}{36}$

$=\dfrac{13}{36}$

Find.
$\dfrac{1}{3} \times \dfrac{-5}{7} \times \dfrac{-21}{10}$

We have

$\dfrac{1}{3} \times \dfrac{-5}{7} \times \dfrac{-21}{10}\\=\dfrac{1}{3} \times \dfrac{-1}{1} \times \dfrac{-3}{2}\\=\dfrac{3}{6}\\ = \dfrac{1}{2}$

Insert $3$ equivalent rational numbers between
$0$ and $-10$

Three equivalent rational numbers between

$0$ and

$-10$ are

$-2,\dfrac{-10}{5},\dfrac{-20}{10}.$

Fill in the blanks to make the statement true:
The reciprocal of $\dfrac{-5}{7}$ is .............

The reciprocal of

$\dfrac{-5}{7}$ is

$\dfrac{-7}{5}$

'a' and 'b' are two different numbers taken from the numbers 1-What is the largest value that $\dfrac{a-b}{a+b}$ can have? What is the largest value that $\left(\dfrac{a+b}{a-b}\right)$ can have?

Let

$a=50$ and

$b=1$
Given a and b are two different numbers between

$1$ and

$50.$

$\therefore \dfrac{a+b}{a-b}=\dfrac{50+49}{50-49}$

$=\dfrac{99}{1}$

Simplify :
$\dfrac {2\sqrt {3}}{3} - \dfrac {\sqrt {3}}{6}$ .

$\dfrac {2\sqrt {3}}{3} - \dfrac {\sqrt {3}}{6} = \sqrt {3} \left (\dfrac {2}{3} - \dfrac {1}{6}\right ) = \sqrt {3} \left (\dfrac {4 - 1}{6}\right ) = \sqrt {3}\times \dfrac {3}{6} = \dfrac {\sqrt {3}}{2}$ .

A body floats $\dfrac{2}{9}$ of its volume above the surface.What is the ratio of the body-submerged volume to its exposed volume? Re-write it as a rational number.

Body's Volume=

$\dfrac{2}{9}$

Volume of body submerged=

$1$ -volume of body exposed =

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

$=\dfrac{9-2}{9}=\dfrac{7}{9}$

Hence, required ratio

$\Rightarrow \dfrac{7}{9}:\dfrac{2}{9}=\dfrac{7}{2}÷\dfrac{2}{9}$

$\dfrac{7}{2}\times\dfrac{9}{2}$

$\dfrac{7}{2}$ i.e.

$7:2$

Subtract:
$-4/9$ and $3\dfrac {5}{8}$ .

$-4/9$ froma

$3\dfrac {5}{8}$
This can be written as,

$- 4 / 9$ from

$29 / 8$

$= 29 / 8 - (- 4 / 9)$

$= 29 / 8 + 4 / 9$
Taking L.C.M. we get,

$= (261 + 32) / 72$

$= 293 / 72$

$= 4\dfrac {5}{72}$ .

Two rational numbers are
called equivalent if they have ________ value

Equivalent rational numbers are rational numbers that have the same value but which are represented in a different form.

Using appropriate properties of addition, find the following:
$3/7 + 4/9 + (-5/21) + (2/3)$ .

$3 / 7 + 4 / 9 + (- 5 / 21) + 2 / 3$

$= 3 / 7 + (- 5 / 21) + 4 / 9 + 2 / 3$
On simplifying, we get,

$= \left \{9 + (-5)\right \} / 21 + (4 + 6) / 9$

$= 4 / 21 + 10 / 9$
Taking L.C.M. we get,

$= (12 + 70) / 63$

$= 82 / 63$

$= 1 \dfrac {19}{63}$ .

Using appropriate properties of addition, find the following:
$4/5 + 11/7 + (-7/5) + (-2/7)$ .

$4 / 5 + 11 / 7 + (- 7 / 5) + (- 2 / 7)$

$= 4 / 5 + (- 7 / 5) + 11 / 7 + (- 2 / 7)$

$= \left \{4 + (-7)\right \}/ 5 + \left \{11 + (-2) \right \}/ 7$

$= (4 - 7) / 5 + (11 - 2) / 7$
On further calculation, we get,

$= - 3 / 5 + 9 / 7$
Now, taking L.C.M. we get,

$= (- 21 + 45) / 35$

$= 24 / 35$ .

Simplify:
$-4/9 + 2\dfrac {12}{13}$ .

Given

$-4/9 + 2\dfrac {12}{13}$
This can be written as,

$-4/9 + 38 / 13$
Taking L.C.M we get,

$- 4 / 9 = (-4 \times 13) / (9 \times 13)$
We get,

$= - 52 / 117$

$38 / 13 = (38 \times 9) / (13 \times 9)$
We get,

$= 342 / 117$
Now,
Adding both the numbers,

$- 52 / 117 + 342 / 117 = (- 52 + 342) / 117$
We get,

$= 290 / 117$

$117$	$290$	$2$
	$234$
	$56$

$= 2\dfrac {56}{117}$ .

Fill in the blanks:
$(-4/9) + (2/7)$ is a ______ number.

$$ (- 4 / 9) + (2 / 7) $$ is a rational number.

Fill in the following blanks:
$2/3\times -4/5$ is a ______ number.

$2/3\times -4/5$ is a rational number.

Verify the following and name the property also:
$3/5\times (-4/7 \times -8/9) = (3/5 \times - 4/7) \times -8/9$ .

$3 / 5 \times (- 4 / 7 \times -8 / 9) = (3 / 5 \times - 4 / 7) \times - 8 / 9$

$L.H.S. = 3 / 5 \times (- 4 / 7 \times - 8 / 9)$

$= 3 / 5 \times (- 4 \times - 8) / 7 \times 9$

$= 3 / 5 \times 32 / 63$

$= (3 \times 32) / (5 ×\times 63)$
We get,

$= 96 / 315$

$R.H.S. = (3 / 5 \times - 4 / 7) \times - 8 / 9$

$= - 12 / 35 \times - 8 / 9$

$= \left \{- 12 \times (- 8)\right \}/ (35 \times 9)$
We get,

$= 96 / 315$

Hence,

$3 / 5 \times (- 4 / 7 \times - 8 / 9) = (3 / 5 \times - 4 / 7) \times - 8 / 9$
The name of the property is Associative property of multiplication.

Multiply and express the result in the lowest form:
$- \dfrac{6}{7} \times \dfrac{{14}}{{30}}$

$\begin{aligned}{} - \frac{6}{7} \times \frac{{14}}{{30}} &= - \frac{6}{7} \times \frac{{7 \times 2}}{{6 \times 5}}\\\\ &= - \left( {\frac{6}{7} \times \frac{7}{6}} \right) \times \frac{2}{5}\\\\ &= - \frac{2}{5}\end{aligned}$

Subtract:
$-3\dfrac {1}{5}$ from $-4\dfrac {7}{9}$ .

$-3\dfrac {1}{5}$ from

$-4\dfrac {7}{9}$
This can be written as,

$= -16/5$ from

$-43/9$

$= -43/9 + 16/5$
Taking L.C.M. we get,

$= (-215 + 144) / 45$
We get,

$= -71 / 45$

$= -1\dfrac {26}{45}$ .

Sum of two rational numbers is $3/5$ . If one of them is $-2/7$ , find the other.

Given
Sum of two rational numbers is

$3 / 5$
One of the number is

$- 2 / 7$
Hence, the other number is calculated as follows:
Other number

$= 3 / 5 - (- 2 / 7)$

$= 3 / 5 + 2 / 7$
Taking L.C.M. we get,

$= (21 + 10) / 35$

$= 31 / 35$
Therefore, the other number is

$31 / 35$ .

Fill in the following blanks:
$16/23 \times$ _____ $= 0$ .

$16/23 \times$ _____

$= 0$

$16/23 \times 0 = 0$ .

Multiply and express the result in the lowest form:
$25/-9\times -3/10$ .

$25 / - 9 × - 3 / 10$

$= \left \{25 \times (- 3)\right \} / \left \{(- 9) \times 10\right \}$

$= - 75 / - 90$

$= \left \{- 75 \div (- 15)\right \} / \left \{- 90 \div (- 15)\right \}$

$\because HCF$ of

$75, 90 = 15$
We get,

$= 5 / 6$ .

What rational number should be subtracted from $-4\dfrac {3}{5}$ to get $-3\dfrac {1}{2}$ ?

The required number can be calculated as follows:

$\left (-4 \dfrac {3}{5}\right ) - \left (-3 \dfrac {1}{2}\right )$
This can be written as,

$(-23/5) + (7/2)$
On further calculation, we get

$= (-46 + 35) / 10$

$= -11/10$

$= -1\dfrac {1}{10}$
Therefore, the required number is

$-1\dfrac {1}{10}$ .

What rational number should be added to $-5/11$ to get $-7/8$ .

Given
According to the statement,
Sum of two numbers

$= - 7 / 8$
One number

$= - 5 / 11$
Hence, the other number is calculated as below:
Other number

$= - 7 / 8 - (- 5 / 11)$

$= - 7 / 8 + 5 / 11$
Taking L.C.M. we get,

$= (- 77 + 40) / 88$

$= - 37 / 88$
Therefore, the other number is

$- 37 / 88$ .

Multiply and express the result in the lowest form:
$6\dfrac {2}{3}\times 1\dfrac {2}{7}$ .

$6\dfrac {2}{3}\times 1\dfrac {2}{7}$
This can be written as,

$= 20 / 3 \times 9 / 7$

$= (20 \times 9) / (3 \times 7)$

$= 180 / 21$

$= (180 \div 3) / (21 \div 3)$

$\because HCF$ of

$180, 21 = 3$
We get,

$= 60 / 7$

$= 8\dfrac {4}{7}$ .

Find the value of the following:
$-8/9\div -3/5$ .

$\begin{aligned}{} - 8/9 \div - 3/5 &= - \frac{8}{9} \div - \frac{3}{5}\\ &= - \frac{8}{9} \times - \frac{5}{3}\\ &= \frac{{40}}{{27}}\end{aligned}$

The product of two rational number is $-11/12$ . If one of them is $2\dfrac {4}{9}$ find the other.

Given
Product of two rational numbers

$= -11/12$
One of the number

$=$

$2\dfrac {4}{9} = 22/9$
The other number is calculated as below

$-11/12 \div 22/9$

$= -11/12\times 9/22$
We get,
Therefore, the other number is

$-3/8$ .

Fill in the following blanks:
The product of a non-zero rational number and its reciprocal is ______.

The product of a non-zero rational number and its reciprocal is

$1$ .

Ex:

$3$ is a non-zero rational number and

$\dfrac{1}{3}$ is it's reciprocal

$\Rightarrow 3\times \dfrac{1}{3} = 1$

Find the value of the following:
$-3/7\div 4$ .

$-3/7 \div 4$

$= -3/7 \times 1/4$
We get,

$= -3/28$
Hence, the value of

$-3/7\div 4 = -3/28$ .

By what rational number should $-3$ is divided to get $-9/13$ ?

The required number can be calculated as follows:

$-3\div -9/13$

$= -3\times 13/-9$
We get,

$= -13/-3$

$= \left \{-3\times (-1) \right \}/ \left \{-3\times (-1)\right \}$

$= 13/3$

$= 4\dfrac {1}{3}$
Therefore, the required number is

$4\dfrac {1}{3}$ .

By what rational number should $-7/12$ be multiplied to get the product as $5/14$ ?

Given
Product

$= 5/14$
The required number can be calculated as below

$5/14 \div -7/12$

$= 5/14\times 12/-7$
We get,

$= 30/-49$

$= \left \{30\times (-1)\right \}/ \left \{-49 \times (-1)\right \}$

$= -30/49$
Hence, the required number is

$-30/49$ .

Compare :
$\dfrac{-3}{8}$ and $\dfrac{2}{5}$

$\dfrac{-3}{8}$ and

$\dfrac{2}{5}$
We know that

$\dfrac{a}{b}$ and

$\dfrac{c}{d} = a \times d$ and

$b \times c$
Substituting the values

$-3 \times 5$ and

$2 \times 8$
So we get

$a \times d < b \times c$

$-15 < 16$
Hence ,

$\dfrac{-3}{8} < \dfrac{2}{5}$ .

Arrange $\displaystyle -\frac{5}{9}, \frac{7}{12}, -\frac{2}{3}$ and $\displaystyle \frac{11}{18}$ in ascending order of their magnitudes. Identify the smallest

Consider the given numbers

$-\dfrac {5}{9},\dfrac {7}{12},-\dfrac {2}{3}$ and

$\dfrac {11}{18}$ .

The LCM of the denominators

$9,12,3,18$ is

$36$ . Therefore, the given numbers will be:

$-\dfrac { 5\times 4 }{ 9\times 4 } =-\dfrac { 20 }{ 36 } \\ \dfrac { 7\times 3 }{ 12\times 3 } =\dfrac { 21 }{ 36 } \\ -\dfrac { 2\times 12 }{ 3\times 12 } =-\dfrac { 24 }{ 36 } \\ \dfrac { 11\times 2 }{ 18\times 2 } =\dfrac { 22 }{ 36 }$

Thus, the numbers in ascending order are shown below:

$-\dfrac { 24 }{ 36 } ,-\dfrac { 20 }{ 36 } ,\dfrac { 21 }{ 36 } ,\dfrac { 22 }{ 36 } \\ \Rightarrow -\dfrac { 2 }{ 3 } ,-\dfrac { 5 }{ 9 } ,\dfrac { 7 }{ 12 } ,\dfrac { 11 }{ 18 }$

Write down ten positive rational numbers such that the sum of the numerator and the denominator of each is $11$ . Write them in decreasing order

Firstly, let us determine the numbers whose sum is

$11$ as follows:

$10+1,9+2,8+3,7+4,6+5,5+6,4+7,3+8,2+9$ and

$1+10$

Now, the ten positive rational numbers in which the sum of the numerator and denominator is

$11$ are:

$\dfrac { 10 }{ 1 } ,\dfrac { 9 }{ 2 } ,\dfrac { 8 }{ 3 } ,\dfrac { 7 }{ 4 } ,\dfrac { 6 }{ 5 } ,\dfrac { 5 }{ 6 } ,\dfrac { 4 }{ 7 } ,\dfrac { 3 }{ 8 } ,\dfrac { 2 }{ 9 } ,\dfrac { 1 }{ 10 }$

Hence, the decreasing order of the numbers is

$\dfrac { 10 }{ 1 } ,\dfrac { 9 }{ 2 } ,\dfrac { 8 }{ 3 } ,\dfrac { 7 }{ 4 } ,\dfrac { 6 }{ 5 } ,\dfrac { 5 }{ 6 } ,\dfrac { 4 }{ 7 } ,\dfrac { 3 }{ 8 } ,\dfrac { 2 }{ 9 } ,\dfrac { 1 }{ 10 }$ .

Write down ten positive rational numbers such that numerator - denominator for each of them is $-2$ . Write them in increasing order

Firstly, let us determine the numbers whose difference is

$-2$ as follows:

$10-12,9-11,8-10,7-9,6-8,5-7,4-6,3-5,2-4$ and

$1-3$

Now, the ten positive rational numbers in which the subtraction of the numerator and denominator is

$-2$ are:

$\dfrac { 10 }{ 12 } ,\dfrac { 9 }{ 11 } ,\dfrac { 8 }{ 10 } ,\dfrac { 7 }{ 9 } ,\dfrac { 6 }{ 8 } ,\dfrac { 5 }{ 7 } ,\dfrac { 4 }{ 6 } ,\dfrac { 3 }{ 5 } ,\dfrac { 2 }{ 4 } ,\dfrac { 1 }{ 3 }$

Hence, the increasing order of the numbers is

$\dfrac { 1 }{ 3 } ,\dfrac { 2 }{ 4 } ,\dfrac { 3 }{ 5 } ,\dfrac { 4 }{ 6 } ,\dfrac { 5 }{ 7 } ,\dfrac { 6 }{ 8 } ,\dfrac { 7 }{ 9 } ,\dfrac { 8 }{ 10 } ,\dfrac { 9 }{ 11 } ,\dfrac { 10 }{ 12 }$ .

Write the following rational numbers in ascending order:
$\dfrac {3}{4}, \dfrac {7}{12}, \dfrac {15}{11}, \dfrac {22}{19}, \dfrac {101}{100}, \dfrac {-4}{5}, \dfrac {-102}{81}, \dfrac {-13}{7}$

We know that numbers are said to be in ascending order when they are arranged from the smallest to the largest number. Therefore, the ascending order of the rational numbers

$\dfrac { 3 }{ 4 } ,\dfrac { 7 }{ 12 } ,\dfrac { 15 }{ 11 } ,\dfrac { 22 }{ 19 } ,\dfrac { 101 }{ 100 } ,\dfrac { -4 }{ 5 } ,\dfrac { -102 }{ 81 } ,\dfrac { -13 }{ 7 }$ or

$0.75,0.5,1.3,1.1,1.01,-0.8,-1.2,-1.8$ is as shown below:

$\dfrac { -13 }{ 7 } ,\dfrac { -102 }{ 81 } ,\dfrac { -4 }{ 5 } ,\dfrac { 7 }{ 12 } ,\dfrac { 3 }{ 4 } ,\dfrac { 101 }{ 100 } ,\dfrac { 22 }{ 19 } ,\dfrac { 15 }{ 11 }$ or

$-1.8,-1.2,-0.8,0.5,0.75,1.01,1.1,1.3$

Which of the rational number is greater in the given pair?
(i) $\frac{-4}{3}$ or $\frac{-8}{7}$
(ii) $\frac{7}{-9}$ or $\frac{-5}{8}$
(iii) $\frac{-1}{3}$ or $\frac{4}{-5}$
(iv) $\frac{9}{-13}$ or $\frac{7}{-12}$
(v) $\frac{4}{-5}$ or $\frac{-7}{10}$
(vi) $\frac{-12}{5}$ or $-3$

1313302_1010195_ans_7b6a29811d62450390c801e47274242b.jpg

Give four rational numbers equivalent to:
i) $\dfrac{-2}{7}$ ii) $\dfrac{5}{-3}$

$i)$ Given rational number is

$\dfrac{-2}{7}$

We know that, multiplying a number by

$1$ gives the same number.

Hence, when we multiply a rational number by

$\dfrac{a}{a}\ (1)$ , it gives an equivalent rational number, without changing the simplest form of the rational number.

$\therefore \$ four rational numbers equivalent to

$\dfrac{-2}{7}$ will be:

$1. \ \dfrac{-2}{7}\times\dfrac{2}{2}=\dfrac{-4}{14}$

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

$3.\ \dfrac{-2}{7}\times\dfrac{4}{4}=\dfrac{-8}{28}$

$4.\ \dfrac{-2}{7}\times\dfrac{5}{5}=\dfrac{-10}{35}$

Hence, four rational numbers equivalent to

$\dfrac{-2}{7}$ are

$\dfrac{-4}{14} ,\dfrac{-6}{21} ,\dfrac{-8}{28} ,\dfrac{-10}{35}$

$ii)$

Similarly, equivalent rational numbers of

$\dfrac{5}{-3}$ will be:

$1.\ \dfrac{5}{-3}\times\dfrac{2}{2}=\dfrac{10}{-6}$

$2.\ \dfrac{5}{-3}\times\dfrac{3}{3}=\dfrac{15}{-9}$

$3.\ \dfrac{5}{-3}\times\dfrac{5}{5}=\dfrac{20}{-12}$

$4.\ \dfrac{5}{-3}\times\dfrac{5}{5}=\dfrac{25}{-15}$

Hence, four rational numbers equivalent to

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

$\dfrac{10}{-6},\dfrac{15}{-9},\dfrac{20}{-12},\dfrac{25}{-15}$

Which of the following pairs represent the same rational number?
$\dfrac {-7}{21}$ and $\dfrac {3}{9}$

$\dfrac{-7}{21}$ and

$\dfrac{3}{9}$

$\dfrac{-7}{21} = \dfrac{-1}{3}$

$\dfrac{3}{9} = \dfrac{1}{3}$

Since

$\dfrac{-1}{3} \neq \dfrac{1}{3}$

So,

$\dfrac{-7}{21}$ and

$\dfrac{3}{9}$

does not represent same rational number.

Are $\dfrac{-1}{2}$ and $\dfrac{-3}{6}$ represented by the same point on the number line?

$\dfrac{-3}{6}=\dfrac{\dfrac{-3}{3}}{\dfrac{6}{3}},\dfrac{-1}{2}$

$\therefore \dfrac{-3}{6}\;\&\;\dfrac{-1}{2}$ are represented by same point on the number line.

Hence, solved.

Write $5$ equivalent rational numbers to:-
$(i)\ \ \dfrac{5}{2}$

$(ii)\ -\dfrac{7}{9}$

$(iii)-\dfrac{3}{7}$

$(i)$

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

$\dfrac{5}{2} \times \dfrac{3}{3} = \dfrac{{15}}{6}$

$\dfrac{5}{2} \times \dfrac{4}{4} = \dfrac{{20}}{8}$

$\dfrac{5}{2} \times \dfrac{5}{5} = \dfrac{{25}}{10}$

$\dfrac{5}{2} \times \dfrac{6}{6} = \dfrac{{30}}{12}$

$(ii)$

$-\dfrac{7}{9} \times \dfrac{2}{2} = -\dfrac{{14}}{18}$

$-\dfrac{7}{9} \times \dfrac{3}{3} = -\dfrac{{21}}{27}$

$-\dfrac{7}{9} \times \dfrac{4}{4} = -\dfrac{{28}}{36}$

$-\dfrac{7}{9} \times \dfrac{5}{5} = -\dfrac{{35}}{45}$

$-\dfrac{7}{9} \times \dfrac{6}{6} = -\dfrac{{42}}{54}$

$(iii)$

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

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

$-\dfrac{3}{7} \times \dfrac{4}{4} = -\dfrac{{12}}{28}$

$-\dfrac{3}{7} \times \dfrac{5}{5} = -\dfrac{{15}}{35}$

$-\dfrac{3}{7} \times \dfrac{6}{6} = -\dfrac{{18}}{42}$

Which of the following pairs represent the same rational number?
$\dfrac {-2}{-3}$ and $\dfrac {2}{3}$

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

$\dfrac{2}{3}$

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

So,

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

$\dfrac{2}{3}$

represent same rational number

Which of the following pairs represent the same rational number?
$\dfrac {-16}{20}$ and $\dfrac {20}{-25}$

$\dfrac{-16}{20}$ and

$\dfrac{20}{-25}$

$\Rightarrow$ consider the fraction

$\dfrac{-16}{20}=\dfrac{-4\times 4}{4\times 5}=\dfrac{-4}{5}$

similarly

$\dfrac{20}{-25}=\dfrac{4\times 5}{-5\times 5}=\dfrac{-4}{5}$

so we see that

$\dfrac{-16}{20}$ and

$\dfrac{20}{25}$ basically represent the same rational number

$\dfrac{-4}{5}.$

Hence, the answer is

$\dfrac{-4}{5}.$

Give four rational numbers equivalent to:
$\dfrac {-2}{7}$

We have,

$\begin{matrix} \frac { { -2 } }{ 7 } \times \frac { 2 }{ 2 } =\frac { { -4 } }{ { 14 } } \\ \frac { { -2 } }{ 7 } \times \frac { 3 }{ 3 } =\frac { { -6 } }{ { 21 } } \\ \frac { { -2 } }{ 7 } \times \frac { 4 }{ 4 } =\frac { { -8 } }{ { 28 } } \\ \frac { { -2 } }{ 7 } \times \frac { 5 }{ 5 } =\frac { { -10 } }{ { 35 } } \\ \end{matrix}$

Therefore,

$4$ equivalent rational numbers are

$\frac{{ - 4}}{{14}},\frac{{ - 6}}{{21}},\frac{{ - 8}}{{21}},\frac{{ - 10}}{{35}}$

Solve
$\dfrac{-3}{5}+\dfrac{-12}{20}$

We have,

$\dfrac { -3 }{ 5 } +\dfrac { -12 }{ 20 }$

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

$\Rightarrow -\dfrac { 6 }{ 5 }$

Hence, this is the answer.

The product of two integers is $105$ .If one of them is $-7$ , then find the other.

1349653_1282071_ans_2cfd76bbce2e41e0813e199e4dd32524.PNG

What should be subtracted from thrice the rational number $\dfrac{-8}{3}$ to get $\dfrac{5}{2}$ .

Let the number be

$x$ .

According to the question,

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

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

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

$x=-\dfrac{21}{2}$

Hence, this is the answer.

The product of two rational numbers is $\dfrac{-28}{75}$ .If one of the numbers is $\dfrac{14}{25}$ .Find the other.

$\because$ Product of two numbers

$=\dfrac{-28}{75}$

Anyone of the rational numbers

$=\dfrac{14}{25}$

$\therefore$ the other number

$=\dfrac{-28}{75}\div\dfrac{14}{25}$

$=\dfrac{-28}{75}\times \dfrac{25}{14}$

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

Thus, the required rational number is

$\dfrac{-2}{3}$

Give four rational numbers equivalent to:
$\dfrac {5}{-3}$

The given rational number is

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

We can get the rational numbers equivalent to the given number by either multiplying or dividing the numeration by the same, non-zero number.

Let's multiply both the numerator and denominator of the given rational number to find four equivalent rational numbers.

$(i)\ \dfrac { 5 }{ { -3 } } \times \dfrac { 2 }{ 2 }$

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

$(ii)\ \dfrac { 5 }{ { -3 } } \times \dfrac { 3 }{ 3 }$

$=\dfrac { { 15 } }{ { -9 } }$

$(iii)\ \dfrac { 5 }{ { -3 } } \times \dfrac { 4 }{ 4 }$

$=\dfrac { { 20 } }{ { -12 } }$

$(iv)\ \dfrac { 5 }{ { -3 } } \times \dfrac { 5 }{ 5 }$

$=\dfrac { { 25 } }{ { -15 } }$

Hence, four equivalent rational numbers of

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

$\dfrac{{10}}{{ - 6}},\dfrac{{15}}{{ - 9}},\dfrac{{20}}{{ - 12}},\dfrac{{25}}{{ - 15}}$ .

Find the sum of the rational numbers $\dfrac{-4}{9},\,\dfrac{15}{12}$ and $\dfrac{-7}{18}$

$\dfrac{-4}{9}+\dfrac{15}{12}+\dfrac{-7}{18}$

$=\left(\dfrac{-4}{9}-\dfrac{7}{18}\right)+\dfrac{15}{12}$

$=\dfrac{-8-7}{18}+\dfrac{15}{12}$

$=\dfrac{-5}{6}+\dfrac{5}{4}$

$=\dfrac{-20}{24}+\dfrac{30}{24}$

$=\dfrac{-20+30}{24}$

$=\dfrac{10}{24}$

$=\dfrac{5}{12}$

Out of the rational numbers $\cfrac { -5 }{ 11 } ,\cfrac { -5 }{ 12 }, \cfrac { -5 }{ 17 }$ which is smallest.

Given rational numbers are

$\dfrac{-5}{11},\dfrac {-5}{12},\dfrac {-5}{17}$ .

Since numerators are equal the fraction with the highest denominator will be the smallest one.

$\dfrac{-5}{17}$ is smallest .

Write the following rational numbers in ascending order:
$\cfrac{-3}{7}$ , $\cfrac{-3}{2}$ , $\cfrac{-3}{4}$

$\cfrac{-3}{7}={\dfrac{-3\times4}{7\times4}}={\dfrac{-12}{28}}$ ,

$\cfrac{-3}{2}={\dfrac{-3\times14}{2\times14}}={\dfrac{-42}{28}}$

$\cfrac{-3}{4}={\dfrac{-3\times7}{4\times7}}={\dfrac{-21}{28}}$

$\dfrac{-42}{28}<\dfrac{-21}{28}<\dfrac{-12}{28}$

$\Rightarrow$

$\cfrac{-3}{2}< \cfrac{-3}{4}< \cfrac{-3}{7}$

Check if the following pair represent the same rational number:
$\cfrac{1}{3}$ and $\cfrac{-3}{9}$

Given:

$\cfrac{1}{3}$ and

$\cfrac{-3}{9}$

$\dfrac{-3}{9}=\dfrac{-3}{3\times3}$

$[\because 9=3\times 3]$

$=\dfrac{-1}{3}$
Now, two numbers are

$\dfrac{1}{3}$ and

$\dfrac{-1}{3}$

Since both the rational numbers have the same denominator and numerator but they are of opposite signs.

$\implies$ they lie on either side of zero on the number line.

$\therefore$

$\cfrac{1}{3}\ne \cfrac{-3}{9}$

Add the following rational numbers:
$\dfrac{-7}{27}$ and $\dfrac{11}{18}$

Since the given two rational numbers have different denominators we need to take LCM of the denominators,
LCM of the denominators

$27$ and

$18$ is

$54.$

Now,

$\displaystyle \frac{{ - 7}}{{27}} + \frac{{11}}{{18}} = \frac{{ - 7 \times 2}}{{27 \times 2}} + \frac{{11 \times 3}}{{18 \times 3}}$

$= \dfrac{{ - 14}}{{54}} + \dfrac{{33}}{{54}}$

$= \dfrac{{ - 14 + 33}}{{54}}$

$= \dfrac{{19}}{{54}}$

Add the rational numbers
$\dfrac {-5}{16}$ and $\dfrac {7}{24}$

Let us add the fractions

$\dfrac {-5}{16} + \dfrac {7}{24}$

Since it has a different denominator we need to take LCM

LCM of the denominators

$16$ and

$24$ is

$48$

So by multiplying the numerator and denominator by

$3$ we get,

$\dfrac {-5}{6} = \dfrac {-5\times 3}{16\times 3} = \dfrac {-15}{48}$

And by multiplying the numerator and denominator by

$2$ we get,

$\dfrac {7}{24} = \dfrac {7\times 2}{24\times 2} = \dfrac {14}{48}$

Now we can add the fractions since it has the same denominators

$\Rightarrow \dfrac {-15}{48} + \dfrac {14}{48} = \dfrac {-15 + 14}{48}$

$=\dfrac {-1}{48}$

Add the following rational numbers:
$\dfrac{31}{-4}$ and $\dfrac{-5}{8}$

Given,

$\dfrac{31}{-4} + \dfrac{-5}{8}$

LCM of

$4$ and

$8$ is

$8$

So,

$\dfrac{-62 - 5}{8}$

$\dfrac{-67}{8}=-8.375$

Simplify:
$\dfrac{5}{26}+\dfrac{11}{-39}$

$\dfrac {5 }{ 26 } +\dfrac { 11 }{ -39 } =\dfrac { 5 }{ 26 } +\dfrac { -11 }{ 39 } =\dfrac { 5\times 3+(-11)\times 2 }{ 78 } =\dfrac { 15-22 }{ 78 } =\dfrac { -7 }{ 78 }$

Add the following rational numbers:
$\dfrac {5}{8}$ and $\dfrac {-7}{12}$ .

Let us add the fractions

$\dfrac {5}{8} + \dfrac {-7}{12}$
Since it has the different denominator we need to take LCM
LCM of the denominators

$8$ and

$12$ is

$24$
So by multiplying the numerator and denominator by

$3$ we get,

$\dfrac {5}{8} = \dfrac {5\times 3}{8\times 3} = \dfrac {15}{24}$
And by multiplying the numerator and denominator by

$2$ we get,

$\dfrac {-7}{12} = \dfrac {-7\times 2}{12\times 2} = \dfrac {-14}{24}$
And by multiplying the numerator and denominator by

$2$ we get,

$\dfrac {-7}{12} = \dfrac {7\times 2}{12\times 2} = \dfrac {-14}{24}$
Now we can add the fractions since it has the same denominators

$\dfrac {15}{24} + \dfrac {-14}{24} = \dfrac {15 + (-14)}{24} = \dfrac {15 - 14}{24}\Rightarrow \dfrac {1}{24}$ .

$\dfrac{-8}{11}$ + $\dfrac{-4}{11}=\dfrac{p}{11}$
then
$p=?$

$\dfrac{-8}{11}$ +

$\dfrac{-4}{11} = \dfrac{-8}{11} + \left(\dfrac{-4}{11}\right) = \dfrac{-8+(-4)}{11} = \dfrac{-8-4}{11} = \dfrac{-12}{11}$

Simplify:
$\dfrac{-8}{19}+\dfrac{-4}{57}$

$\dfrac{-8}{19}+ \dfrac{-4}{57} = \dfrac{(-8\times 3) + (-4\times 1)}{57} = \dfrac{-24 - 4}{57} = \dfrac{-28}{57}$

Add the following rational numbers:
$\dfrac{-15}{4}$ and $\dfrac{7}{4}$

Add

$\dfrac{-15}{4}$ &

$\dfrac{7}{4} = \dfrac{-15}{4} + \dfrac{7}{4} = \dfrac{-15+7}{4} = \dfrac{-8}{4} = -2$

Multiply:
$\dfrac{7}{11}$ by $\dfrac{5}{4}$ .

If the answer is $\dfrac ab$ , then the value of $(a+b)$ will be:-

So comparing with

$\dfrac ab$ mentioned in question, the value of (a+b) becomes 79.
1508803_1671369_ans_4cb90dbdbd6b4b57b8f25a82042d4236.jpg

What should be subtracted from $\left(\dfrac{3}{4}-\dfrac{2}{3}\right)$ to get $\dfrac{-1}{6}$ ? Hence, find the sum of numerator and denominator.

Let

$\left(\dfrac{3}{4} - \dfrac{2}{3} \right) - x = \dfrac{-1}{6}$

$\Rightarrow \left(\dfrac{3\times 3 - 2\times 4}{12}\right) -x = \dfrac{-1}{6}$

$\Rightarrow \dfrac{9-8}{12} -x = \dfrac{-1}{6}$

$\Rightarrow x = \dfrac{1}{12} + \dfrac{1}{6}$

$\Rightarrow x =\dfrac{1+2}{12}$

$\Rightarrow x = \dfrac{3}{12} = \dfrac{1}{4}$

Find the value of $p+q$

$\dfrac{-11}{9}\times \dfrac{-81}{-88}=\,\dfrac{-p}{q}$

where $\dfrac{-p}{q}$ is in lowest form

Given,

$\dfrac{-11}{9}\times \dfrac{-81}{-88}=\,\dfrac{-p}{q}$

$\implies \dfrac{-9}{8}=\,\dfrac{-p}{q}$

Thus, we get

$p=9,q=8$

$\therefore p+q=9+8=17$

Select those rational numbers which can be written as a rational number with numerator $6$ :
$\displaystyle \frac{1}{22}, \frac{2}{3}, \frac{3}{4}, \frac{4}{-5}, \frac{5}{6}, \frac{-6}{7}, \frac{-7}{8}$

1591468_1676309_ans_145b0ca064e84380b819c28a29fb0b40.jpg

Verify commutativity of addition of rational number for each of the following pairs of rational numbers:
$4$ and $\dfrac{-3}{5}$

Commutativity of addition of rational number means that if

$\dfrac ab$ and

$\dfrac cd$ are two rational numbers,

Then

$\dfrac ab+\dfrac cd=\dfrac cd+\dfrac ab$

Here we have,

$4$ and

$\dfrac{-3}{5}=-\dfrac35$

$\\\therefore 4+\dfrac{-3}{5}=4-\dfrac35 \\=\dfrac{4\times5}{1\times 5}-\dfrac{3\times1}{5\times1}=\dfrac{20}{5}-\dfrac{3}{5}\\=\dfrac{20-3}{5}=\dfrac{17}{5}\\$

$\text{And, } \dfrac{-3}{5}+4=-\dfrac35 +4\\=-\dfrac{3\times1}{5\times1}+\dfrac{4\times5}{1\times 5}=-\dfrac{3}{5}+\dfrac{20}{5}\\=\dfrac{-3+20}{5}=\dfrac{17}{5}\\$

$\therefore 4+\dfrac{-3}{5}=\dfrac{-3}{5}+4$

Hence verified.

Write down the rational number whose numerator is $(-3) \times 4$ , and whose denominator is $(34-23) \times (7-4)$ .

1590785_1676068_ans_cfd00043b8f94177b5db109daa1216cf.jpg

What number should be subtracted from $\dfrac{-5}{3}$ to get $\dfrac{5}{6}$ ?

Given

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

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

$x = \dfrac{-10 - 5}{6}$

$x = \dfrac{-15}{6}$

$x= \dfrac{-3\times 5}{3\times 5}$

$x = \dfrac{-5}{2}$

$\dfrac{-5}{2}$ should be subtracted from

$\dfrac{-5}{3}$ to get

$\dfrac{5}{6}$

Number should be subtracted from $\dfrac{3}{7}$ to get $\dfrac{5}{4}$ is $\dfrac{-23}{x}$ .Find 'x'.

Let

$\dfrac{3}{7} -x = \dfrac{5}{4}$

$\Rightarrow x = \dfrac{3}{7} - \dfrac{5}{4}$

$\Rightarrow x = \dfrac{3\times 4 - 5 \times 7}{28}$

$= \dfrac{12-35}{28} = \dfrac{-23}{28}$

Express $\dfrac{3}{5}$ as a rational number with numerator:
$6$

Given fraction :

$\dfrac{3}{5}$

Multiply by

$2$ to both numerator and denominator

$\dfrac{3\times 2}{5\times 2}=\dfrac{6}{10}$

Express $\dfrac{5}{7}$ as a rational number with denominator:
$-14$

1590923_1676187_ans_44bff71456584fad80885b6914eaa41c.jpg

Express $\dfrac{5}{7}$ as a rational number with denominator:
$-84$

1590928_1676202_ans_056932f1ae70436781019585c90db592.jpg

Express $\dfrac{3}{4}$ as a rational number with denominator:
$44$

1516283_1676221_ans_dc8a5cf35a224d0ab8dae6eb6b3fbb4b.jpg

Express $\dfrac{3}{5}$ as a rational number with numerator:
$21$

Given number is

$\dfrac35$

To find out,

The rational number with numerator

$21$ , equivalent to the given fraction.

We will get a numerator

$21$ only when we multiply the numerator of the given rational number by

$7$ .

Hence, we will have to multiply the denominator by

$7$ as well.

So, the equivalent fraction will be:

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

$=\dfrac{21}{35}$ .

Hence,

$\dfrac35=\dfrac{21}{35}$ .

Express $\dfrac{3}{4}$ as a rational number with denominator:
$20$

1590931_1676211_ans_3505b11c25d64ecd88ab1a86d36e8eab.jpg

Express $\dfrac{3}{4}$ as a rational number with denominator:
$36$

Given rational number is

$\dfrac34$ .

We have to represent it as a rational number having denominator

$36$ .

The denominator of the original number is

$4$ .

To make it

$36$ , we need to multiply it with

$9$ .

Hence, multiplying both the numerator and denominator of the given rational number by

$9$ .

$\Rightarrow \dfrac{3\times 9}{4\times 9}$

$=\dfrac{27}{36}$

Hence,

$\dfrac{27}{36}$ is the required rational number.

Express $\dfrac{5}{7}$ as a rational number with denominator:
$70$

1516260_1676192_ans_60773dfc93104a61a0cab726aafdb04e.jpg

Express $\dfrac{2}{5}$ as a rational number with numerator :
$154$

1590934_1676234_ans_3845a0c04a9047e88020884a6c23d00d.jpg

Express $\dfrac{5}{7}$ as a rational number with denominator:
$-28$

1516261_1676197_ans_fffe540806e546b19bb15e5a808c409f.jpg

Express $\dfrac{3}{5}$ as a rational number with numerator:
$-15$

2039845_1676173_ans_4d52fc64819f420e8321f5e928b48e6c.png

Fill in the blanks to make the statement true:
Two rational numbers are said to be equal, if they have the same _____ form.

In order to determine the equality of two rational numbers, we express both the rational numbers in the standard form. If they have the same standard form they are equal, otherwise they are not equal.

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.

Two rational numbers are said to be equal, if they have the same standard form.

Fill in the blanks so as to make the statement true:
Two rational numbers with different numerators are equal, if their numerators are in the same ______ as their denominators.

Two rational numbers with different numerators are equal, if their numerators are in the same ratio as their denominators.

$\dfrac {3}{4}$ and $\dfrac {-3}{5}$ .

Let us add the fractions

$\dfrac {3}{4} + \dfrac {-3}{5}$
Since it has the different denominator we need to take LCM
LCM of the denominators

$4$ and

$5$ is

$20$
So by multiplying with the denominator we get,

$\dfrac {3}{4} = \dfrac {3\times 5}{4\times 5} = \dfrac {15}{20}$
And,

$\dfrac {-3}{5} = \dfrac {-3\times 4}{5\times 4} = \dfrac {-12}{20}$
Now we can add the fractions since it has the same denominators

$\dfrac {15}{20} + \dfrac {-12}{20} = \dfrac {15 + (-12)}{20} = \dfrac {15 - 12}{20}\Rightarrow \dfrac {3}{20}$ .

Add the following rational numbers:
$\dfrac{-15}{4}$ and $\dfrac{7}{4}$

1516270_1676631_ans_0529568f450b4533aa1e455291de569f.jpg

Add the following rational numbers:
$-3$ and $\dfrac{3}{5}$

1516883_1676642_ans_fb0d0d34923a45f3aba38159063e56e7.jpg

Add the following rational numbers:
$\dfrac{31}{-4}$ and $\dfrac{-5}{8}$

1514863_1676648_ans_8abeaee081504c0d94e6482f03529c05.jpg

Find the value and express as a rational number in standard form:
$\dfrac{2}{5} \div \dfrac{26}{15}$

Given,

$\dfrac25\div \dfrac{26}{15}=\dfrac25\times \dfrac{15}{26}\\$

$=\dfrac{30}{130}=\dfrac{30\div10}{130\div10}\qquad[\because$ LCM of

$30$ and

$130$ is

$10]$

$\\=\dfrac3{13}$

Hence, the simplest form is

$\dfrac3{13}$ .

Subtract the first rational number from the second in each of the following:
$\dfrac{-2}{11}, \dfrac{-9}{11}$

1517400_1676690_ans_4bb5d39e43574b1a8f374da4c5c4fe1d.jpg

State whether the given statement is true or false.
(i) the sum of two rationals is always rational.
(ii) the product of two rational is always rational.
(iii) The sum of two irrational is always an irrational.
(iv) the product of two irrational is always an irrational.
(v) the sum of a rational and an irrational is irrational.
(vi) The product of a rational and an irrational is irrational.

(i) True
(ii) True
(iii) False
Reason :
Sum of two irrational can be rational let us take an example,
Consider two irrational numbers :

$( 3 + \surd {2}) and (3 - \surd {2} )$

Sum :

$( 3 + \surd {2})+ (3 - \surd {2} )= 3 + \surd {2} + 3 - \surd{2} = 6$ which is rational
(iv) false
Reason :
Product of two irrational numbers can be rational. let us take an example, consider two rational numbers :

$( 3 +\surd {2} ) and ( 3 - \surd{2})$
Product :

$( 3 +\surd {2} ) ( 3 - \surd{2}) = (3)^2 - ( \surd{2})^2 = 9-2=7$ ; which is rational
(v) True
(vi) True

Subtract the first rational number from the second in each of the following:
$\dfrac{11}{13}, \dfrac{-4}{13}$

1517405_1676693_ans_2e86412828a44933b0a06d2d4f66257a.jpg

Subtract the first rational number from the second in each of the following:
$\dfrac{-7}{9}, \dfrac{4}{9}$

1514945_1676687_ans_e6166af7f1434b7ca9686affcc4e3d9f.jpg

The product of two rational numbers is $\dfrac{-8}{9}$ . If one of the numbers is $\dfrac{-4}{15}$ , find the other.

Given

Product of two rational numbers

$=\dfrac{-8}{9}$ .

One of the numbers

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

Let, the other number be

$x$

ATQ,

$\dfrac{-4}{15}\times x=\dfrac{-8}{9}$

$\Rightarrow \dfrac{4}{15}\times x=\dfrac{8}{9}$

$\Rightarrow x=\dfrac{8\times 15}{9\times 4}$

$\Rightarrow x=\dfrac{10}{3}$

Hence, the other rational number is

$\dfrac{10}{3}$ .

Express $\dfrac {-3}{5}$ as a rational number whose denominators are $35$ .

Similarly,

$\dfrac {-3}{5} = \dfrac {x}{35}$
By using cross multiplication

$5x = -105$

$x = \dfrac {105}{5} = -21$

$\therefore$ By replacing the value of

$x$ we get,

$\dfrac {-3}{5} = \dfrac {-21}{35}$ .

Express $\dfrac {-48}{60}$ as a rational number whose denominator is $5$ .

Let us consider,

$\dfrac {-48}{60} = \dfrac {x}{5}$

By using cross multiplication

$60x = -240$

$x = \dfrac {240}{60} = -4$

$\therefore$ By replacing the value of

$x$ we get,

$\dfrac {-48}{60} = \dfrac {-4}{5}$ .

Express $\dfrac {-42}{98}$ as a rational number whose denominator is $7$ .

Let us consider,

$\dfrac {-42}{98} = \dfrac {x}{7}$

By using cross multiplication

$98x = -294$

$x = \dfrac {-294}{98} = -3$

$\therefore$ By replacing the value of

$x$ we get,

$\dfrac {-42}{98} = \dfrac {-3}{7}$ .

Express $\dfrac {-3}{5}$ as a rational number whose denominators is $20$ .

Let us consider

$\dfrac {-3}{5} = \dfrac {x}{20}$
By using cross multiplication

$5x = -60$

$x = \dfrac {-60}{5} = -12$

$\therefore$ By replacing the value of

$x$ we get,

$\dfrac {-3}{5} = \dfrac {-12}{20}$ .

Express $\dfrac {-3}{5}$ as a rational number whose denominator is $-40$ .

Similarly,

$\dfrac {-3}{5} = \dfrac {x}{-40}$
By using cross multiplication

$5x = 120$

$x = \dfrac {120}{5} = 24$

$\therefore$ By replacing the value of

$x$ we get,

$\dfrac {-3}{5} = \dfrac {24}{-40}$ .

Express $\dfrac {-3}{5}$ as a rational number whose denominator is $-30$ .

Let us consider

$\dfrac {-3}{5} = \dfrac {x}{-30}$

By using cross multiplication

$5x = 90$

$x = \dfrac {90}{5} = 18$

$\therefore$ By replacing the value of

$x$ we get,

$\dfrac {-3}{5} = \dfrac {18}{-30}$ .

Fill in the blanks:
$\left (\dfrac {-8}{13} + \dfrac {3}{7}\right ) + \left (\dfrac {-13}{4}\right ) = (....) + \left [\dfrac {3}{7} + \left (\dfrac {-13}{4}\right )\right ]$ .

As we all know associative property states,

$(a + b) + c = a + (b + c)$
We can simply solve by using the property

$\therefore \left (\dfrac {-8}{13} + \dfrac {3}{7}\right ) + \left (\dfrac {-13}{4}\right ) = \left (\dfrac {-8}{13}\right ) + \left [\dfrac {3}{7} + \left (\dfrac {-13}{4}\right )\right ]$ .

Add the following rational numbers:
$\dfrac {-8}{9}$ and $\dfrac {11}{6}$ .

Let us add the fractions

$\dfrac {-8}{9} + \dfrac {11}{6}$

Since it has the different denominator we need to take LCM

LCM of the denominators

$9$ and

$6$ is

$18$

So by multiplying the numerators and denominator by

$2$ we get,

$\dfrac {-8}{9} = \dfrac {-8\times 2}{9\times 2} = \dfrac {-16}{18}$

And by multiplying the numerator and denominator by

$3$ we get,

$\dfrac {11}{6} = \dfrac {11\times 3}{6\times 3} = \dfrac {33}{18}$

Now we can add the fractions since it has the same denominators

$\dfrac {-16}{18} + \dfrac {33}{18} = \dfrac {-16 + 33}{18}\Rightarrow \dfrac {17}{18}$ .

Add the following rational numbers:
$\dfrac {1}{-12}$ and $\dfrac {2}{-15}$ .

Let us add the fractions

$\dfrac {1}{-12} + \dfrac {2}{-15}$

Since the denominators is negative we need to change to positive by multiplying with

$-1$

$\dfrac {1}{-12} = \dfrac {1\times -1}{-12\times -1} = \dfrac {-1}{12}$ and

$\dfrac {2}{-15} = \dfrac {2\times -1}{-15\times -1} = \dfrac {-2}{15}$

Since it has the different denominator we need to take LCM

LCM of the denominators

$12$ and

$15$ is

$60$

So by multiplying the numerator and denominator by

$5$ we get,

$\dfrac {-1}{12} = \dfrac {-1\times 5}{12\times 5} = \dfrac {-5}{60}$

And by multiplying the numerator and denominator by

$4$ we get,

$\dfrac {-2}{15} = \dfrac {-2\times 4}{15\times 4} = \dfrac {-8}{60}$

Now we can add the fractions since it has the same denominators

$\dfrac {-5}{60} + \dfrac {-8}{60} = \dfrac {-5 + (-8)}{60} = \dfrac {-5 - 8}{60} = \dfrac {-13}{60}$ .

Add the following rational numbers:
$2$ and $\dfrac {-5}{4}$ .

We can write

$2$ as

$\dfrac {2}{1}$

Let us add the fractions

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

Since it has the different denominator we need to take LCM

LCM of the denominators

$1$ and

$4$ is

$4$

So by multiplying with the denominator we get,

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

And by multiplying with the denominator we get,

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

Now we can add the fractions since it has the same denominators

$\dfrac {8}{4} + \dfrac {-5}{4} = \dfrac {8 + (-5)}{4} =\dfrac {8 - 5}{4} =\dfrac {3}{4}$ .

Add the following rational numbers:
$\dfrac {-17}{15}$ and $\dfrac {-1}{15}$ .

Let us add the fractions

$\dfrac {-17}{15} + \dfrac {-1}{15}$

Since it has the same denominator we can add directly,

$\dfrac {-17 + (-1)}{15}$

The multiplication of

$(+)(-)$ is

$(-)$ and we add the numbers.

$\therefore$ We get,

$\dfrac {-17 - 1}{15} = \dfrac {-18}{15}$ further we can divide with the common divisor i.e. by

$3$ ,

We get,

$\dfrac {-18}{15}\Rightarrow \dfrac {-6}{5}$ .

Add the following rational numbers:
$-1$ and $\dfrac {3}{4}$ .

We can write

$-1$ as

$\dfrac {-1}{1}$

Let us add the fractions

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

Since it has the different denominator we need to take LCM

LCM of the denominators

$1$ and

$4$ is

$4$

So by multiplying with the denominator we get,

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

And by multiplying with the denominator we get,

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

Now we can add the fractions since it has the same denominators

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

Verify the following:
$3 + \dfrac {-7}{12} = \dfrac {-7}{12} + 3$ .

Firstly let us consider the left hand side (LHS)

$3$ can be written as

$\dfrac {3}{1}, \dfrac {3}{1} + \dfrac {-7}{12}$

Since the denominators are different we need to take the LCM

LCM of the denominators

$1$ and

$12$ is

$12$

So by multiplying the numerator and denominator by

$12$ we get,

$\dfrac {3}{1} = \dfrac {3\times 12}{1\times 12} = \dfrac {36}{12}$

And by multiplying the numerator and denominator by

$1$ we get,

$\dfrac {-7}{12} = \dfrac {-7\times 1}{12\times 1} = \dfrac {-7}{12}$

$\therefore LHS= \dfrac {3}{1} + \dfrac {-7}{12} = \dfrac {36}{12} + \dfrac {-7}{12}$

Since it has the same denominators we can add directly

$\dfrac {36}{12} + \dfrac {-7}{12} = \dfrac {36 + (-7)}{12} = \dfrac {36 - 7}{12}\Rightarrow \dfrac {29}{12}$

Similarly,

We have right hand side (RHS):

$\dfrac {-7}{12} + \dfrac {3}{1}$

Since the denominators are different we need to take the LCM

LCM of the denominators

$12$ and

$1$ is

$12$

So by multiplying the numerator and denominator by

$1$ we get,

$\dfrac {-7}{12} = \dfrac {-7\times 1}{12\times 1} = \dfrac {-7}{12}$

And by multiplying the numerator and denominator by

$12$ we get,

$\dfrac {3}{1} = \dfrac {3\times 12}{1\times 12} = \dfrac {36}{12}$

$\therefore RHS= \dfrac {-7}{12} + \dfrac {3}{1} = \dfrac {-7}{12} + \dfrac {36}{12}$

Since it has the same denominators we can add directly

$\dfrac {-7}{12} + \dfrac {36}{12} = \dfrac {-7 + 36}{12}\Rightarrow \dfrac {29}{12}$

$LHS = RHS$ i.e.

$\dfrac {29}{12} = \dfrac {29}{12} \therefore 3 + \dfrac {-7}{12} = \dfrac {-7}{12} + 3$ is verified.

Verify the following:
$\left (\dfrac {3}{4} + \dfrac {-2}{5}\right ) + \dfrac {-7}{10} = \dfrac {3}{4} + \left (\dfrac {-2}{5} + \dfrac {-7}{10}\right )$ .

Firstly considering the LHS

$=\left (\dfrac {3}{4} + \dfrac {-2}{5}\right ) + \dfrac {-7}{10}$

Solving inside the bracket firstly by taking the LCM of

$4$ and

$5$ i.e.

$20$ and cross multplying using the denominators.

$\left (\dfrac {3\times 5 + (-2) \times 4}{20}\right ) + \dfrac {-7}{10} = \left (\dfrac {15 - 8}{20}\right ) + \dfrac {-7}{10} = \dfrac {7}{20} + \dfrac {-7}{10}$

Now considering

$\dfrac {7}{20} + \dfrac {-7}{10}$ , the LCM of

$20$ and

$10$ is

$20$

$\dfrac {7}{20} + \dfrac {-7}{10} = \dfrac {7\times 1 + (-7) \times 2}{20} = \dfrac {7 - 14}{20}\Rightarrow \dfrac {-7}{20}$

$RHS= \dfrac {3}{4} + \left (\dfrac {-2}{5} + \dfrac {-7}{10}\right )$

Solving inside the bracket firstly by taking the LCM of

$5$ and

$10$ i.e.

$10$ and cross multiply using the denominators.

$\dfrac {3}{4} + \left (\dfrac {(-2) \times 2 + (-7) \times 1}{10}\right ) = \dfrac {3}{4} + \left (\dfrac {-4 - 7}{10}\right ) = \dfrac {3}{4} + \dfrac {-11}{10}$

Now considering

$\dfrac {3}{4} + \dfrac {-11}{10}$ , the LCM of

$4$ and

$10$ is

$20$

$\dfrac {3}{4} + \dfrac {-11}{10} = \dfrac {3\times 5 + (-11)\times 2}{20} = \dfrac {15 - 22}{20}\Rightarrow \dfrac {-7}{20}$

$\therefore LHS = RHS$ i.e.

$\left (\dfrac {3}{4} + \dfrac {-2}{5}\right ) + \dfrac {-7}{10} = \dfrac {3}{4} + \left (\dfrac {-2}{5} + \dfrac {-7}{10}\right )$ is verified.

Verify the following:
$\dfrac {-5}{8} + \dfrac {-9}{13} = \dfrac {-9}{13} + \dfrac {-5}{8}$ .

Firstly let us consider the left hand side (LHS)

$\dfrac {-5}{8} + \dfrac {-9}{13}$

Since the denominators are different we need to take the LCM

LCM of the denominators

$8$ and

$13$ is

$104$

So by multiplying the numerator and denominator by

$13$ we get,

$\dfrac {-5}{8} = \dfrac {-5\times 13}{8\times 13} = \dfrac {-65}{104}$

And by multiplying the numerator and denominator by

$8$ we get,

$\dfrac {-9}{13} = \dfrac {-9\times 8}{13\times 8} = \dfrac {-72}{104}$

$\therefore LHS \dfrac {-5}{8} + \dfrac {-9}{13} = \dfrac {-65}{104} + \dfrac {-72}{104}$

Since it has the same denominators we can add directly

$\dfrac {-65}{104} + \dfrac {-72}{104} = \dfrac {-65 + (-72)}{104} = \dfrac {-65 - 72}{104}$

$\Rightarrow \dfrac {-137}{104}$

Similarly,

We have right hand side (RHS):

$\dfrac {-9}{13} + \dfrac {-5}{8}$

Since the denominators are different we need to take the LCM

LCM of the denominators

$13$ and

$8$ is

$104$

So by multiplying the numerator and denominator by

$8$ we get,

$\dfrac {-9}{13} = \dfrac {-9\times 8}{13\times 8} = \dfrac {-72}{104}$

And by multiplying the numerator and denominator by

$13$ we get,

$\dfrac {-5}{8} = \dfrac {-5\times 13}{8\times 13} = \dfrac {-65}{104}$

$\therefore RHS= \dfrac {-9}{13} + \dfrac {-5}{8} = \dfrac {-72}{104} + \dfrac {-65}{104}$

Since it has the same denominators we can add directly

$\dfrac {-72}{104} + \dfrac {-65}{104} = \dfrac {-72 + (-65)}{104} = \dfrac {-72 - 65}{104}$

$\Rightarrow \dfrac {-137}{104}$

$LHS = RHS$ i.e.

$\dfrac {-137}{104} = \dfrac {-137}{104} \therefore \dfrac {-5}{8} + \dfrac {-9}{13} = \dfrac {-9}{13} + \dfrac {-5}{8}$ is verified.

Verify the following:
$\dfrac {-12}{5} + \dfrac {2}{7} = \dfrac {2}{7} + \dfrac {-12}{5}$ .

Firstly let us consider the left hand side (LHS)

$\dfrac {-12}{5} + \dfrac {2}{7}$

Since the denominators are different we need to take the LCM

LCM of the denominators

$5$ and

$7$ is

$35$

So by multiplying the numerator and denominator by

$7$ we get,

$\dfrac {-12}{5} = \dfrac {-12\times 7}{5\times 7} = \dfrac {-84}{35}$

And by multiplying the numerator and denominator by

$5$ we get,

$\dfrac {2}{7} = \dfrac {2\times 5}{7\times 5} = \dfrac {10}{35}$

$\therefore LHS= \dfrac {-12}{5} + \dfrac {2}{7} = \dfrac {-84}{35} + \dfrac {10}{35}$

Similarly,

We have right hand side (RHS):

$\dfrac {2}{7} + \dfrac {-12}{5}$

Since the denominators are different we need to take the LCM

LCM of the denominators

$7$ and

$5$ is

$35$

So by multiplying the numerator and denominator by

$5$ we get,

$\dfrac {2}{7} = \dfrac {2\times 5}{7\times 5} = \dfrac {10}{35}$

And by multiplying the numerator and denominator by

$7$ we get,

$\dfrac {-12}{5} = \dfrac {-12\times 7}{5\times 7} = \dfrac {-84}{35}$

$\therefore RHS =\dfrac {2}{7} + \dfrac {-12}{5} = \dfrac {10}{35} + \dfrac {-84}{35}$

Since it has the same denominators we can add directly

$\dfrac {10}{35} + \dfrac {-84}{35} = \dfrac {10 + (-84)}{35} = \dfrac {10 - 84}{35}$

$\Rightarrow \dfrac {-74}{35}$

$LHS = RHS$ i.e.

$\dfrac {-75}{35} = \dfrac {-74}{35}\therefore \dfrac {-12}{5} + \dfrac {2}{7} = \dfrac {2}{7} + \dfrac {-12}{5}$ is verified.

Fill in the blanks:
$\left (\dfrac {-13}{17}\right ) + \left (\dfrac {-12}{5}\right ) = \left (\dfrac {-12}{5}\right ) + (.....)$ .

As we all know commulative property states,

$a + b = b + a$
We can simply solve by using the property

$\therefore \left (\dfrac {-13}{17}\right ) + \left (\dfrac {-12}{5}\right ) = \left (\dfrac {-12}{5}\right ) + \left (\dfrac {-13}{17}\right )$ .

Verify the following:
$\left (\dfrac {-7}{11} + \dfrac {2}{-5}\right ) + \dfrac {-13}{22} = \dfrac {-7}{11} + \left (\dfrac {2}{-5} + \dfrac {-13}{22}\right )$ .

Firstly considering the LHS

$\left (\dfrac {-7}{11} + \dfrac {2}{-5}\right ) + \dfrac {-13}{22}$

$\dfrac {2}{-5}$ Is a negative number it has to be converted to positive number by multiplying

$-1$ .

$\dfrac {2\times -1}{-5\times -1} = \dfrac {-2}{5}\therefore LHS$ is

$\left (\dfrac {-7}{11} + \dfrac {-2}{5}\right ) + \dfrac {-13}{22}$

Solving inside the bracket firstly by taking the LCM of

$11$ and

$5$ i.e.

$55$ and cross multiplying using the denominators.

$\left (\dfrac {-7\times 5 + (-2) \times 11}{55}\right ) + \dfrac {-13}{22} = \left (\dfrac {-35 - 22}{55}\right ) + \dfrac {-13}{22} = \dfrac {-57}{55} + \dfrac {-13}{22}$

Now considering

$\dfrac {-57}{55} + \dfrac {-13}{22}$ , the LCM of

$55$ and

$22$ is

$110$ .

$\dfrac {-57 \times 2 + (-13) \times 5}{110} = \dfrac {-114 - 65}{110} \Rightarrow \dfrac {-179}{110}$

$RHS= \dfrac {-7}{11} + \left (\dfrac {2}{-5} + \dfrac {-13}{22}\right )$

$\dfrac {2}{-5}$ Is a negative number it has to be converted to positive number by multiplying

$-1$ .

$\dfrac {2\times -1}{-5\times -1} = \dfrac {-2}{5} \therefore RHS= \dfrac {-7}{11} + \left (\dfrac {-2}{5} + \dfrac {-13}{22}\right )$

Solving inside the bracket firstly by taking the LCM of

$5$ and

$22$ i.e.

$110$ and cross multiply using the denominators.

$\dfrac {-7}{11} + \left (\dfrac {(-2) \times 22 + (-13) \times 5)}{110} \right ) = \dfrac {-7}{11} + \dfrac {-44 - 65}{110} = \dfrac {-7}{11} + \dfrac {-109}{110}$

Now considering

$\dfrac {-7}{11} + \dfrac {-109}{110}$ , the LCM of

$11$ and

$110$ is

$110$ .

$\dfrac {-7\times 10 + (-109) \times 1}{110} = \dfrac {-70 - 109}{110}\Rightarrow \dfrac {-179}{110}$

$\therefore LHS = RHS$ i.e.

$\left (\dfrac {-7}{11} + \dfrac {2}{-5}\right ) + \dfrac {-13}{22} = \dfrac {-7}{11} + \left (\dfrac {2}{-5} + \dfrac {-13}{22}\right )$ is verified.

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

Firstly considering the

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

Solving inside the bracket firstly by taking the LCM of

$3$ and

$4$ i.e.

$12$ and cross multiply using the denominators.

$\dfrac {-1}{1} + \left (\dfrac {-2}{3} + \dfrac {-3}{4}\right ) = \dfrac {-1}{1} + \left (\dfrac {-2\times 4 + (-3)\times 3}{12}\right ) = \dfrac {-1}{1} + \dfrac {-8 - 9}{12} = \dfrac {-1}{1} + \dfrac {-17}{12}$

Now considering

$\dfrac {-1}{1} + \dfrac {-17}{12}$ , the LCM of

$1$ and

$12$ is

$12$ .

$\dfrac {-1\times 12 + (-17) \times 1}{12} = \dfrac {-12 - 17}{12}\Rightarrow \dfrac {-29}{12}$

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

Solving inside the bracket firstly by taking the LCM of

$1$ and

$3$ i.e.

$3$ and cross multiply using the denominators.

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

Now considering

$\dfrac {-5}{3} + \dfrac {-3}{4}$ , the LCM of

$3$ and

$4$ is

$12$ .

$\dfrac {-5}{3} + \dfrac {-3}{4} = \dfrac {-5\times 4 + (-3) \times 3}{12} = \dfrac {-20 - 9}{12}\Rightarrow \dfrac {-29}{12}$

$\therefore LHS = RHS$ i.e.

$-1 + \left (\dfrac {-2}{3} + \dfrac {-3}{4}\right ) = \left (-1 + \dfrac {-2}{3}\right ) + \dfrac {-3}{4}$ is verified.

Verify the following:
$\dfrac {2}{-7} + \dfrac {12}{-35} = \dfrac {12}{-35} + \dfrac {2}{-7}$ .

1688753_1771637_ans_b645acb38883446791c30ea5a4f1cc75.jpg

The sum of two rational numbers is $\dfrac {-1}{2}$ . If one the number is $\dfrac {5}{6}$ , find the other.

Let us consider one of the number as

$x$

Other number is

$\dfrac {5}{6}$

So,

$\dfrac {5}{6} + x = \dfrac {-1}{2}$

Now solve for

$x$

$x = \dfrac {-1}{2} - \dfrac {5}{6}$

By taking LCM of

$2$ and

$6$ is

$6$

$\dfrac {(-1\times 3) -5}{6} = \dfrac {-3 - 5}{6} = \dfrac {-8}{6}$ we further can divide to lowest terms,

$\dfrac {-8 \div 2}{6\div 2} = \dfrac {-4}{3}$

$x = \dfrac {-4}{3}\therefore$ the other rational number is

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

The sum of two rational numbers is $-2$ . If one of the numbers is $\dfrac {-14}{5}$ , find the other.

Let us consider one of the number as

$x$

Other number is

$\dfrac {-14}{5}$

According to the statement,

$\dfrac {-14}{5} + x = -2$

Now solve for

$x$ :

$x = -2 - \dfrac {-14}{5} = \dfrac {-2}{1} - \dfrac {-14}{5}$

By taking LCM of

$1$ and

$5$ is

$5$

$\dfrac {-10 - (-14)}{5} = \dfrac {-10 + 14}{5} = \dfrac {4}{5}$

$x = \dfrac {4}{5}$

$\therefore$ the other rational number is

$\dfrac {4}{5}$ .

Fill in the blanks:
$\dfrac {19}{-5} + \left (\dfrac {-3}{11} + \dfrac {-7}{8}\right ) = \left \{\dfrac {19}{-5} + (....)\right \} + \dfrac {-7}{8}$ .

As we all know associative property states,

$(a + b) + c = a + (b + c)$
We can simply solve by using the property

$\therefore \dfrac {19}{-5} + \left (\dfrac {-3}{11} + \dfrac {-7}{8}\right ) = \left \{\dfrac {19}{-5} + \left (\dfrac {-3}{11}\right )\right \} + \dfrac {-7}{8}$ .

Fill in the blanks:
$-12 + \left (\dfrac {7}{12} + \dfrac {-9}{11}\right ) = \left (12 + \dfrac {7}{12}\right ) + (....)$ .

As we all know associative property states,

$(a + b) + c = a + (b + c)$
We can simply solve by using the property

$\therefore -12 + \left (\dfrac {7}{12} + \dfrac {-9}{11}\right ) = \left (12 + \dfrac {7}{12}\right ) + \left (\dfrac {-9}{11}\right )$ .

Which of the two rational number is greater in each of the following pairs?
$\cfrac7{-9}$ or $\cfrac{-5}8$

The

$LCM$ of the denominators

$9$ and

$8$ id

$72$

$\therefore \cfrac7{-9}=\cfrac{(7\times 8)}{(-9\times 8)}=\cfrac{56}{-72}=\cfrac{-56}{72}$

and

$\cfrac{-5}{8}=\cfrac{(-5\times 9)}{(8\times 9)}=\cfrac{-45}{72}$

Now,

$-56< -45$

$\Rightarrow \cfrac{-56}{72}<\cfrac{-45}{72}$

Hence,

$\cfrac{7}{-9}<\cfrac{-5}{8}$

$\therefore \cfrac{-5}{8}$ greater.

Which of the two rational numbers is greater in each of the following pairs?
$\cfrac{-15}4$ or $\cfrac{-17}4$

Since both denominators are same therefore compare the numerators only.

So, we have

$-15 > -17$

$\therefore \cfrac{-15}4>\cfrac{-17}4$

Which of the two rational number is greater in each of the following pairs?
$\cfrac{9}{-13}$ or $\cfrac{7}{-12}$

First we write each of the given numbers with a positive denominator.

$\cfrac{9}{-13}$

$=\cfrac{(9\times (-1))}{(-13\times -1)}$

$=\cfrac{-9}{13}$

$\cfrac{7}{-12}$

$=\cfrac{(7\times (-1))}{(-12\times -1)}$

$=\cfrac{-7}{12}$

The

$LCM$ of the denominators

$13$ and

$12$ is

$156$

$\therefore\cfrac{-9}{13}=\cfrac{(-9\times (12))}{(13\times 12)}=\cfrac{-108}{156}$

and

$\cfrac{-7}{12}=\cfrac{(-7\times (13))}{(12\times 13)}=\cfrac{-91}{156}$

Now,

$-108 < -91$

$\Rightarrow \cfrac{-108}{156} < \cfrac{-91}{156}$

$\Rightarrow\cfrac{-9}{13} <\cfrac{-7}{12}$

$\Rightarrow\cfrac{9}{-13} <\cfrac{7}{-12}$

$\therefore$

$\cfrac{7}{-12}$ is greater.

Which of the two rational numbers is greater in each of the following pairs?
$\cfrac{-6}{11}$ or $\cfrac5{11}$

Since both denominators are same therefore compare the numerators only.

So, we have

$-6 < 5$

$\therefore \cfrac{-6}{11}<\cfrac5{11}$

Which of the two rational numbers is greater in each of the following pairs?
$\cfrac 58$ or $\cfrac38$

Since both denominators are same therefore we compare the numerators only.

So, we have

$5 > 3$

$\therefore \cfrac58>\cfrac38$

Which of the two rational numbers is greater in the following pair?
$\cfrac{4}{-3}$ or $\cfrac{-8}{7}$

The given rational numbers are

$\cfrac{4}{-3}$ and

$\cfrac{-8}{7}$ .

For comparing the given rational numbers, we need to make their denominators same.

The

$LCM$ of the denominators

$3$ and

$7$ is

$21$

$\therefore \cfrac4{-3}=\cfrac{(4\times 7)}{(-3\times 7)}\\$

$=\cfrac{28}{-21}\\$

$=\cfrac{-28}{21}\\$

and

$\cfrac{-8}{7}=\cfrac{(-8\times 3)}{(7\times 3)}$

$=\cfrac{-24}{21}\\$

Since,

$-28< -24, \ \cfrac{-28}{21}<\cfrac{-28}{21}$

Hence,

$\cfrac{4}{-3}<\cfrac{-8}{7}$

$\therefore \cfrac{-8}{7}$ is greater.

Which of the two rational number is greater in each of the following pairs?
$\cfrac{4}{-5}$ or $\cfrac{-7}{8}$

The

$LCM$ of the denominators

$5$ and

$8$ id

$40$

$\therefore \cfrac4{-5}=\cfrac{(4\times 8)}{(-5\times 8)}=\cfrac{32}{-40}=\cfrac{-32}{40}$

and

$\cfrac{-7}{8}=\cfrac{(-7\times 5)}{(8\times 5)}=\cfrac{-35}{40}$

Now,

$-32>-35$

$\Rightarrow \cfrac{-32}{40}>\cfrac{-35}{40}$

Hence,

$\cfrac{4}{-5}>\cfrac{-7}{8}$

$\therefore \cfrac{4}{-5}$ greater.

Fill in the blanks with the correct symbol out of $> , <$ and $=$
$\cfrac{-3}{7} ....\cfrac{6}{-13}$

First, we write each of the given numbers with a positive denominator.

$\therefore \cfrac{6}{-13}$

$=\cfrac{(6\times (-1))}{(-13\times -1)}$

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

The

$LCM$ of the denominators

$7$ and

$13$ is

$91$

$\therefore \cfrac{-3}{7}=\cfrac{(-3\times 13)}{(7\times 13)}=\cfrac{-39}{91}$

and

$\cfrac{-6}{13}=\cfrac{(-6\times 7)}{(13\times 7)}=\cfrac{-42}{91}$

Now,

$-39 > -42$

$\Rightarrow \cfrac{-39}{91} > \cfrac{-42}{91}$

$\Rightarrow \cfrac{-3}{7}> \cfrac{-6}{13}$

Hence,

$\cfrac{-3}{7}\boxed> \cfrac{6}{-13}$

Which of the two rational number is greater in each of the following pairs?
$\cfrac59$ or $\cfrac{-3}8$

The

$LCM$ of the denominators

$9$ and

$8$ id

$72$

$\therefore \cfrac59=\cfrac{(5\times 8)}{(9\times 8)}=\cfrac{40}{72}$

and

$\cfrac{-3}{8}=\cfrac{(-3\times 9)}{(8\times 9)}=\cfrac{-27}{72}$

Now,

$40 > -27$

$\Rightarrow \cfrac{40}{72}>\cfrac{-27}{72}$

Hence,

$\cfrac59>\cfrac{-3}8$

$\therefore \cfrac59$ greater.

Fill in the blanks with the correct symbol out of $> , <$ and $=$
$-2...\cfrac{13}5$

$-2$ can be written as

$\cfrac{-2}{1}$

The

$LCM$ of the denominators

$1$ and

$5$ is

$5$

$\therefore \cfrac{-2}{1}=\cfrac{(-2\times 5)}{(1\times 5)}=\cfrac{-10}{5}$

Now,

$-10<13$

$\Rightarrow \cfrac{-10}{5} < \cfrac{13}{5}$

Hence,

$-2\boxed< \cfrac{13}{5}$

Fill in the blanks with the correct symbol out of $> , <$ and $=$
$\cfrac{5}{-13}$ _____ $\cfrac{-35}{91}$

First we write each of the given numbers with a positive denominator.

$\therefore \cfrac{5}{-13}$

$=\cfrac{(5\times (-1))}{(-13\times -1)}$

$=\cfrac{-5}{13}$

The

$LCM$ of the denominators

$13$ and

$91$ is

$91$

$\therefore \cfrac{-5}{13}=\cfrac{(-5\times 7)}{(13\times 7)}=\cfrac{-35}{91}$

Hence,

$\cfrac{5}{-13} = \cfrac{-35}{91}$

Add the following rational numbers:
$\cfrac{12}7$ and $\cfrac37$

We have to add

$\cfrac{12}7$ and

$\cfrac37$

As, their denominator is same.

So, denominator will become common denominator of addition and terms in numerator will be added.

$\therefore \cfrac{12}7+\cfrac37=\cfrac{12+3}7=\cfrac{15}7$

Fill in the blanks with the correct symbol out of $> , <$ and $=$
$0...\cfrac{-3}{-15}$

$\cfrac{-3}{-15}==\cfrac{(-3\times (-1))}{(-15\times(-1))}=\cfrac{3}{15}$ , which is positive number

All positive numbers are greater than

$0$

$\therefore 0\ \boxed<\ \cfrac{-3}{-15}$

Fill in the blanks with the correct symbol out of $> , <$ and $=$
$\cfrac{-8}{9}...\cfrac{-9}{10}$

The

$LCM$ of the denominators

$9$ and

$10$ is

$90$

$\therefore\cfrac{-8}{9}=\cfrac{(-8\times 10)}{(9\times 10)}=\cfrac{-80}{90}$

and

$\cfrac{-9}{10}=\cfrac{(-9\times 9)}{(10\times 9)}=\cfrac{-81}{90}$

Now,

$-80 > -81$

$\Rightarrow \cfrac{-80}{90} > \cfrac{-81}{90}$

$\Rightarrow\cfrac{-8}{9}\ \boxed> \ \cfrac{-9}{10}$

Fill in the blanks with the correct symbol out of $> , <$ and $=$ :-
$\cfrac{-2}{3}.... \cfrac{5}{-8}$

The given numbers are,

$\cfrac{-2}{3}$ and

$\cfrac{5}{-8}$

First we have to write each of the given numbers with a positive denominator.

$\therefore \cfrac{5}{-8}$

$=\cfrac{5\times (-1)}{-8\times (-1)}$

$=\cfrac{-5}{8}$

The

$LCM$ of the denominators

$3$ and

$8$ is

$24$

$\therefore \cfrac{-2}{3}=\cfrac{(-2\times 8)}{(3\times 8)}=\cfrac{-16}{24}$

and

$\cfrac{-5}{8}=\cfrac{(-5\times 3)}{(8\times 3)}=\cfrac{-15}{24}$

Now,

$-16 < -15$

$\Rightarrow \cfrac{-16}{24} < \cfrac{-15}{24}$

$\Rightarrow \cfrac{-2}{3}< \cfrac{-5}{8}$

Hence,

$\cfrac{-2}{3}< \cfrac{5}{-8}$

Add the following rational numbers:
$\cfrac{-2}5$ and $\cfrac15$

We have to add

$\cfrac{-2}5$ and

$\cfrac15$

As, their denominator is same.

So, denominator will become common denominator of addition and terms in numerator will be added.

$\therefore \cfrac{-2}5+\cfrac15=\cfrac{-2+1}5=\cfrac{-1}5$

Add the following rational numbers:
$\cfrac{-3}{7}\ and \ \cfrac{5}{-7}$

To, add rational numbers their denominators should be same, so it will become common and numerators will be added to get answer.

$\therefore \cfrac{5}{-7}=\cfrac{(5\times (-1))}{(-7\times(-1))}=\cfrac{-5}{7}$ and

$\cfrac{-3}{7}$

Now, their denominator is same.

So,

$\cfrac{-3}{7}+\cfrac{-5}{7}=\cfrac{-3-5}{7}=\cfrac{-8}{7}$

Add the following rational numbers:
$\cfrac{-2}9$ and $\cfrac{-5}9$

To, add rational numbers their denominators should be same, so it will become common and numerators will be added to get answer.

As, their denominator is already same.

So,

$\cfrac{-2}{9}+\cfrac{-5}{9}=\cfrac{-2-5}{9}=\cfrac{-7}{9}$

Add the following rational numbers:
$-4$ and $\cfrac12$

To, add rational numbers their denominators should be same, so it will become common and numerators will be added to get answer.

$-4$ can also be written as

$\cfrac{-4}1$

$LCM$ of

$1$ and

$2$ is

$2$

$\therefore \cfrac{-4}{1}=\cfrac{(-4\times (2))}{(1\times(2))}=\cfrac{-8}{2}$

Now, their denominator is same.

So,

$\cfrac{-8}{2}+\cfrac{1}{2}=\cfrac{-8+1}{2}=\cfrac{-7}{2}$

Add the following rational numbers:
$\cfrac{-5}{11}$ and $\cfrac{7}{-11}$

To, add rational numbers their denominators should be same, so it will become common and numerators will be added to get answer.

$\therefore \cfrac{7}{-11}=\cfrac{(7\times (-1))}{(-11\times(-1))}=\cfrac{-7}{11}$ and

$\cfrac{-5}{11}$

Now, their denominator is same.

So,

$\cfrac{-5}{11}+\cfrac{-7}{11}=\cfrac{-5-7}{11}=\cfrac{-12}{11}=\cfrac{-12}{11}$

Add the following rational numbers:
$\cfrac{9}{-13}\ and \ \cfrac{-11}{-13}$

To, add rational numbers their denominators should be same, so it will become common and numerators will be added to get answer.

As, their denominator is already same.

So,

$\cfrac{9}{-13}+\cfrac{-11}{-13}=\cfrac{9-11}{-13}=\cfrac{-2}{-13}=\cfrac{2}{13}$

Add the following rational numbers:
$\cfrac3{-8}$ and $\cfrac18$

To, add rational numbers their denominators should be same, so it will become common and numerators will be added to get answer.

$\therefore \cfrac{3}{-8}=\cfrac{(3\times (-1))}{-8\times(-1))}=\cfrac{-3}{8}$ and

$\cfrac18$

Now, their denominator is same.

So,

$\cfrac{-3}8+\cfrac18=\cfrac{-3+1}{8}=\cfrac{-2}{8}=\cfrac{-1}4$

Express $\cfrac58$ as a rational number with numerator $15$

To get

$15$ in the numerator multiply by

$3$ for both numerator and denominator.

Then we get,

$\cfrac58=\cfrac{5\times 3}{8\times 3}=\cfrac{15}{24}$

Add the following rational numbers:
$\cfrac{-2}5$ and $\cfrac 34$

To, add rational numbers their denominators should be same, so it will become common and numerators will be added to get answer.

$LCM$ of

$5$ and

$4$ is

$20$

$\therefore \cfrac{-2}{5}=\cfrac{(-2\times (4))}{(5\times(4))}=\cfrac{-8}{20}$

and

$\cfrac{3}{4}=\cfrac{(3\times (5))}{(4\times(5))}=\cfrac{15}{20}$

Now, their denominator is same.

So,

$\cfrac{-8}{20}+\cfrac{15}{20}=\cfrac{-8+15}{20}=\cfrac{7}{20}$

Add the following rational numbers:
$\cfrac{-5}{9}\ and \ \cfrac{2}{3}$

To, add rational numbers their denominators should be same, so it will become common and numerators will be added to get answer.

$LCM$ of

$9$ and

$3$ is

$9$

$\therefore \cfrac{-5}{9}=\cfrac{(-5\times (1))}{(9\times(1))}=\cfrac{-5}{9}$

and

$\cfrac{2}{3}=\cfrac{(2\times (3))}{(3\times(3))}=\cfrac{6}{9}$

Now, their denominator is same.

So,

$\cfrac{-5}{9}+\cfrac{6}{9}=\cfrac{-5+6}{9}=\cfrac{1}{9}$

Add the following rational numbers:
$\cfrac{-17}9$ and $\cfrac{-11}9$

To, add rational numbers their denominators should be same, so it will become common and numerators will be added to get answer.

As, their denominator is already same.

So,

$\cfrac{-17}{9}+\cfrac{-11}{9}=\cfrac{-17-11}{9}=\cfrac{-28}{9}$

Write each of the following rational numbers with positive denominators: $\dfrac{5}{-8}, \dfrac{15}{-28}, \dfrac{-17}{-13}$

Rational numbers with positive denominators

$\Rightarrow \dfrac{5}{-8}=\dfrac{5\times-1}{-8\times -1}=\dfrac{-5}{8}$

$\Rightarrow \dfrac{15}{-28}=\dfrac{15\times-1}{-28\times-1}=\dfrac{-15}{28}$

$\Rightarrow \dfrac{-17}{-13}=\dfrac{-17\times-1}{-13\times-1}=\dfrac{17}{13}$

Subtract:
$\cfrac{-8}{9}$ from $\cfrac{-3}{5}$

We have to subtract

$\cfrac{-8}{9}$ from

$\cfrac{-3}{5}$ i.e

$\cfrac{-3}{5}-$

$\left(\cfrac{-8}{9}\right)$

To subtract rational numbers their denominator should be same, so we can simply subtract their numerator and denominator will remain common.

$LCM$ of

$9$ and

$5$ is

$45$

Let's express each of the given rational numbers with the above

$LCM$ as the common denominator.

Now,

$\cfrac{-3}{5}=\cfrac{(-3\times (9))}{(5\times(9))}=\cfrac{-27}{45}$

$\cfrac{-8}{9}=\cfrac{(-8\times (5))}{(9\times(5))}=\cfrac{-40}{45}$

So,

$\cfrac{-3}{5}-$

$\left(\cfrac{-8}{9}\right)$

$=$

$\cfrac{-27}{45}-$

$\left(\cfrac{-40}{45}\right)=\cfrac{-27+40}{45}$

$=\cfrac{40-27}{45}$

$=\cfrac{13}{45}$

Add the following rational numbers:
$\cfrac{27}{-4}$ and $\cfrac{-15}8$

let's make numbers with positive denominator

$\therefore \cfrac{27}{-4}=\cfrac{(27\times (-1))}{(-4\times(-1))}=\cfrac{-27}{4}$

To, add rational numbers their denominators should be same, so it will become common and numerators will be added to get answer.

$LCM$ of

$4$ and

$8$ is

$8$

$\therefore \cfrac{-27}{4}=\cfrac{(-27\times (2))}{(4\times(2))}=\cfrac{-54}{8}$

Now, their denominator is same.

So,

$\cfrac{-54}{8}+\cfrac{-15}{8}=\cfrac{-54-15}{8}=\cfrac{-69}{8}$

Subtract:
$\cfrac{-9}{7}$ from $-1$

$-1$ can be written as

$\cfrac{-1}{1}$

We have to subtract

$\cfrac{-9}{7}$ from

$-1$ i.e

$-1-$

$\left(\cfrac{-9}{7}\right)=$

$\cfrac{-1}{1}-$

$\left(\cfrac{-9}{7}\right)$

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

$7$ is

$7$

Let's express each of the given rational numbers with the above

$LCM$ as the common denominator.

Now,

$\cfrac{-1}{1}=\cfrac{(-1\times (7))}{(1\times(7))}=\cfrac{-7}{7}$

So,

$-1-$

$\left(\cfrac{-9}{7}\right)=$

$\cfrac{-1}{1}-$

$\left(\cfrac{-9}{7}\right)$

$=\cfrac{-7}{7}-$

$\left(\cfrac{-9}{7}\right)$

$=\cfrac{-7+9}{7}$

$=\cfrac{2}{7}$

Add the following rational numbers:
$\cfrac{-7}{27}$ and $\cfrac5{18}$

To, add rational numbers their denominators should be same, so it will become common and numerators will be added to get answer.

$LCM$ of

$27$ and

$18$ is

$54$

$\therefore \cfrac{-7}{27}=\cfrac{(-7\times (2))}{(27\times(2))}=\cfrac{-14}{54}$

and

$\cfrac{5}{18}=\cfrac{(5\times (3))}{(18\times(3))}=\cfrac{15}{54}$

Now, their denominator is same.

So,

$\cfrac{14}{54}+\cfrac{15}{54}=\cfrac{-14+15}{54}=\cfrac{1}{54}$

Add the following rational numbers:
$\cfrac{-9}{24}$ and $\cfrac{-1}{18}$

To add rational numbers their denominators should be same, so it will become common and numerators will be added to get answer.

$LCM$ of

$18$ and

$24$ is

$72$

$\therefore \cfrac{-9}{24}=\cfrac{(-9\times (3))}{(24\times(3))}=\cfrac{-27}{72}$

and

$\cfrac{-1}{18}=\cfrac{(-1\times (4))}{(18\times(4))}=\cfrac{-4}{72}$

Now, their denominator is same.

So,

$\cfrac{-27}{72}+\cfrac{-4}{72}=\cfrac{-27-4}{72}=\cfrac{-31}{72}$

Add the following rational numbers:
$\cfrac{-5}{36}$ and $\cfrac{-7}{12}$

To, add rational numbers their denominators should be same, so it will become common and numerators will be added to get answer.

$LCM$ of

$36$ and

$12$ is

$36$

$\therefore \cfrac{-7}{12}=\cfrac{(-7\times (3))}{(12\times(3))}=\cfrac{-21}{36}$

Now, their denominator is same.

So,

$\cfrac{-5}{36}+\cfrac{-21}{36}=\cfrac{-5-21}{36}=\cfrac{-26}{36}$

Subtract:
$\cfrac{-18}{11}$ from $1$

$1$ can be written as

$\cfrac{1}{1}$

We have to subtract

$\cfrac{-18}{11}$ from

$1$ i.e

$1-$

$\left(\cfrac{-18}{11}\right)=$

$\cfrac{1}{1}-$

$\left(\cfrac{-18}{11}\right)$

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

$11$ is

$11$

Let's express each of the given rational numbers with the above

$LCM$ as the common denominator.

Now,

$\cfrac{1}{1}=\cfrac{(1\times (11))}{(1\times(11))}=\cfrac{11}{11}$

So,

$1-$

$\left(\cfrac{-18}{11}\right)=$

$\cfrac{1}{1}-$

$\left(\cfrac{-18}{11}\right)$

$=\cfrac{11}{11}-$

$\left(\cfrac{-18}{11}\right)$

$=\cfrac{11+18}{11}$

$=\cfrac{29}{11}$

Subtract:
$\cfrac{-5}{6}$ from $\cfrac13$

We have to subtract

$\cfrac{-5}{6}$ from

$\cfrac{1}{3}$ i.e

$\cfrac{1}{3}-$

$\left(\cfrac{-5}{6}\right)$

To subtract rational numbers their denominator should be same, so we can simply subtract their numerator and denominator will remain common.

$LCM$ of

$6$ and

$3$ is

$6$

Let's express each of the given rational numbers with the above

$LCM$ as the common denominator.

Now,

$\cfrac{1}{3}=\cfrac{(1\times (2))}{(3\times(2))}=\cfrac{2}{6}$

So,

$\cfrac{1}{3}-$

$\left(\cfrac{-5}{6}\right)$

$=\cfrac{2}{6}-$

$\left(\cfrac{-5}{6}\right)$

$=\cfrac{2+5}{6}$

$=\cfrac{7}{6}$

Subtract:
$\cfrac{3}{4}$ from $\cfrac{1}{3}$

We have to subtract

$\cfrac{3}{4}$ from

$\cfrac{1}{3}$ i.e

$\cfrac{1}{3}-$

$\cfrac{3}{4}$

To subtract rational numbers their denominator should be same, so we can simply subtract their numerator and denominator will remain common.

$LCM$ of

$4$ and

$3$ is

$12$

Let's express each of the given rational numbers with the above

$LCM$ as the common denominator.

Now,

$\cfrac{3}{4}=\cfrac{(3\times (3))}{(4\times(3))}=\cfrac{9}{12}$

$\cfrac{1}{3}=\cfrac{(1\times (4))}{(3\times(4))}=\cfrac{4}{12}$

So,

$\cfrac{1}{3}-$

$\cfrac{3}{4}$

$=\cfrac{4}{12}-\cfrac{9}{12}$

$=\cfrac{4-9}{12}$

$=\cfrac{-5}{12}$

Write the next three rational numbers to complete the pattern:
$\dfrac{4}{-5},\dfrac{8}{-10},\dfrac{12}{-15},\dfrac{16}{-20},$ , , , .

Next three equivalent rational numbers are

$\Rightarrow \dfrac{4}{-5}$

$\Rightarrow \dfrac{4}{-5}=\dfrac{4\times5}{-5\times5}=\dfrac{20}{-25}$

$\Rightarrow \dfrac{4}{-5}=\dfrac{4\times6}{-5\times6}=\dfrac{28}{-35}$

$\Rightarrow \dfrac{4}{-5}=\dfrac{4\times7}{-5\times7}=\dfrac{28}{-35}$
Therefore, next equivalent rational numbers of given number are

$\dfrac{20}{-25},\dfrac{24}{-30}$ and

$\dfrac{28}{-35}.$

Add the following rational numbers:
$\cfrac1{-9}$ and $\cfrac4{-27}$

let's make numbers with positive denominator

$\therefore \cfrac{1}{-9}=\cfrac{(1\times (-1))}{(-9\times(-1))}=\cfrac{-1}{9}$

$\therefore \cfrac{4}{-27}=\cfrac{(4\times (-1))}{(-27\times(-1))}=\cfrac{-4}{27}$

To, add rational numbers their denominators should be same, so it will become common and numerators will be added to get answer.

$LCM$ of

$9$ and

$27$ is

$27$

$\therefore \cfrac{-1}{9}=\cfrac{(-1\times (3))}{(9\times(3))}=\cfrac{-3}{27}$

Now, their denominator is same.

So,

$\cfrac{-3}{27}+\cfrac{-4}{27}=\cfrac{-3-4}{27}=\cfrac{-7}{27}$

Subtract:
$\cfrac{5}{1}$ from $\cfrac{-3}{5}$

$5$ can also be written as

$\cfrac51$

We have to subtract

$\cfrac{5}{1}$ from

$\cfrac{-3}{5}$

i.e

$\cfrac{-3}{5}$

$-\left(\cfrac{5}{1}\right)$

To subtract rational numbers their denominator should be same, so we can simply subtract their numerator and denominator will remain common.

$LCM$ of

$5$ and

$1$ is

$5$

Let's express each of the given rational numbers with the above

$LCM$ as the common denominator.

Now,

$\cfrac{5}{1}=\cfrac{(5\times (5))}{(1\times(5))}=\cfrac{25}{5}$

So,

$\cfrac{-3}{5}$

$-\left(\cfrac{5}{1}\right)=\cfrac{-3}{5}-\left(\cfrac{25}{5}\right)$

$=\cfrac{-3-25}{5}$

$=\cfrac{-28}{5}$

Evaluate:
$\cfrac{3}{4}$ $-\left(\cfrac{4}{5}\right)$

$\cfrac{3}{4}$

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

To subtract rational numbers their denominator should be same, so we can simply subtract their numerator and denominator will remain common.

$LCM$ of

$4$ and

$5$ is

$20$

Let's express each of the given rational numbers with the above

$LCM$ as the common denominator.

Now,

$\cfrac{3}{4}=\cfrac{(3\times (5))}{(4\times(5))}=\cfrac{15}{20}$

$\cfrac{4}{5}=\cfrac{(4\times (4))}{(5\times(4))}=\cfrac{16}{20}$

So,

$\cfrac{3}{4}$

$-\left(\cfrac{4}{5}\right)=\cfrac{15}{20}-\left(\cfrac{16}{20}\right)$

$=\cfrac{15-16}{20}$

$=\cfrac{-1}{20}$

Evaluate:
$\cfrac{7}{24}$ $-\left(\cfrac{19}{36}\right)$

$\cfrac{7}{24}$

$-\left(\cfrac{19}{36}\right)$

To subtract rational numbers their denominator should be same, so we can simply subtract their numerator and denominator will remain common.

$LCM$ of

$24$ and

$36$ is

$72$

Let's express each of the given rational numbers with the above

$LCM$ as the common denominator.

Now,

$\cfrac{7}{24}=\cfrac{(7\times (3))}{(24\times(3))}=\cfrac{21}{72}$

$\cfrac{19}{36}=\cfrac{(19\times (2))}{(36\times(2))}=\cfrac{38}{72}$

So,

$\cfrac{7}{24}$

$-\left(\cfrac{19}{36}\right)$

$=\cfrac{21}{72}-\cfrac{38}{72}$

$=\cfrac{21-38}{72}$

$=\cfrac{-17}{72}$

Subtract:
$\cfrac{5}{9}$ from $\cfrac{-2}{3}$

We have to subtract

$\cfrac{5}{9}$ from

$\cfrac{-2}{3}$

i.e

$\cfrac{-2}{3}$

$-\left(\cfrac{5}{9}\right)$

To subtract rational numbers their denominator should be same, so we can simply subtract their numerator and denominator will remain common.

$LCM$ of

$9$ and

$3$ is

$9$

Let's express each of the given rational numbers with the above

$LCM$ as the common denominator.

Now,

$\cfrac{-2}{3}=\cfrac{(-2\times (3))}{(3\times(3))}=\cfrac{-6}{9}$

So,

$\cfrac{-2}{3}$

$-\left(\cfrac{5}{9}\right)=\cfrac{-6}{9}-\left(\cfrac{5}{9}\right)$

$=\cfrac{-6-5}{9}$

$=\cfrac{-11}{9}$

Subtract:
$7$ from $\cfrac{-4}{7}$

$7$ can be written as

$\cfrac{7}{1}$

We have to subtract

$7$ from

$\cfrac{-4}{7}$

$\left(\cfrac{-4}{7}\right)-$

$7=\cfrac{-4}7-\cfrac{7}{1}$

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

$7$ is

$7$

Let's express each of the given rational numbers with the above

$LCM$ as the common denominator.

Now,

$\cfrac{7}{1}=\cfrac{(7\times (7))}{(1\times(7))}=\cfrac{49}{7}$

So,

$\left(\cfrac{-4}{7}\right)-$

$7=\cfrac{-4}7-\cfrac{7}{1}$

$=\cfrac{-4}{7}-$

$\left(\cfrac{49}{7}\right)$

$=\cfrac{-4-49}{7}$

$=\cfrac{-53}{7}$

Evaluate:
$\cfrac{4}{9}$ $-\left(\cfrac{2}{-3}\right)$

$\cfrac{4}{9}$

$-\left(\cfrac{2}{-3}\right)$

$\cfrac{2}{-3}$ can also be written as

$\cfrac{-2}{3}$ i.e

$\cfrac{4}{9}$

$-\left(\cfrac{-2}{3}\right)$

To subtract rational numbers their denominator should be same, so we can simply subtract their numerator and denominator will remain common.

$LCM$ of

$9$ and

$3$ is

$9$

Let's express each of the given rational numbers with the above

$LCM$ as the common denominator.

Now,

$\cfrac{-2}{3}=\cfrac{(-2\times (3))}{(3\times(3))}=\cfrac{-6}{9}$

So,

$\cfrac{4}{9}$

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

$\cfrac{4}{9}$

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

$\cfrac{4}{9}-\left(\cfrac{-6}{9}\right)$

$=\cfrac{4+6}{9}$

$=\cfrac{10}{9}$

Subtract:
$\cfrac{-32}{13}$ from $\cfrac{-6}{5}$

We have to subtract

$\cfrac{-32}{13}$ from

$\cfrac{-6}{5}$

i.e

$\cfrac{-6}{5}$

$-\left(\cfrac{-32}{13}\right)$

To subtract rational numbers their denominator should be same, so we can simply subtract their numerator and denominator will remain common.

$LCM$ of

$5$ and

$13$ is

$65$

Let's express each of the given rational numbers with the above

$LCM$ as the common denominator.

Now,

$\cfrac{-6}{5}=\cfrac{(-6\times (13))}{(5\times(13))}=\cfrac{-78}{65}$

$\cfrac{-32}{13}=\cfrac{(-32\times (5))}{(13\times(5))}=\cfrac{-160}{65}$

So,

$\cfrac{-6}{5}$

$-\left(\cfrac{-32}{13}\right)=\cfrac{-78}{65}-\left(\cfrac{-160}{65}\right)$

$=\cfrac{-78+160}{65}$

$=\cfrac{82}{65}$

Find the sum of
$\dfrac{8}{13}$ and $\dfrac{3}{11}$

$\Rightarrow \dfrac{8}{13}$ and

$\dfrac{3}{11}$

$=\dfrac{8}{13}+\dfrac{3}{11}$

$=\dfrac{8\times11}{13\times11}+\dfrac{3\times11}{11\times13}$

$=\dfrac{88}{143}+\dfrac{39}{143}$

$=\dfrac{127}{143}$
Therefore, the sum of

$\dfrac{8}{13}$ and

$\dfrac{3}{11}$ is

$\dfrac{127}{143}$

Evaluate:
$\cfrac{14}{15}$ $-\left(\cfrac{13}{20}\right)$

$\cfrac{14}{15}$

$-\left(\cfrac{13}{20}\right)$

To subtract rational numbers their denominator should be same, so we can simply subtract their numerator and denominator will remain common.

$LCM$ of

$15$ and

$20$ is

$60$

Let's express each of the given rational numbers with the above

$LCM$ as the common denominator.

Now,

$\cfrac{14}{15}=\cfrac{(14\times (4))}{(15\times(4))}=\cfrac{56}{60}$

$\cfrac{13}{20}=\cfrac{(13\times (3))}{(20\times(3))}=\cfrac{39}{60}$

So,

$\cfrac{14}{15}$

$-\left(\cfrac{13}{20}\right)$

$=\cfrac{56}{60}-\cfrac{33}{60}$

$=\cfrac{56-39}{60}$

$=\cfrac{17}{60}$

Evaluate:
$-3-\left(\cfrac{-4}{7}\right)$

$-3-\left(\cfrac{-4}{7}\right)$

$-3$ ca be written as

$\cfrac{-3}{1}$

$-\left(\cfrac{-4}{7}\right)$

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

$7$ is

$7$

Let's express each of the given rational numbers with the above

$LCM$ as the common denominator.

Now,

$\cfrac{-3}{1}=\cfrac{(-3\times (7))}{(1\times(7))}=\cfrac{-21}{7}$

So,

$-3-\left(\cfrac{-4}{7}\right)$ =

$\cfrac{-3}{1}$

$-\left(\cfrac{-4}{7}\right)=\cfrac{-21}{7}-\left(\cfrac{-4}{7}\right)$

$=\cfrac{-21+4}{7}$

$=\cfrac{-17}{7}$

Subtract:
$\cfrac{-13}9$ from $0$

We have to subtract

$\cfrac{-13}9$ from

$0$ i.e

$0-\left(\cfrac{-13}9\right)$

$=0+\left(\cfrac{13}9\right)$

$=\cfrac{13}9$

Evaluate:
$\cfrac{-5}{-8}$ $-\left(\cfrac{-3}{4}\right)$

$\cfrac{-5}{-8}$

$-\left(\cfrac{-3}{4}\right)$

$\cfrac{-5}{-8}$ can also be written as

$\cfrac{5}{8}$ i.e

$\Rightarrow\cfrac{5}{8}$

$-\left(\cfrac{-3}{4}\right)$

To subtract rational numbers their denominator should be same, so we can simply subtract their numerator and denominator will remain common.

$LCM$ of

$8$ and

$4$ is

$8$

So, let's make numbers with same denominator equal to their

$LCM$

$\cfrac{-3}4=\cfrac{(-3\times (2))}{(4\times (2))}=\cfrac{-6}{8}$

So,

$\cfrac{-5}{-8}$

$-\left(\cfrac{-3}{4}\right)=\cfrac{5}{8}$

$-\left(\cfrac{-6}{8}\right)$

$=\cfrac{5+6}{8}$

$=\cfrac{11}{8}$

Evaluate the following.
$| 5 - 13 |$

$| 5 - 13 | = |-8|$

Since

$|-8| = 8$

$\therefore \, | 5 - 13 | = 8$

Subtract the sum of $\cfrac{-36}{11}$ and $\cfrac{49}{22}$ from the sum of $\cfrac{33}{8}$ and $\cfrac{-19}{4}$

First we have to find the sum of

$\cfrac{-36}{11}$ and

$\cfrac{49}{22}$

$=\cfrac{-36}{11}+\cfrac{49}{22}$

$LCM$ of

$11$ and

$22$ is

$22$

Express each of the given rational numbers with the above

$LCM$ as the common denominator.

Now,

$\cfrac{-36}{11}=\cfrac{-36\times 2}{11\times 2}=\cfrac{-72}{22}$

As their denominator are now same.

So,

$\cfrac{-36}{11}+\cfrac{49}{22}$

$=\cfrac{-72}{22}+\cfrac{49}{22}$

$=\cfrac{-72+49}{22}$

$=\cfrac{-23}{22}$

Now, we have to find the sum of

$\cfrac{33}{8}$ and

$\cfrac{-19}{4}$

$=\cfrac{33}{8}+\left(\cfrac{-19}{4}\right)$

$LCM$ of

$8$ and

$4$ is

$8$

Express each of the given rational numbers with the above

$LCM$ as the common denominator.

Now,

$\cfrac{-19}{4}=\cfrac{-19\times 2}{4\times 2}=\cfrac{-38}{8}$

As their denominator are now same.

So,

$\cfrac{33}{8}+\left(\cfrac{-19}{4}\right)$

$=\cfrac{33}{8}+\left(\cfrac{-38}{8}\right)$

$=\cfrac{33-38}{8}$

$=\cfrac{-5}{8}$

Now, we have to subtract

$1^{st}$ total from

$2^{nd}$ total i.e,

$\cfrac{-5}{8}-\left(\cfrac{-23}{22}\right)$

$=\cfrac{-5}{8}+\cfrac{23}{22}$

$LCM$ of

$8$ and

$22$ is

$88$

Express each of the given rational numbers with the above

$LCM$ as the common denominator.

Now,

$\cfrac{-5}{8}=\cfrac{-5\times 11}{8\times 11}=\cfrac{-55}{88}$

$\cfrac{23}{22}=\cfrac{23\times 4}{22\times 4}=\cfrac{92}{88}$

So,

$\cfrac{-5}{8}+\cfrac{23}{22}$

$=\cfrac{-55}{88}+\left(\cfrac{92}{88}\right)$

$=\cfrac{-55+92}{88}$

$=\cfrac{37}{88}$

The sum of two rational numbers is $(-3)$ . If one of them is $(-15/7)$ , find the other.

Let the required number be

$x$ . Then,

$=(-15/7)+x=(-3)$

By sending

$(-15/7)$ from left hand side to the right hand side it change to

$-(15/7)$

$x=(-3)-(-15/7)$

$LCM$ of

$1$ and

$7$ is

$7$

Express each of the given rational numbers with the above

$LCM$ as the common denominator.

Now,

$\dfrac{(-3\times 7)}{(1\times 7)}=\dfrac{-21}{7}$

$=\dfrac{(-15\times 1)}{(7\times 1)}=\dfrac{-15}{7}$

Then,

$=(-21/7)-(15/7)$

$=-(21+15)/7$

$=-6/7$

Hence the required number is

$(-6/7)$

The sum of two rational number is $(-3/8)$ . If one of them is $(3/16)$ , find the other.

Let the required number be

$x$ . Then,

$=(3/16)+x=(-3/8)$
By sending

$(3/16)$ from left hand side to the right hand side it change to

$-(3/16)$

$x=(-3/8)-(3/16)$

$LCM$ of

$8$ and

$16$ is

$16$
Express each of the given rational numbers with the above

$LCM$ as the common denominator.
Now,

$\dfrac{(-3\times 2)}{(8\times 2)}=\dfrac{-6}{16}$

$=\dfrac{(3\times 1)}{(16\times 1)}=\dfrac{3}{16}$
Then,

$=\dfrac{-6}{16}-\dfrac{3}{16}$

$=\dfrac{-6-3}{16}$

$=\dfrac{-9}{16}$
Hence the required number is

$=\dfrac{-9}{16}$

Evaluate:
$\cfrac{7}{11}$ $-\left(\cfrac{-4}{-11}\right)$

$\cfrac{7}{11}$

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

$\cfrac{-4}{-11}$ can also be written as

$\cfrac{4}{11}$ i.e

$\Rightarrow\cfrac{7}{11}$

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

To subtract rational numbers their denominator should be same, so we can simply subtract their numerator and denominator will remain common.

As, their denominator is already same,

So,

$\cfrac{7}{11}$

$-\left(\cfrac{-4}{-11}\right)=$

$\cfrac{7}{11}$

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

$=\cfrac{7-4}{11}$

$=\cfrac{3}{11}$

Evaluate the following.
$|-8| + |-6|$

$| -8 | + | -6 |$
Since

$|-8| = 8$ and

$|-6| = 6$

$\therefore \, 8 + 6 = 14$

Evaluate the following.
$| -11 | + | 9 |$

$| -11 | + | 9 |$

Since

$| -11 | = 11$ and

$| 9 | = 9$

$\therefore \, 11 + 9 = 20$

Evaluate:
$\cfrac{-5}{14}-\left(\cfrac{-2}{7}\right)$

$\cfrac{-5}{14}$

$-\left(\cfrac{-2}{7}\right)$

To subtract rational numbers their denominator should be same, so we can simply subtract their numerator and denominator will remain common.

$LCM$ of

$14$ and

$7$ is

$14$

Let's express each of the given rational numbers with the above

$LCM$ as the common denominator.

Now,

$\cfrac{-2}{7}=\cfrac{(-2\times (2))}{(7\times(2))}=\cfrac{-4}{14}$

So,

$\cfrac{-5}{14}$

$-\left(\cfrac{-2}{7}\right)=\cfrac{-5}{14}-\left(\cfrac{-4}{14}\right)$

$=\cfrac{-5+4}{14}$

$=\cfrac{-1}{14}$

$\dfrac{-3}{7} \div \dfrac{-7}{3}=$ __________

$\dfrac{-3}{7}\times\dfrac{-3}{7}=\dfrac{9}{49}$

The sum of two rational number is $(4/21)$ . If one of them is $(5/7)$ , find the other.

Let the required number be

$x$ . Then,

$=(5/7)+x=(4/21)$
By sending

$(5/7)$ from left hand side to the right hand side it change to

$-(5/7)$

$x=(4/21)-(5/7)$

$LCM$ of

$21$ and

$7$ is

$21$
Express each of the given rational numbers with the above

$LCM$ as the common denominator.
Now,

$\dfrac{(4\times 1)}{(21\times 1)}=\dfrac{4}{21}$

$=\dfrac{(5\times 3)}{(7\times 3)}=\dfrac{15}{21}$
Then,

$=(4/21)-(15/21)$

$=(4-15)/21$

$=(-11/21)$
Hence the required number is

$(-11/21)$

Evaluate the following.
$|-19| - |-13|$

$|-19| - |-13|$
Since

$|-19| = 19$ and

$|-13| = 13$

$\therefore \, 19 - 13 = 6$

Evaluate :
$|35 - 41| - |7 - (-2)|$

$|35 - 41| - |7 - (-2)|$

$=|-6|-|9|$

$= 6 - 9 = -3$

Evaluate the following.
$|7| - |-3|$

$|7| - |-3|$
Since

$|7| = 7$ and

$|-3| = 3$

$\therefore \, 7 - 3 = 4$

Evaluate :
$|-13| - |-15|$

$|-13| - |-15|$

$= +13 - 15 = -2$

Fill in the blanks to make the statements true.
$\dfrac{-2}{9}-\dfrac{7}{9}=$ ________

$\Rightarrow \dfrac{-2}{9}-\dfrac{7}{9}$

$\Rightarrow \dfrac{-2-7}{9}$

$\Rightarrow \dfrac{-9}{9}$

$=-1$

Fill in the blanks to make the statements true.
$\dfrac{-3}{5}+\dfrac{2}{5}=$ ________.

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

$\Rightarrow \dfrac{-3+2}{5}\qquad ...$ LCM=

$5$

$\Rightarrow \dfrac{-1}{5}$

Fill in the boxes with the correct symbol 0 , > , < or =.
$\dfrac{5}{8}\square \dfrac{8}{4}$

We convert the rational numbers with the same denominators.

$\therefore \dfrac{5\times2}{6\times2}=\dfrac{10}{10}$ and

$\dfrac{8\times3}{4\times3}=\dfrac{24}{12}$

i.e.

$24>10\Rightarrow \dfrac{24}{12}>\dfrac{10}{12}$

Hence,

$\dfrac{5}{6}<\dfrac{8}{4}$

Fill in the blanks to make the statements true.
$\dfrac{-3}{4}\times\left( \dfrac{-2}{3}\right)=$ ________.

$\Rightarrow \dfrac{-3}{4}\times\left( \dfrac{-2}{3}\right)=$

$=\dfrac{3\times(-2)}{4\times3}$

$=\dfrac{-6}{12} \qquad$ ...divided by

$6$

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

Fill in the blanks to make the statements true.
$\dfrac{-5}{6}+\dfrac{-1}{6}=$ ________.

$\Rightarrow \dfrac{-5}{6}+\dfrac{-1}{6}$

$\Rightarrow \dfrac{-5-1}{6}$

$\Rightarrow \dfrac{-6}{6}$

$\Rightarrow -1$

Fill in the boxes with the correct symbol 0 , > , < or =.
$\dfrac{7}{-8} \square \dfrac{8}{9}$

Since,

$\dfrac{7}{-8}$ is a negative rational number.

Hence,

$\dfrac{7}{-8}<\dfrac{8}{9}$

Fill in the boxes with the correct symbol 0 , > , < or =.
$\dfrac{-9}{7}\square \dfrac{4}{-7}$

Both fractions have the same denominator but their numerators are different,
Hence,

$\dfrac{-9}{7}<\dfrac{4}{-7}$

Fill in the boxes with the correct symbol 0 , > , < or =.
$\dfrac{8}{8}\square \dfrac{2}{2}$

Given,

$\Rightarrow \dfrac{8}{8}=1$

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

Hence,

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

Fill in the blanks to make the statements true.
$-\dfrac{1}{2}$ is ________ than $\dfrac{1}{5}$ .

Here,

$-\dfrac{1}{2}$ is a negative integer and

$\dfrac{1}{5}$ is a positive integer,
Hence,

$-\dfrac{1}{2}$ is smaller than

$\dfrac{1}{5}$ .

Fill in the blanks to make the statements true.
$\dfrac{-27}{45}$ and $\dfrac{-3}{5}$ represent ________ rational numbers.

As we know that two rational numbers said to be equivalent if they are equal to each other.
HCF of

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

HCF of

$\dfrac{-27}{45}=9$
Dividing denominator and numerator of

$\dfrac{-27}{45}$ by

$9$

$\Rightarrow \dfrac{-27÷9}{45÷9}=\dfrac{-3}{5}$

$\Rightarrow \dfrac{-3}{5}=\dfrac{-3}{5}$

$\therefore \dfrac{-27}{45}=\dfrac{-3}{5}$
Therefore, both are equivalent rational numbers.

Fill in the blanks to make the statements true.
The reciprocal of 1 is ______.

The reciprocal of 1 is 1.

$\Rightarrow 1=\dfrac{1}{1}=1$

Fill in the blanks to make the statements true.
The reciprocal of ______ does not exist.

Zero

Fill in the blanks to make the statements true.
If m is a common divisor of a and b, then $\dfrac{a}{b}=\dfrac{am}{}$

Here, m is a common divisor,
Now,

$\Rightarrow \dfrac{a÷m}{b÷m}=\dfrac{a}{b}$

Fill in the blanks to make the statements true.
$\dfrac{-3}{7} ÷ \left(\dfrac{-7}{3}\right)=$ ______.

$\Rightarrow \dfrac{-3}{7}÷ \left(\dfrac{-7}{3}\right)$

$=\dfrac{-3\times3}{7\times-7}$

$=\dfrac{-9}{-49}$

$=\dfrac{9}{49}$

Write the next three rational numbers to complete the pattern:
$\dfrac{-8}{7},\dfrac{-16}{14},\dfrac{-24}{21},\dfrac{-32}{28},$ , , , .

Next three equivalent rational numbers are
$\Rightarrow \dfrac{-8}{7}$

$\Rightarrow \dfrac{-8}{7}=\dfrac{-8\times5}{7\times5}=\dfrac{-40}{35}$

$\Rightarrow \dfrac{-8}{7}=\dfrac{-8\times6}{7\times6}=\dfrac{-48}{42}$

$\Rightarrow \dfrac{-8}{7}=\dfrac{-8\times7}{7\times7}=\dfrac{-56}{49}$

Therefore, next equivalent rational numbers of given number are $\dfrac{-40}{35},\dfrac{-48}{42}$ and $\dfrac{-56}{49}.$

Which is greater in the following?
$\dfrac{3}{4},\dfrac{7}{8}$

$\dfrac{3}{4},\dfrac{7}{8}$
Here, make the denominator same of both rational numbers.

$\Rightarrow \dfrac{3}{4}=\dfrac{3\times2}{4\times2}=\dfrac{6}{8}$

$\Rightarrow \dfrac{7}{8}=\dfrac{7\times1}{8\times1}=\dfrac{7}{8}$

$\therefore \dfrac{3}{4}<\dfrac{7}{8}$
Therefore, the greater number is

$\dfrac{7}{8}.$

Simplify:
$\dfrac{13}{11}\times \dfrac{-14}{5}+\dfrac{13}{11}\times \dfrac{-7}{5}+\dfrac{-13}{11}\times \dfrac{34}{5}$

$\dfrac{13}{11}\times \dfrac{-14}{5}+\dfrac{13}{11}\times \dfrac{-7}{5}+\dfrac{-13}{11}\times \dfrac{34}{5}$

$=\dfrac{13\times(-14)}{11\times5}+\dfrac{13\times(-7)}{11\times5}+\dfrac{-13\times 34}{11\times5}$

$=\dfrac{-182}{55}+\dfrac{-91}{55}+\dfrac{-442}{55}$

$=\dfrac{-182-91-442}{55}$

$\dfrac{-715}{55}$

$-13$

Find the product of:
$\dfrac{-22}{11}$ and $\dfrac{-21}{11}$

$=\dfrac{-22}{11}$ and

$\dfrac{-21}{11}$

$=\dfrac{-22}{11}\times\dfrac{-21}{11}$

$=\dfrac{462}{121}$

$=\dfrac{42}{11}$

Find the sum of
$\dfrac{7}{3}$ and $\dfrac{-4}{3}$

$\dfrac{7}{3}$ and

$\dfrac{-4}{3}$

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

$=\dfrac{7}{3}-\dfrac{4}{3}$

$=\dfrac{7-4}{3}$

$=\dfrac{3}{3}=1$
Therefore, the sum of

$\dfrac{7}{3}$ and

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

$1$ .

Find the product of:
$\dfrac{-4}{5}$ and $\dfrac{-5}{12}$

$\dfrac{-4}{5}$ and

$\dfrac{-5}{12}$

$=\dfrac{-4}{5}\times\dfrac{-5}{12}$

$=\dfrac{-4\times-5}{12\times5}$

$=\dfrac{20}{60}$

$=\dfrac{1}{3}$

Solve:
$\dfrac{29}{4}-\dfrac{30}{7}$

$\dfrac{29}{4}-\dfrac{30}{7}$

$=\dfrac{(29\times 7)-(30\times 4)}{28}$

$=\dfrac{203-120}{28}$

$=\dfrac{83}{28}$

Given that $\dfrac{p}{q}$ and $\dfrac{r}{s}$ are two rational numbers with different denominators and both of them are in standard form. To compare these rational numbers we say that:
$\dfrac{p}{q}=\dfrac{r}{s},$ if =

$\dfrac{p}{q}=\dfrac{r}{s}$ ,then,
By cross-multiplication

$\Rightarrow p\times s=r\times q$

Taking $x=\dfrac{-4}{9},y=\dfrac{5}{12}$ and $z=\dfrac{7}{18},$ find:
the rational number which subtracted from y gives z.

Given, $x=\dfrac{-4}{9},y=\dfrac{5}{12}$ and $z=\dfrac{7}{18},$
We Subtract $A$ from $y$ to get $z$ .

$\Rightarrow y-A=z$

$\Rightarrow \dfrac{5}{12}-A=\dfrac{7}{18}$

$\Rightarrow A=\dfrac{7}{18}-\dfrac{5}{12}$

$\Rightarrow A=\dfrac{14-15}{36}$

$\Rightarrow A=\dfrac{-1}{36}$

Find the reciprocal of the following:
$\left(\dfrac{1}{2}\times \dfrac{1}{4}\right)+\left(\dfrac{1}{2}\times 6\right)$

$\left(\dfrac{1}{4}\times \dfrac{1}{4}\right)+\left(\dfrac{1}{2}\times 6\right)$

$=\left(\dfrac{1}{4}\times \dfrac{1}{4}\right)+\left(\dfrac{1}{2}\times 6\right)$

$=\dfrac{1}{8}+\dfrac{6}{2}$

$=\dfrac{1+24}{8}$

$=\dfrac{25}{8}$
Therefore, the reciprocal is

$\dfrac{8}{25}$

Complete the following table by finding the sums:
+ $-\dfrac{1}{9}$ $\dfrac{4}{11}$ $\dfrac{-5}{6}$
$\dfrac{2}{3}$
$-\dfrac{5}{4}$ $-\dfrac{39}{44}$
$-\dfrac{1}{3}$

1900273_1791463_ans_9947c9a6bf5548efab3a45bbf3803cf2.jpg

Taking $x=\dfrac{-4}{9},y=\dfrac{5}{12}$ and $z=\dfrac{7}{18},$ find the rational number which when added to x gives y.

Given,

$x=\dfrac{-4}{9},y=\dfrac{5}{12}$ and

$z=\dfrac{7}{18},$

We add

$A$ to

$x$ we get

$y$ .

$\Rightarrow A+x=y$

$\Rightarrow A+\dfrac{-4}{9}=\dfrac{5}{12}$

$\Rightarrow A=\dfrac{5}{12}+\dfrac{4}{9}$

$\Rightarrow A=\dfrac{15+16}{36}$

$\Rightarrow A=\dfrac{31}{36}$

Taking
$x=\dfrac{-4}{9},y=\dfrac{5}{12}$ and $z=\dfrac{7}{18},$ find:
$(x-y)+z$

Given,

$x=\dfrac{-4}{9},y=\dfrac{5}{12}$ and

$z=\dfrac{7}{18},$

To find:

$(x-y)+z$

$\Rightarrow (x-y)+z$

$=\left( \dfrac{-4}{9}-\dfrac{5}{12}\right) +\dfrac{7}{18}$

$=\left(\dfrac{-16-15}{36}\right)+\dfrac{7}{18}$

$=\dfrac{-31}{36}+\dfrac{7}{18}$

$=\dfrac{-31+14}{36}$

$=\dfrac{-17}{36}$

Fill in the blanks to make the statement true:
Between the numbers $\dfrac{15}{20}$ and $\dfrac{34}{40},$ the greater number is .........

Multiplying the numerator and denominator of

$\dfrac{15}{20}$ by

$2$ We get

$\dfrac{15}{20}=\dfrac{15\times 2}{20\times 2}=\dfrac{30}{40}$

Now, we compare

$\dfrac{30}{40}$ and

$\dfrac{34}{40}$

Clearly,

$\dfrac{30}{40}<\dfrac{34}{40}$

$\therefore \dfrac{15}{20} <\dfrac{34}{40}$

What should be added $\dfrac{-1}{2}$ to obtain the nearest natural number?

As we know that nearest number of

$\dfrac{-1}{2}$ is

$1$
Let y be added

$\dfrac{-1}{2}$ to obtain

$1.$
Then,

$\Rightarrow \dfrac{-1}{2}+y=1$

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

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

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

Therefore,

$\dfrac{3}{2}$ should be added to

$\dfrac{-1}{2}$ to obtain nearest natural number.

Divide the sum of $8/3$ and $4/7$ by the product of $-3/7$ and $14/9$ .

Sum of

$8 / 3$ and

$4 / 7$ is calculated as below

$8 / 3 + 4 / 7 = (56 + 12) / 21$
We get,

$= 68 / 21$
Product of

$- 3 / 7$ and

$14 / 9$ is calculated as follows:

$- 3 / 7 \times 14 / 9 = - 2 / 3$
Hence,

$68 / 21 \div - 2 / 3 = 68 / 21 \times 3 / - 2$
We get,

$= 34 / - 7$

$= \left \{34 \times (- 1)\right \} / \left \{- 7 \times (- 1)\right \}$

$= - 34 / 7$

$= -4\dfrac {6}{7}$ .

Taking $x=\dfrac{-4}{9},y=\dfrac{5} {12}$ and $z=\dfrac{7}{18},$ find:
the rational number which when added to z gives us x.

Given, $x=\dfrac{-4}{9},y=\dfrac{5}{12}$ and $z=\dfrac{7}{18},$

We add $A$ to $z$ to get $x$ .

$\Rightarrow A+z=x$
$\Rightarrow A+\dfrac{7}{18}=\dfrac{-4}{9}$

$\Rightarrow A=\dfrac{-4}{9}-\dfrac{7}{18}$

$\Rightarrow A=\dfrac{-8-7}{18}$

$\Rightarrow A=\dfrac{-5}{6}$

What should be subtracted from $\dfrac{-2}{3}$ to obtain nearest integer?

As we know that nearest number of

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

$-1.$
Then,

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

$\Rightarrow y=\dfrac{-2}{3}+1$

$\Rightarrow y=\dfrac{-2+3}{3}$

$\Rightarrow y=\dfrac{1}{3}$

Therefore,

$\dfrac{1}{3}$ should be added

$\dfrac{-2}{3}$ to obtain nearest natural number.

Fill in the blanks to make the statement true:
$\dfrac{-5}{7}$ is ............. than $-3.$

Let's multiply and divide

$-3$ by

$7$ to make the denominator

$7$

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

Comparing

$\dfrac{-5}7$ and

$\dfrac{-21}7$

Clearly,

$\dfrac{-5}7>\dfrac{-21}7$

$\therefore \dfrac{-5}{7} > -3$

Verify $- (-x) = x$ for $x = \dfrac{-7}{9}$

Here,

$LHS=-(-x )=-\left(-\left( \dfrac{-7}9\right)\right) =\left(\dfrac{-7}9\right) =\dfrac{-7}{9} = x=RHS$

Simplify-
$\dfrac{7}{8} + \dfrac{1}{16} - \dfrac{1}{12}$

$\dfrac{7}{8} + \dfrac{1}{16} - \dfrac{1}{12}$

$= \dfrac{42 + 3 - 4}{48}$

$= \dfrac{41}{48}$

A train travels $\dfrac{1445}{2}$ km in $\dfrac{17}{2}$ hours. Find the speed of the train in km/h.

Given

Distance

$= \dfrac{1445}{2}\ \ \text{km}$

Time

$= \dfrac{17}{2}\ \ \text{hr}$

$\therefore Speed = \dfrac{Distance}{Time}$

$= \dfrac{\dfrac{1445}{2}}{\dfrac{17}{2}}$

$= \dfrac{1445}{2} \times \dfrac{2}{17}$

$= 85\ \text{kmph}$

The product of two rational numbers is $\dfrac{-14}{27}.$ If one of the numbers be $\dfrac{7}{9},$ find the other.

Let the other number be

$x$

According to question,

$\dfrac{7}{9} \times x=\dfrac{-14}{27}$

$\Rightarrow x = \dfrac{-14 \times 9}{27 \times 7}$

$\Rightarrow x = \dfrac{-2}{3}$

Hence, The other number is

$-\dfrac23$ .

By what numbers should we multiply $\dfrac{-15}{20}$ so that the product may be $\dfrac{-5}{7}$ ?

Let the number be

$x$

According to question,

$x \times \dfrac{-15}{20} = \dfrac{-5}{7}$

$\Rightarrow x = \dfrac{(-5 )\times 20}{7 \times (-15)}$

$\Rightarrow x = \dfrac{22}{21}$

Hence, The other number is

$\dfrac{22}{21}$ .

Simplify-
$\dfrac{32}{5} + \dfrac{23}{11} \times \dfrac{22}{15}$

We have

$\dfrac{32}{5} + \dfrac{23}{11} \times \dfrac{22}{15}\\$

$= \dfrac{32}{5} + \dfrac{46}{15}\qquad\\$ {by multiplication]

$= \dfrac{96 + 46}{15}\\$

$= \dfrac{142}{15}$

Simplify
$\dfrac{3}{7} \times \dfrac{28}{15} \div \dfrac{14}{5}$

$\dfrac{3}{7} \times \dfrac{28}{15} \div \dfrac{14}{5}$

$= \dfrac{3}{7} \times \dfrac{28}{15} \times \dfrac{5}{14}$

$= \dfrac{2}{7}$

Verify $- (-x) = x$ for $x = \dfrac{13}{-15}$

Here,

$LHS=-(-x )=-\left(-\left( \dfrac{13}{-15}\right)\right) =\left(\dfrac{13}{-15}\right)= x=RHS$

The table shows the portion of some common materials that are recycled.
Material Recycled
Paper $\dfrac{5}{11}$
Aluminium Cans $\dfrac{5}{8}$
Glass $\dfrac{2}{5}$
Scap $\dfrac{3}{4}$
(a) Is the rational number expressing the amount of paper recycled more than $\dfrac{1}{2}$ or less than $\dfrac{1}{2}$ ?
(b) Which items have a recycled amount less than $\dfrac{1}{2}$ ?
(c) Is the quantity of aluminium cans recycled more (or less) than half of the quantity of aluminium cans ?
(d) Arrange the rate of recycling the materials from the greatest to the smallest.

(a) Proportion of paper recycled

$= \dfrac{5}{11}$

Comparing it with

$\dfrac{1}{2}$ , we get

$\dfrac5{11}<\dfrac12$

(b) Paper and Glass

Proportion of aluminium recycled

$=\dfrac58$

Comparing it with

$\dfrac{1}{2}$ , we get

$\dfrac5{8}>\dfrac12$

(d) Descending order of the rate of recycling the materials is:

$\dfrac34> \dfrac{5}{8} > \dfrac{5}{11}> \dfrac{2}{5}$

i.e., Scap > Aluminium cans > Paper > Glass.

Which of the
following pairs represent the same rational number?

(i) $-\dfrac{7}{21}, \dfrac{3}{9}$

(ii) $\dfrac{-16}{20}, \dfrac{20}{-25}$

(iii) $\dfrac{-3}{5}, \dfrac{-12}{20}$

(iv) $\dfrac{8}{-5}, \dfrac{-24}{15}$

$(i)-\dfrac{7}{21}, \dfrac{3}{9}$

$-\dfrac{7}{21}=\dfrac{-1}{3} \quad \dots (i)$

$\dfrac{3}{9}=\dfrac{1}{3} \quad \dots (ii)$

From (i) and
(ii),

$\dfrac{-1}{3} \neq \dfrac{1}{3}$

$\therefore \dfrac{-7}{21}$ and $\dfrac{3}{9}$ are not same rational numbers.

(ii) $\dfrac{-16}{20}, \dfrac{20}{-25}$

$\therefore \dfrac{-16}{20}=\dfrac{-4}{5} \quad \dots (i)$

$\dfrac{20}{-25}=\dfrac{4}{-5}=\dfrac{-4}{5} \quad \dots (ii)$

From

(i) and (ii)

$\dfrac{-16}{20}$
and $\dfrac{20}{-25}$ are same rational numbers.

(iii) $\dfrac{-3}{5}, \dfrac{-12}{20}$

$\dfrac{-12}{20}=\dfrac{-3}{5} \quad \dots (i)$

$\because \dfrac{-3}{5}=\dfrac{-3}{5} \quad \dots (ii)$

From (i) and
(ii)

$\dfrac{-3}{5}$
and $\dfrac{-12}{20}$ are same rational numbers.

(iv) $\dfrac{8}{-5}, \dfrac{-24}{15}$

$\dfrac{8}{-5} ;=\dfrac{-8}{5} \quad \dots (i)$

$\dfrac{-24}{15}=\dfrac{-8}{5} \quad \dots (ii)$

From (i) and
(ii)

$\therefore \dfrac{8}{-5}=\dfrac{-24}{15}$

$\dfrac{8}{-5}$
and $\dfrac{-24}{15}$ are same ratinoal numbers.

Which of the
following pairs of rational numbers are equal?

(i) $\dfrac{-3}{-7}$
and $\dfrac{15}{35}$

(ii) $\dfrac{-6}{8}$
and $\dfrac{10}{-15}$

(iii) $\dfrac{6}{-10}$
and $\dfrac{-12}{20}$

(i) $\dfrac{-3}{-7}$
and $\dfrac{15}{35} \Rightarrow \dfrac{15}{35}=\dfrac{3}{7}$

$\therefore \dfrac{-3}{-7}=\dfrac{15}{35}$

(ii) $\dfrac{-6}{8}$
and $\dfrac{10}{-15}$ if $-6 \times(-15)=10 \times 8$

$\Rightarrow 90=80$ which is not true.

$\therefore \dfrac{-6}{8} \neq \dfrac{10}{-15}$

(iii) $\dfrac{6}{-10}$
and $\dfrac{-12}{20}$ if $6 \times 20=-12 \times(-10)$

$\Rightarrow 120=120$ which is true.

$\therefore \dfrac{6}{-10}=\dfrac{-12}{20}$

Fill the blanks to make the statement true:
The denominator of a rational number is greater than the numerator by $10$ . If the numerator is increased by $1$ and denominator is decreased by $1$ , then expression for new denominator is _________.

$x+9$
If the numerator is

$x$ , the denominator will be =

$x+10$ .
Therefore, Rational number=

$\dfrac{x}{x+10}$
As per the question,
The new rational number is=

$\dfrac{Numerator+1}{Denominator-1}=\dfrac{x+1}{x+10-1}=\dfrac{x+1}{x+9}$
Hence,

$x+9$ is the new denominator.

Add the
following pairs of rational numbers

(i) $\dfrac{3}{11}, \dfrac{-5}{11}$

(ii) $\dfrac{4}{9}, \dfrac{5}{-9}$

(iii) $\dfrac{5}{-7}, \dfrac{-2}{-7}$

(iv) $\dfrac{-2}{5}, \dfrac{3}{4}$

$(i) \dfrac{3}{11}, \dfrac{-5}{11}$

$=\dfrac{3}{11}+\dfrac{-5}{11}=\dfrac{3-5}{11}=\dfrac{-2}{11}$

(ii) $\dfrac{4}{9}, \dfrac{5}{-9}$

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

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

(iii) $\dfrac{5}{-7}, \dfrac{-2}{-7}$

$=\dfrac{5}{-7}+\dfrac{-2}{-7}=\dfrac{5 \times(-1)}{-7 \times(-1)}+\dfrac{2}{7}$

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

$(i v) \dfrac{-2}{5}, \dfrac{3}{4}$

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

$=\dfrac{(-2 \times 4)+(3 \times 5)}{20}$

$=\dfrac{-8+15}{20} \quad(\mathrm{LCM} \text { of } 5,4=20)$

$=\dfrac{7}{20}$

The sum of two rational
numbers is $\dfrac{2}{5}$ if one of them is $\dfrac{-4}{7}$ , find the other

Sum of two
rational numbers $=\dfrac{2}{5}$

One of them $=\dfrac{-4}{7}$

$\therefore \text { Other }=\dfrac{2}{5}-\left(\dfrac{-4}{7}\right)=\dfrac{2}{5}+\dfrac{4}{7}$

$=\dfrac{14+20}{35} \text { (LCM of }5,7=35)$

$=\dfrac{34}{35}$

What
rational number should be added to $\dfrac{-5}{12}$ to get $\dfrac{-7}{8} ?$

Sum of two

rational numbers $=\dfrac{-7}{8}$ One number $=\dfrac{-5}{12}$

$$ \therefore

\text { Required number }=\dfrac{-7}{8}-\left(\dfrac{-5}{12}\right) $$

$$\quad=\dfrac{-7}{8}+\dfrac{5}{12}

$$\quad=\dfrac{-21+10}{24}

\quad \quad(\mathrm{LCM} \text { of } 8,12=24) $$

$\quad=\dfrac{-11}{24}$

Which of the
two rational numbers is greater in each of the following pairs?

(i) $\dfrac{-4}{7}, 0$

(ii) $\dfrac{5}{-9}, \dfrac{3}{7}$

(iii) $\dfrac{-9}{-5}, 0$

$(i v) \dfrac{7}{-5}, \dfrac{-21}{-23}$

(i) $$\dfrac{-4}{7},

0$$

0 is greater

than $\dfrac{-4}{7}$

(ii) $$\dfrac{5}{-9},

\dfrac{3}{7}$$

$\dfrac{3}{7}$

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

(iii) $$\dfrac{-9}{-5},

0$$

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

0$$

$=\dfrac{9}{5}$

is greater than 0

(iv) $$\dfrac{7}{-5},

\dfrac{-21}{-23} $or$ \dfrac{21}{23}$$

$=\dfrac{21}{23}$

is greater than $\dfrac{7}{-5}$

What
rational number should be subtracted from $\dfrac{-2}{3}$ to get $\dfrac{-5}{6} ?$

$$ \text {

Required number }=\dfrac{-2}{3}-\left(\dfrac{-5}{6}\right) $$

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

$=\dfrac{-4+5}{6}=\dfrac{1}{6}$

Subtract:

(i) $\dfrac{-6}{13}$
from $\dfrac{4}{13}$

(ii) $\dfrac{-1}{2}$
from $\dfrac{-2}{3}$

(iii) $\dfrac{5}{9}$
from $\dfrac{-2}{3}$

(i) $\dfrac{-6}{13}$
from $\dfrac{4}{13}$

$=\dfrac{4}{13}-\dfrac{-6}{13}=\dfrac{4}{13}+\dfrac{6}{13}$

$=\dfrac{4+6}{13}=\dfrac{10}{13}$

(ii) $\dfrac{-1}{2}$
from $\dfrac{-2}{3}$

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

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

(iii) $\dfrac{5}{9}$
from $\dfrac{-2}{3}$

$=\dfrac{-2}{3}-\dfrac{5}{9}$

$=\dfrac{-6-5}{9}=\dfrac{-11}{9}$

Compare :
$\dfrac{-7}{8}$ and $\dfrac{5}{-6}$

$\dfrac{-7}{8}$ and

$\dfrac{5}{-6}$
It can be written as

$\dfrac{-7}{8}$ and

$\dfrac{-5}{6}$
We know that

$\dfrac{a}{b}$ and

$\dfrac{c}{d} = a\times d$ and

$b \times c$
Substituting the values

$-7 \times 6$ and

$-5 \times 8$
So we get

$a \times d < b \times c$

$-42 < - 40$
Hence ,

$\dfrac{-7}{8} < \dfrac{5}{-6}$ .

Add:
$\dfrac{-4}{9}$ and $\dfrac{2}{9}$

$\dfrac{-4}{9}$ and

$\dfrac{2}{9}$
It can be be written as

$= \dfrac{-4}{9} + \dfrac{2}{9}$
By further calculation

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

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

Find:

(i) $\dfrac{5}{63}-\left(\dfrac{-6}{21}\right)$

(ii) $\dfrac{-6}{13}-\left(\dfrac{-7}{15}\right)$

$(\text { in }) 3 \dfrac{1}{8}-\left(-1 \dfrac{5}{6}\right)$

$(i) \dfrac{5}{63}-\left(\dfrac{-6}{21}\right)$

$=\dfrac{5}{63}+\dfrac{6}{21}$

$=\dfrac{5+18}{63}=\dfrac{23}{63} \quad(\mathrm{LCM} \text { of } 63,21=63)$

$(1) \dfrac{-6}{13}-\left(\dfrac{-7}{15}\right)$

$=\dfrac{-6}{13}+\dfrac{7}{15}$

$=\dfrac{-90+91}{195}=\dfrac{1}{195}$

$(\mathrm{LCM} \text { of } 13,15=195)$

(iii) $3 \dfrac{1}{8}-\left(-1 \dfrac{5}{6}\right)$

$=\dfrac{25}{8}+\dfrac{11}{6}$

$=\dfrac{75+44}{24} \quad(\mathrm{LCM} \text { of } 6,8=24)$

$=\dfrac{119}{24}=4 \dfrac{23}{24}$

Find the
sum:

(i) $\dfrac{27}{-4}+\dfrac{-15}{8}$

(ii) $\dfrac{-1}{18}+\dfrac{-3}{8}$

$(i i i)-3 \dfrac{1}{6}+2 \dfrac{3}{8}$

$(i v)-2 \dfrac{4}{5}+4 \dfrac{3}{10}$

(i) $\dfrac{27}{-4}+\dfrac{-15}{8}$

$=\dfrac{27 \times(-1)}{-4 \times(-1)}+\dfrac{-15}{8}$

$=\dfrac{-27}{4}+\dfrac{-15}{8}$

$=\dfrac{(-27 \times 2)+(-15)}{8}$

$=\dfrac{-54+(-15)}{8}=\dfrac{-54-15}{8}=\dfrac{-69}{8}$

(ii) $\dfrac{-1}{18}+\dfrac{-3}{8}$

$=\dfrac{(-1 \times 4)+(-3 \times 9)}{72}$

$\quad(\mathrm{LCM} \text { of } 18,8=72)$

$=\dfrac{-4+(-27)}{72}$

$=\dfrac{-4-27}{72}=\dfrac{-31}{72}$

$(i i i)-3 \dfrac{1}{6}+2 \dfrac{3}{8}$

$=\dfrac{-19}{6}+\dfrac{19}{8}$

$=\dfrac{(-19 \times 4)+(19 \times 3)}{24} \quad \text{(LCM of 6,8 =24)}$

$=\dfrac{-76+57}{24}$

$=\dfrac{-19}{24}$

$(i v)-2 \dfrac{4}{5}+4 \dfrac{3}{10}$

$=\dfrac{-14}{5}+\dfrac{43}{10}$

$=\dfrac{(-14 \times 2)+43}{10} \quad(\mathrm{LCM} \text { of } 5,10=10)$

$=\dfrac{-28+43}{10}$

$=\dfrac{15}{10}=\dfrac{3}{2}=1 \dfrac{1}{2}$

Find the
product:

(i) $\dfrac{2}{3} \times \dfrac{-7}{8}$

(ii) $\dfrac{-6}{7} \times \dfrac{5}{7}$

(iii) $\dfrac{-2}{9} \times(-5)$

$(i v) \dfrac{-5}{11} \times\left(\dfrac{11}{-5}\right)$

(v) $\dfrac{8}{35} \times \dfrac{21}{-32}$

$(v i) \dfrac{-105}{128} \times\left(-1 \dfrac{29}{35}\right)$

(i) $\dfrac{2}{3} \times \dfrac{-7}{8}$

$=\dfrac{1 \times(-7)}{3 \times 4}=\dfrac{-7}{12}$

(ii) $\dfrac{-6}{7} \times \dfrac{5}{7}$

$=\dfrac{-6 \times 5}{7 \times 7}=\dfrac{-30}{49}$

(iii) $\dfrac{-2}{9} \times(-5)$

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

$=\dfrac{10}{9}=1 \dfrac{1}{9}$

$(i v) \dfrac{-5}{11} \times\left(\dfrac{11}{-5}\right)$

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

(v) $\dfrac{8}{35} \times \dfrac{21}{-32}$

$=\dfrac{1 \times 3}{5 \times(-4)}=\dfrac{3}{-20}=\dfrac{-3}{20}$

$(v i) \dfrac{-105}{128} \times\left(-1 \dfrac{29}{35}\right)$

$=\dfrac{-105}{128} \times \dfrac{-64}{35}=\dfrac{-3 \times(-1)}{2}$

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

Subtract:
(i) $\dfrac{-11}{24}$ from $\dfrac{-5}{36}$
(ii) $\dfrac{-8}{15}$ from $-1 \dfrac{2}{5}$
$(i i i)-2 \dfrac{2}{9}$ from $-3 \dfrac{5}{12}$

Subtract

(i) $\dfrac{-11}{24}$
from $\dfrac{-5}{36}$

$=\dfrac{-5}{36}-\left(\dfrac{-11}{24}\right)$

$=\dfrac{-5}{36}+\dfrac{11}{24}$

$=\dfrac{-10+33}{72} \quad(\mathrm{LCM} \text { of } 36,24=72)$

$=\dfrac{23}{72}$

(ii) $\dfrac{-8}{15}$
from $-1 \dfrac{2}{5}$

$=-1 \dfrac{2}{5}-\left(\dfrac{-8}{15}\right)$

$=\dfrac{-21+8}{15}=\dfrac{-13}{15}$

$(i i i)-2 \dfrac{2}{9}$ from $-3 \dfrac{5}{12}$

$=-3 \dfrac{5}{12}-\left(-2 \dfrac{2}{9}\right)$

$=\dfrac{-41}{12}+\dfrac{20}{9}$

$=\dfrac{-123+80}{36}=\dfrac{-43}{36}=-1 \dfrac{7}{36}$

Find the
reciprocal of the following:

(i) $\dfrac{3}{13} \div \dfrac{-4}{65}$

$(i i)\left(-5 \times \dfrac{12}{15}\right)-\left(-3 \times \dfrac{2}{9}\right)$

(i) $\dfrac{3}{13} \div \dfrac{-4}{65}$

$=\dfrac{3}{13} \times \dfrac{65}{-4}=\dfrac{3 \times 5}{-4}=\dfrac{15}{-4}$

$\therefore$
Reciprocal of it $=\dfrac{-4}{15}$

$(i i)\left(-5 \times \dfrac{12}{15}\right)-\left(-3 \times \dfrac{2}{9}\right)$

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

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

$=\dfrac{-12+2}{3}=\dfrac{-10}{3}$

$\therefore$
Reciprocal of it $=\dfrac{-3}{10}$

Find the sum:
(i) $\dfrac{-2}{3}+\left(\dfrac{5}{-7}\right)$
$(i i)-1 \dfrac{1}{12}+\dfrac{-5}{9}$
$(i i i) 2 \dfrac{2}{5}+\left(-4 \dfrac{3}{10}\right)$

(i) $\dfrac{-2}{3}+\left(\dfrac{5}{-7}\right)$

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

$=\dfrac{-P 4-15}{21}=\dfrac{-29}{21}=-1 \dfrac{8}{21}$

$(i i)-1 \dfrac{1}{12}+\dfrac{-5}{9}$

$=\dfrac{-13}{12}-\dfrac{5}{9} \quad(\mathrm{LCM} \text { of } 9,12=36)$

$=\dfrac{-39-20}{36}=\dfrac{-59}{36}=-1 \dfrac{23}{36}$

(iii) $2 \dfrac{2}{5}+\left(-4 \dfrac{3}{10}\right)$

$=\dfrac{12}{5}-\dfrac{43}{10}$

$=\dfrac{24-43}{10}=\dfrac{-19}{10}=-1 \dfrac{9}{10}$

The product of two
rational numbers is $\dfrac{18}{35}$ If one of them is $\dfrac{-2}{5}$ ,
find the other.

Product of
two rational numbers $=\dfrac{18}{35}$

One of them $=\dfrac{-2}{5}$

$\therefore \text { Second number }=\dfrac{18}{35} \div\left(\dfrac{-2}{5}\right)$

$\qquad =\dfrac{18}{35} \times \dfrac{5}{-2}=\dfrac{9 \times 1}{7 \times(-1)}=\dfrac{9}{-7}$

$=\dfrac{-9}{7}=-1 \dfrac{2}{7}$

Subtract the sum of $-\dfrac57$ and $-\dfrac83$ from the sum of $\dfrac52$ and $-\dfrac{11}{12}$ .

Given numbers are

$-\dfrac57,\ -\dfrac83,\ \dfrac52\ and\ -\dfrac{11}{12}$

To find out: The result when we subtract the sum of

$-\dfrac57$ and

$-\dfrac83$ from the sum of

$\dfrac52$ and

$-\dfrac{11}{12}$

Sum of

$-\dfrac57$ and

$-\dfrac83$ :

$-\dfrac57+\left(-\dfrac83\right)\\$

$\Rightarrow -\dfrac57-\dfrac83\\$

$\Rightarrow \dfrac{-5(3)-8(7)}{21}\quad [LCM(3,7)=21]\\$

$\Rightarrow \dfrac{-15-56}{21}\\$

$\Rightarrow \dfrac{-71}{21}\quad \quad ...(1)\\$

Similarly, sum of

$\dfrac52$ and

$-\dfrac{11}{12}$ :

$\dfrac52+\left(-\dfrac{11}{12}\right)\\$

$\Rightarrow \dfrac52-\dfrac{11}{12}\\$

$\Rightarrow \dfrac{5(6)-11(1)}{12}\quad [LCM(2,12)=12]\\$

$\Rightarrow \dfrac{30-11}{12}\\$

$\Rightarrow \dfrac{19}{12}\quad \quad ...(2)\\$

Now, subtracting

$(1)$ from

$(2)$ , we get:

$\dfrac{19}{12}-\left(\dfrac{-71}{21}\right)\\$

$\Rightarrow \dfrac{19}{12}+\dfrac{71}{21}\\$

$\Rightarrow \dfrac{19(7)+71(4)}{84}\quad [LCM(12,21)=84]\\$

$\Rightarrow \dfrac{133+284}{84}\\$

$\Rightarrow \dfrac{417}{84}\\$

Dividing both numerator and denominator by

$3$ , we get,

$\dfrac{139}{28}\ or \ 4\dfrac{27}{28}\\$

Hence, the result when we subtract the sum of

$-\dfrac57$ and

$-\dfrac83$ from the sum of

$\dfrac52$ and

$-\dfrac{11}{12}$ is

$\dfrac{139}{28}$ or

$4\dfrac{27}{28}$ .

If $p = \dfrac{-3}{2},\ q = \dfrac 45$ and $r = \dfrac{-7}{12},$ then verify that $(p\div q)\div r\neq p\div (q\div r)$

Given:

$p = \dfrac{-3}{2},\ q = \dfrac 45$ and

$r = \dfrac {- 7}{12}$

$(p \div q) \div r \neq p \div (q \div r)$

$LHS = (p\div q) \div r$

$= \bigg(\dfrac {- 3 }{ 2} \div \dfrac 45\bigg) \div \bigg(\dfrac{- 7 }{ 12}\bigg)$

$= \bigg(\dfrac {- 3 }{2} \times \dfrac{5}{ 4}\bigg) \div \bigg(\dfrac {- 7}{12}\bigg)$

$= \dfrac{- 15 }{8} \div \dfrac{- 7}{12}$

$= \dfrac{- 15}{8} \times \dfrac{12}{- 7}$

We get,

$= \dfrac{- 45}{ - 14}$

$= \dfrac{- 45 \times (- 1) }{- 14 \times (- 1)}$

$= \dfrac {45}{14}$

Now,

$RHS = p \div (q \div r)$

$= \dfrac{- 3}{ 2}\div \bigg(\dfrac 4 5\bigg) \div \bigg(\dfrac{- 7 }{ 12}\bigg)$

$= \dfrac {- 3 }{ 2} \div \bigg(\dfrac 45 \times \dfrac{12 }{ - 7}\bigg)$

We get,

$= \dfrac{- 3 }{ 2} \div \dfrac{48 }{ - 35}$

$= \dfrac{- 3 }{ 2} \times \dfrac{- 35 }{ 48}$

We get,

$= \dfrac{35 }{ 32}$

Therefore,

$LHS \neq RHS$

Express $\dfrac{3}{5}$ as a rational number with denominator :
$-20$

Denominator

$=-20$
It can be written as

$\dfrac{3}{5} = \dfrac{3 \times -4}{5 \times -4} = \dfrac{-12}{-20}$

Express $\dfrac{3}{5}$ as a rational number with denominator :
$25$

The existing denominator of the given rational number is

$5$ . We need to multiply the denominator and numerator by

$5$ , in order to get

$25$ in the denominator.
It can be written as:

$\dfrac{3}{5} = \dfrac{3 \times 5}{5 \times 5} = \dfrac{15}{25}$

Express each of the following integers as a rational number with denominator $7$ :
$0$

$0$
By multiplying and dividing by

$7$

$= \dfrac{0 \times 7}{7}$

$= \dfrac{0}{7}$

Express $\dfrac{3}{5}$ as a rational number with denominator :
$-35$

Denominator

$=-35$
It can be written as

$\dfrac{3}{5} = \dfrac{3 \times -7}{5 \times -7} = \dfrac{-21}{-35}$

Express each of the following integers as a rational number with denominator $7$ :
$-16$

$-16$
By multiplying and dividing by

$7$

$= \dfrac{-16 \times 7}{7}$

$= \dfrac{-112}{7}$

Express each of the following integers as a rational number with denominator $7$ :
$-8$

$-8$
By multiplying and dividing by

$7$

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

$= \dfrac{-56}{7}$

Express each of the following integers as a rational number with denominator $7$ :
$7$

$7$
By multiplying and dividing by

$7$

$= \dfrac{7 \times 7}{7}$

$= \dfrac{49}{7}$

Express $\dfrac{3}{5}$ as a rational number with denominator :
$45$

Denominator

$=45$
It can be written as

$\dfrac{3}{5} = \dfrac{3 \times 9}{5 \times 9} = \dfrac{27}{45}$

Express each of the following integers as a rational number with denominator $7$ :
$5$

$5$
By multiplying and dividing by

$7$

$= \dfrac{5 \times 7}{7}$

$= \dfrac{35}{7}$

Compare :
$\dfrac{5}{-9}$ and $\dfrac{-5}{-9}$

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

$\dfrac{-5}{-9}$
We know that

$\dfrac{a}{b}$ and

$\dfrac{c}{d} = a \times d$ and

$b \times c$
Substituting the values

$5 \times -9$ and

$-5 \times -9$
So we get

$a \times d < b \times c$

$-45 < 45$
Hence ,

$\dfrac{5}{-9} < \dfrac{-5}{9}$ .

Add:
$\dfrac{7}{5}$ and $\dfrac{2}{5}$

$\dfrac{7}{5}$ and

$\dfrac{2}{5}$
It can be written as

$= \dfrac{7}{5} + \dfrac{2}{5}$
By further calculation

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

$= \dfrac{9}{5}$

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

Compare :
$\dfrac{-5}{8}$ and $\dfrac{7}{-12}$

$\dfrac{-5}{8}$ and

$\dfrac{7}{-12}$
It can be written as

$\dfrac{-5}{8}$ and

$\dfrac{-7}{12}$
We know that

$\dfrac{a}{b}$ and

$\dfrac{c}{d} = a \times d$ and

$b \times c$
Substituting the values

$-5 \times 12$ and

$-7 \times 8$
So we get

$a \times d < b \times c$

$-60 < -56$
Hence,

$\dfrac{-5}{8} < \dfrac{7}{-12}$ .

Compare :
$\dfrac{3}{5}$ and $\dfrac{5}{7}$

Compare :
$\dfrac{2}{7}$ and $\dfrac{-3}{-8}$

$\dfrac{2}{7}$ and

$\dfrac{-3}{-8}$
It can be written as

$\dfrac{2}{7}$ and

$\dfrac{3}{8}$
We know that

$\dfrac{a}{b}$ and

$\dfrac{c}{d} = a \times d$ and

$b \times c$
Substituting the values

$2 \times 8$ and

$7 \times 3$
So we get

$a \times d < b \times c$

$16 < 21$
Hence,

$\dfrac{2}{7} < \dfrac{-3}{-8}$ .

Add:
$\dfrac{-5}{6}$ and $\dfrac{4}{9}$

$\dfrac{-5}{6}$ and

$\dfrac{4}{9}$
It can be written as

$= \dfrac{-5}{6} + \dfrac{4}{9}$

LCM of

$6$ and

$9$ is

$36$

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

By further calculation

$= \dfrac{-30}{36} + \dfrac{16}{36}$

So we get

$= \dfrac{-30 + 16}{36}$

$= \dfrac{-14}{36}$

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

Add:
$\dfrac{4}{-15}$ and $\dfrac{-7}{-15}$

$\dfrac{4}{-15}$ and

$\dfrac{-7}{-15}$
It can be written as

$= \dfrac{-4}{15} + \dfrac{7}{15}$
By further calculation

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

$= \dfrac{3}{15}$

$= \dfrac{1}{5}$

Add:
$\dfrac{-2}{5}$ and $\dfrac{3}{7}$

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

$\dfrac{3}{7}$

It can be written as

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

LCM of

$5$ and

$7$ is

$35$

$= \dfrac{-14}{35} + \dfrac{15}{35}$

By further calculation

$= \dfrac{-14 + 15}{35}$

$= \dfrac{1}{35}$

Add:
$\dfrac{-7}{26}$ and $\dfrac{7}{-26}$

$\dfrac{-7}{26}$ and

$\dfrac{7}{-26}$
It can be written as

$= \dfrac{-7}{26} + \dfrac{-7}{26}$
By further calculation

$= \dfrac{(-7) + (-7)}{26}$

$= \dfrac{-14}{26}$

$= \dfrac{-7}{13}$

Add:
$\dfrac{-7}{25}$ and $\dfrac{9}{-25}$

$\dfrac{-7}{25}$ and

$\dfrac{9}{-25}$
It can be written as

$= \dfrac{-7}{25} + \dfrac{-9}{25}$

By further calculation

$= \dfrac{(-7) + (-9)}{25}$

$= \dfrac{-16}{25}$

Add:
$\dfrac{1}{-18}$ and $\dfrac{5}{-27}$

$\dfrac{1}{-18}$ and

$\dfrac{5}{-27}$
It can be written as

$= \dfrac{-1}{18} + \dfrac{-5}{27}$
LCM of

$18$ and

$27$ is

$54$

$= \dfrac{-1 \times 3}{18 \times 3} + \dfrac{-5 \times 2}{27 \times 2}$

By further calculation

$= \dfrac{-3}{54} + \dfrac{-10}{54}$

So we get

$= \dfrac{-3 - 10}{54}$

$= \dfrac{-13}{54}$

Add:
$\dfrac{-7}{24}$ and $\dfrac{-5}{48}$

$\dfrac{-7}{24}$ and

$\dfrac{-5}{48}$
It can be written as

$= \dfrac{-7}{24} + \dfrac{-5}{48}$
LCM of

$24$ and

$48$ is

$48$

$= \dfrac{-7 \times 2}{24 \times 2} + \dfrac{-5 \times 1}{48 \times 1}$

By further calculation

$= \dfrac{-14}{48} + \dfrac{-5}{48}$

So we get

$= \dfrac{-14 - 5}{48}$

$= \dfrac{-19}{48}$

Add:
$\dfrac{5}{-12}$ and $\dfrac{1}{12}$

$\dfrac{5}{-12}$ and

$\dfrac{1}{12}$
It can be written as

$= \dfrac{-5}{12} + \dfrac{1}{12}$
By further calculation

$= \dfrac{-5 + 1}{12}$

$= \dfrac{-4}{12}$

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

Add :
$\dfrac{-5}{9}$ and $\dfrac{7}{18}$

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

$\dfrac{7}{18}$
It can be written as

$= \dfrac{-5}{9} + \dfrac{7}{18}$

LCM of

$9$ and

$18$ is

$18$

$= \dfrac{-5 \times 2}{9 \times 2} + \dfrac{7 \times 1}{18 \times 1}$

By further calculation

$= \dfrac{-10}{18} + \dfrac{7}{18}$

So we get

$= \dfrac{-10 + 7}{18}$

$= \dfrac{-3}{18}$

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

Add :
$-3$ and $\dfrac{2}{3}$

$-3$ and

$\dfrac{2}{3}$
It can be written as

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

LCM of

$1$ and

$3$ is

$3$

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

By further calculation

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

So we get

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

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

Evaluate :
$-2 + \dfrac{2}{5} + \dfrac{-2}{15}$

$-2 + \dfrac{2}{5} + \dfrac{-2}{15}$
It can be written as

$\dfrac{-2}{1}+ \dfrac{2}{5} + \dfrac{-2}{15}$

LCM of

$1 , 5$ and

$15$ is

$15$

$= \dfrac{-2 \times 15}{1 \times 15} + \dfrac{2 \times 3}{5 \times 3} + \dfrac{-2 \times 1}{15 \times 1}$

By further calculation

$= \dfrac{-30}{15} + \dfrac{6}{15} + \dfrac{-2}{15}$

So we get

$= \dfrac{-30 + 6 - 2}{15}$

$= \dfrac{-32 + 6}{15}$

$= \dfrac{-26}{15}$

Evaluate :
$\dfrac{-7}{6} + \dfrac{4}{-15} + \dfrac{-4}{-30}$

$\dfrac{-7}{6} + \dfrac{4}{-15} + \dfrac{-4}{-30}$

It can be written as

$= \dfrac{-7}{6} + \dfrac{-4}{15} + \dfrac{4}{30}$

LCM of

$6 , 15$ and

$30$ is

$30$

$= \dfrac{-7 \times 5}{6 \times 5} + \dfrac{-4 \times 2}{15 \times 2} + \dfrac{4 \times 1}{30 \times 1}$

By further calculation

$= \dfrac{-35}{30} + \dfrac{-8}{30} + \dfrac{4}{30}$

So we get

$= \dfrac{-35 - 8 + 4}{30}$

$= \dfrac{-43 + 4}{30}$

$= \dfrac{-39}{30}$

$= \dfrac{-13}{10}$

Evaluate :
$\dfrac{5}{-24} + \dfrac{-1}{8} + \dfrac{3}{16}$

$\dfrac{5}{-24} + \dfrac{-1}{8} + \dfrac{3}{16}$
It can be written as

$= \dfrac{-5}{24} + \dfrac{-1}{8} + \dfrac{3}{16}$
LCM of

$8 , 16$ and

$24$ is

$48$

$= \dfrac{-5 \times 2}{24 \times 2} + \dfrac{-1 \times 6}{8 \times 6} + \dfrac{3 \times 3}{16 \times 3}$

By further calculations

$= \dfrac{-10}{48} + \dfrac{-6}{48} + \dfrac{9}{48}$

So we get

$= \dfrac{-10 - 6 + 9}{48}$

$= \dfrac{-16 + 9}{48}$

$= \dfrac{-7}{48}$

Add :
$\dfrac{13}{-16}$ and $\dfrac{-11}{24}$

$\dfrac{13}{-16}$ and

$\dfrac{-11}{24}$
It can be written as

$= \dfrac{-13}{16} + \dfrac{-11}{24}$
LCM of

$16$ and

$24$ is

$48$

$= \dfrac{-13 \times 3}{16 \times 3} + \dfrac{-11 \times 2}{24 \times 2}$

By further calculation

$= \dfrac{-39}{48} + \dfrac{-22}{48}$

So we get

$= \dfrac{-39 - 22}{48}$

$= \dfrac{-61}{48}$

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

$\dfrac{-2}{5} + \dfrac{3}{5} + \dfrac{-1}{5}$
It can be written as

$= \dfrac{-2 + 3 - 1}{5}$
By further calculation

$= \dfrac{0}{5}$

$= 0$

Evaluate :
$\dfrac{-11}{12} + \dfrac{5}{16} + \dfrac{-3}{8}$

$\dfrac{-11}{12} + \dfrac{5}{16} + \dfrac{-3}{8}$
It can be written as

$= \dfrac{-11}{12} + \dfrac{5}{16} + \dfrac{-3}{8}$

LCM of

$12 , 16$ and

$8$ is

$48$

$= \dfrac{-11 \times 4}{12 \times 4} + \dfrac{5 \times 3}{16 \times 3} + \dfrac{-3 \times 6}{8 \times 6}$

By further calculation

$= \dfrac{-44}{48} + \dfrac{15}{48} + \dfrac{-18}{48}$

So we get

$= \dfrac{-44 + 15 - 18}{48}$

$= \dfrac{-62 + 15}{48}$

$= \dfrac{-47}{48}$

Add:
$\dfrac{-9}{-16}$ and $\dfrac{-11}{8}$

$\dfrac{-9}{-16}$ and

$\dfrac{-11}{8}$
It can be written as

$= \dfrac{9}{16} + \dfrac{-11}{8}$
LCM of

$16$ and

$8$ is

$16$

$= \dfrac{9 \times 1}{16 \times 1} + \dfrac{-11 \times 2}{8 \times 2}$

By further calculation

$= \dfrac{9}{16} + \dfrac{-22}{16}$

So we get

$= \dfrac{9 - 22}{16}$

$= \dfrac{-13}{16}$

Add :
$\dfrac{-9}{25}$ and $\dfrac{1}{-75}$

$\dfrac{-9}{25}$ and

$\dfrac{1}{-75}$
It can be written as

$= \dfrac{-9}{25} + \dfrac{-1}{75}$
LCM of

$24$ and

$75$ is

$75$

$= \dfrac{-9 \times 3}{25 \times 3} + \dfrac{-1 \times 1}{75 \times 1}$

By further calculation

$= \dfrac{-27}{75} + \dfrac{-1}{75}$

So we get

$= \dfrac{-27 - 1}{75}$

$= \dfrac{-28}{75}$

Evaluate :
$\dfrac{-11}{18} + \dfrac{-3}{9} + \dfrac{2}{-3}$

$\dfrac{-11}{18} + \dfrac{-3}{9} + \dfrac{2}{-3}$
It can be written as

$\dfrac{-11}{18} + \dfrac{-3}{9} + \dfrac{-2}{3}$

LCM of

$3 , 9$ and

$18$ is

$18$

$= \dfrac{-11 \times 1}{18 \times 1} + \dfrac{-3 \times 2}{9 \times 2} + \dfrac{-2 \times 6}{3 \times 6}$

By further calculation

$= \dfrac{-11}{18} + \dfrac{-6}{18} + \dfrac{-12}{18}$

So we get

$= \dfrac{-11 - 6 - 12}{18}$

$= \dfrac{-29}{18}$

Evaluate :
$\dfrac{-8}{9} + \dfrac{4}{9} + \dfrac{-2}{9}$

$\dfrac{-8}{9} + \dfrac{4}{9} + \dfrac{-2}{9}$
It can be written as

$= \dfrac{-8 + 4 - 2}{9}$
By further calculation

$= \dfrac{-10 + 4}{9}$

$= \dfrac{-6}{9}$

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

Subtract :
$\dfrac{5}{8}$ from $\dfrac{-5}{16}$

$\dfrac{5}{8}$ from

$\dfrac{-5}{16}$
It can be written as

$= \dfrac{-5}{16} - \dfrac{5}{8}$
LCM of

$8$ and

$16$ is

$16$

$= \dfrac{-5}{16} - \dfrac{5 \times 2}{8 \times 2}$
By further calculation

$= \dfrac{-5}{16} - \dfrac{10}{16}$
So we get

$= \dfrac{-5 - 10}{16}$

$= \dfrac{-15}{16}$

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

$\dfrac{-2}{3} + \dfrac{5}{2} + 2$
It can be written as

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

LCM of

$3 , 2$ and

$1$ is

$6$

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

By further calculation

$= \dfrac{-4}{6} + \dfrac{15}{6} + \dfrac{12}{6}$

So we get

$= \dfrac{-4 + 15 + 12}{6}$

$= \dfrac{23}{6}$

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

Subtract :
$\dfrac{2}{9}$ from $\dfrac{5}{9}$

$\dfrac{2}{9}$ from

$\dfrac{5}{9}$
It can be written as

$= \dfrac{5}{9} - \dfrac{2}{9}$
By further calculation

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

$= \dfrac{3}{9}$

$= \dfrac{1}{3}$

Subtract :
$\dfrac{-2}{15}$ from $\dfrac{-8}{15}$

$\dfrac{-2}{15}$ from

$\dfrac{-8}{15}$
It can be written as

$=\dfrac{-8}{15} - \left (\dfrac{-2}{15} \right )$
By further calculation

$= \dfrac{-8}{15} + \dfrac{2}{15}$
So we get

$= \dfrac{-8 + 2}{15}$

$= \dfrac{-6}{15}$

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

Subtract :
$\dfrac{-3}{10}$ from $\dfrac{1}{5}$

$\dfrac{-3}{10}$ from

$\dfrac{1}{5}$
It can be written as

$= \dfrac{1}{5} - \left ( \dfrac{-3}{10} \right )$
LCM of

$5$ and

$10$ is

$10$

$= \dfrac{1 \times 2}{5 \times 2} + \dfrac{3}{10}$
By further calculation

$= \dfrac{2}{10} + \dfrac{3}{10}$
So we get

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

$= \dfrac{5}{10}$

$= \dfrac{1}{2}$

Subtract :
$\dfrac{-4}{11}$ from $-2$

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

$-2$
It can be written as

$= \dfrac{-2}{1} - \left ( \dfrac{-4}{11} \right )$

LCM of

$1$ and

$11$ is

$11$

$= \dfrac{-2 \times 11}{1 \times 11} + \dfrac{4 \times 1}{11 \times 1}$

By further calculation

$= \dfrac{-22}{11} + \dfrac{4}{11}$

So we get

$= \dfrac{-22 + 4}{11}$

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

Subtract :
$\dfrac{-6}{11}$ from $\dfrac{-3}{-11}$

$\dfrac{-6}{11}$ from

$\dfrac{-3}{-11}$
It can be written as

$= \dfrac{3}{11} - \left ( \dfrac{-6}{11} \right )$
By further calculation

$= \dfrac{3}{11} + \dfrac{6}{11}$
So we get

$= \dfrac{3 + 6}{11}$

$= \dfrac{9}{11}$

Subtract :
$\dfrac{11}{18}$ from $\dfrac{-5}{18}$

$\dfrac{11}{18}$ from

$\dfrac{-5}{18}$
It can be written as

$= \dfrac{-5}{18} - \dfrac{11}{18}$
By further calculation

$= \dfrac{-5 - 11}{18}$

So we get

$= \dfrac{-16}{18}$

$= \dfrac{-8}{9}$

Evaluate :
$-5 + \dfrac{5}{-8} + \dfrac{-5}{-12}$

1887839_1825969_ans_dada90904fab4b5b8197ccc6e0b6939f.jpg

Evaluate :
$\dfrac{-9}{4} + \dfrac{13}{3} + \dfrac{25}{6}$

$\dfrac{-9}{4} + \dfrac{13}{3} + \dfrac{25}{6}$
It can be written as

$\dfrac{-9}{4} + \dfrac{13}{3} + \dfrac{25}{6}$

LCM of

$4 , 3$ and

$6$ is

$12$

$= \dfrac{-9 \times 3}{4 \times 3} + \dfrac{13 \times 4}{3 \times 4} + \dfrac{25 \times 2}{6 \times 2}$

By further calculation

$= \dfrac{-27}{12} + \dfrac{52}{12} + \dfrac{50}{12}$

So we get

$= \dfrac{-27 + 52 + 50}{12}$

$= \dfrac{75}{12}$

$=\dfrac{25}{4}$

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

Evaluate :
$5 + \dfrac{-3}{4} + \dfrac{-5}{8}$

$5 + \dfrac{-3}{4} + \dfrac{-5}{8}$
It can be written as

$\dfrac{5}{1} + \dfrac{-3}{4} + \dfrac{-5}{8}$

LCM of

$1 , 4$ and

$8$ is

$8$

$= \dfrac{5 \times 8}{1 \times 8} + \dfrac{-3 \times 2}{4 \times 2} + \dfrac{-5 \times 1}{8 \times 1}$

By further calculation

$= \dfrac{40}{8} + \dfrac{-6}{8} + \dfrac{-5}{8}$

So we get

$= \dfrac{40 - 6 - 5}{8}$

$= \dfrac{40 - 11 }{8}$

$= \dfrac{29}{8}$

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

The sum of two rational numbers is $\dfrac{-7}{12}$ . If one of them is $\dfrac{13}{24}$ , find the other.

It is given that
Sum of two rational numbers

$= \dfrac{-7}{12}$
One of the rational number

$= \dfrac{13}{24}$
Other rational number

$= x$

$\dfrac{13}{24}+x=\dfrac{-7}{12}$

$x=\dfrac{-7}{12} - \dfrac{13}{24}$

LCM of

$12$ and

$24$ is

$24$

$= \dfrac{-7 \times 2}{12 \times 2} - \dfrac{13}{24}$

By further calculation

$= \dfrac{-14}{24} - \dfrac{13}{24}$

So we get

$= \dfrac{-14 - 13}{24}$

$= \dfrac{-27}{24}$

$= \dfrac{-9}{8}$

Subtract :
$-2$ from $\dfrac{-3}{10}$

$-2$ from

$\dfrac{-3}{10}$
It can be written as

$= \dfrac{-3}{10} - \left ( \dfrac{-2}{1} \right )$
By further calculation

$= \dfrac{-3}{10} + \dfrac{2}{1}$
So we get

$= \dfrac{-3 + 2 \times 10}{10}$

$= \dfrac{17}{10}$

$= 1 \dfrac{7}{10}$

The sum of two rational number is $\dfrac{11}{24}$ . If one of them is $\dfrac{3}{8}$ , find the other.

It is given that
Sum of two rational numbers

$= \dfrac{11}{24}$
One of the rational number

$= \dfrac{3}{8}$
Other rational number

$=x$

$\dfrac{3}{8}+x=\dfrac{11}{24}$

$x=\dfrac{11}{24} - \dfrac{3}{8}$

LCM of

$24$ and

$8$ is

$24$

$= \dfrac{11}{24} - \dfrac{3 \times 3}{8 \times 3}$

By further calculation

$= \dfrac{11}{24} - \dfrac{9}{24}$

So we get

$= \dfrac{11 - 9}{24}$

$= \dfrac{2}{24}$

$= \dfrac{1}{12}$

Subtract :
$\dfrac{-6}{25}$ from $\dfrac{-8}{5}$

$\dfrac{-6}{25}$ from

$\dfrac{-8}{5}$
It can be written as

$= \left (\dfrac{-8}{5} \right ) - \left ( \dfrac{-6}{25} \right )$
LCM of

$5$ and

$25$ is

$25$

$= \dfrac{-8 \times 5}{5 \times 5} + \dfrac{6}{25}$
By further calculation

$= \dfrac{-40}{25} + \dfrac{6}{25}$
So we get

$= \dfrac{-40 + 6}{25}$

$= \dfrac{-34}{25}$

Subtract :
$4$ from $\dfrac{-3}{13}$

$4$ from

$\dfrac{-3}{13}$
It can be written as

$= \dfrac{-3}{13} - \dfrac{4}{1}$
LCM of

$13$ and

$1$ is

$13$

$= \dfrac{-3 - 4 \times 13}{13}$
By further calculation

$= \dfrac{-3 - 52}{13}$

$= \dfrac{-55}{13}$

Subtract :
$\dfrac{-16}{21}$ from $1$

$\dfrac{-16}{21}$ from

$1$
It can be written as

$= \dfrac{1}{1} - \left ( -\dfrac{16}{21} \right )$

$= \dfrac{1}{1} + \dfrac{16}{21}$

By further calculation

$= \dfrac{21 + 16}{21}$
So we get

$= \dfrac{37}{21}$

$= 1 \dfrac{16}{21}$

Subtract :
$\dfrac{-7}{4}$ from $-2$

$\dfrac{-7}{4}$ from

$-2$
It can be written as

$= \dfrac{-2}{1} - \dfrac{-7}{4}$
LCM of

$1$ and

$4$ is

$4$

$= \dfrac{-2 \times 4}{1 \times 4} + \dfrac{7}{4}$

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

By further calculation

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

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

Subtract :
$\dfrac{-8}{15}$ from $0$

$\dfrac{-8}{15}$ from

$0$
It can be written as

$= 0 - \left ( \dfrac{-8}{15} \right )$
By further calculation

$= 0 + \dfrac{8}{15}$

$= \dfrac{8}{15}$

The sum of two numbers is $\dfrac{-6}{5}$ . If one of them is $-2$ , find the other.

Consider

$x$ as the required rational number
It is given that
Sum of two numbers

$= \dfrac{-6}{5}$
One of the numbers

$=-2$
From the question

$-2+x=\dfrac{-6}{5}$

$x = \dfrac{-6}{5} - \left ( \dfrac{-2}{1} \right )$
LCM of

$1$ and

$5$ is

$5$

$= \dfrac{-6}{5} + \dfrac{2 \times 5}{1 \times 5}$
By further calculation

$= \dfrac{-6 + 10}{5}$

$= \dfrac{4}{5}$

Evaluate :
$\dfrac{-12}{5} \times \dfrac{10}{-3}$

$\dfrac{-12}{5} \times \dfrac{10}{-3}$
It can be written as

$= \dfrac{-12 \times 10}{5 \times -3}$
By further calculation

$= 4 \times 2$

$= 8$

Evaluate :
$\dfrac{-45}{39} \times \dfrac{-13}{15}$

$\dfrac{-45}{39} \times \dfrac{-13}{15}$
It can be written as

$= \dfrac{-45 \times -13}{39 \times 15}$
By further calculation

$= \dfrac{-3 \times -1}{3 \times 1}$
So we get

$= \dfrac{3}{3}$

$= 1$

What should be subtracted from $\dfrac{-4}{5}$ to get $1$ ?

Consider

$x$ as the required rational number
Other number

$= \dfrac{-4}{5}$
Here the difference between two numbers is

$1$
From the question

$\dfrac{-4}{5} - x = 1$
By further calculation

$\dfrac{-4}{5} - 1 = x$
So we get

$x = \dfrac{-4 - 1 \times 5}{5}$

$x = \dfrac{-4 - 5}{5} = \dfrac{-9}{5}$

Evaluate :
$\dfrac{2}{3} \times \dfrac{-6}{7}$

$\dfrac{2}{3} \times \dfrac{-6}{7}$
It can be written as

$= \dfrac{2 \times -6}{3 \times 7}$
By further calculation

$= \dfrac{2 \times -2}{7}$

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

Evaluate :
$\dfrac{5}{4} \times \dfrac{3}{7}$

$\dfrac{5}{4} \times \dfrac{3}{7}$
It can be written as

$= \dfrac{5 \times 3}{4 \times 7}$

$= \dfrac{15}{28}$

What should be added to $\dfrac{-7}{12}$ to get $\dfrac{3}{8}$ ?

Consider

$x$ as the required rational number
Other number

$= \dfrac{-7}{12}$
Sum of two numbers

$= \dfrac{3}{8}$
Using the question

$\dfrac{-7}{12} + x = \dfrac{3}{8}$
So we get

$x = \dfrac{3}{8} - \left (-\dfrac{7}{12} \right )$

LCM of

$8$ and

$12$ is

$24$

$x = \dfrac{3 \times 3}{8 \times 3} + \dfrac{7 \times 2}{12 \times 2}$

By further calculation

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

So we get

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

What should be added to $\dfrac{-3}{5}$ to get $2$ ?

Let

$x$ be the rational number to be added.
Other number

$= \dfrac{-3}{5}$
Here the sum of two numbers

$=2$

From the question

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

$\Rightarrow x = 2 + \dfrac{3}{5}$

$\Rightarrow x = \dfrac{2 \times 5 + 3\times 1}{5}$

$\Rightarrow x = \dfrac{10 + 3}{5}$

$\Rightarrow x = \dfrac{13}{5}$

$\Rightarrow x = 2\dfrac{3}{5}$

Hence,

$2\dfrac{3}{5}$ should be added.

What should be added to $\dfrac{-3}{16}$ to get $\dfrac{11}{24}$

Consider

$x$ as the required rational number
Other number

$= \dfrac{-3}{16}$
Sum of two numbers

$= \dfrac{11}{24}$
From the question

$\dfrac{-3}{16} + x = \dfrac{11}{24}$
By further calculation

$x = \dfrac{11}{24} + \dfrac{3}{16}$
LCM of

$16$ and

$24$ is

$48$

$x = \dfrac{11 \times 2}{24 \times 2} + \dfrac{3 \times 3}{16 \times 3}$

So we get

$x = \dfrac{22}{48} + \dfrac{9}{48}$

$x = \dfrac{22 + 9}{48} = \dfrac{31}{48}$

What should be subtracted from $\dfrac{5}{9}$ to get $\dfrac{9}{5}$ ?

Consider

$x$ as the first number
Other number is

$\dfrac{5}{9}$
Here the difference between two numbers is

$\dfrac{9}{5}$
Using the question

$\dfrac{5}{9} - x = \dfrac{9}{5}$
So we get

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

LCM of

$9$ and

$5$ is

$45$

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

By further calculation

$x = \dfrac{25}{45} - \dfrac{81}{45}$

$x = \dfrac{25 - 81}{45} = \dfrac{-56}{45}$

Multiply :
$\dfrac{3}{25}$ and $\dfrac{4}{5}$

$\dfrac{3}{25}$ and

$\dfrac{4}{5}$
It can be written as

$= \dfrac{3}{25} \times \dfrac{4}{5}$

By further calculation

$= \dfrac{3 \times 4}{25 \times 5}$

$= \dfrac{12}{125}$

Evaluate :
$3\dfrac{1}{8} \times \left ( -2 \dfrac{2}{5} \right)$

$3\dfrac{1}{8} \times \left ( -2 \dfrac{2}{5} \right)$
It can be written as

$= \dfrac{3 \times 8 + 1}{8} \times \dfrac{-(2 \times 5 + 2)}{5}$
By further calculation

$= \dfrac{25}{8} \times \dfrac{-12}{5}$
So we get

$= \dfrac{25 \times -12}{8 \times 5}$
On further simplification

$= \dfrac{5 \times -3}{2 \times 1}$

$= \dfrac{-15}{2}$

Multiply:
$35$ and $\dfrac{-18}{25}$

$35$ and

$\dfrac{-18}{25}$
It can be written as

$= 35 \times \dfrac{-18}{25}$
By further calculation

$= \dfrac{35 \times -18}{25}$
So we get

$= \dfrac{7 \times -18}{5}$

$= \dfrac{-126}{5}$

Multiply:
$1\dfrac{1}{6}$ and $18$

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

$18$
It can be written as

$= \dfrac{7}{6} \times 18$
By further calculation

$= 7 \times 3$

$= 21$

Multiply:
$2 \dfrac{1}{14}$ and $-7$

$2 \dfrac{1}{14}$ and

$-7$
It can be written as

$= \dfrac{2 \times 14 + 1}{14} \times (-7)$
By further calculation

$= \dfrac{29}{14} \times (-7)$
So we get

$= \dfrac{29 \times -1}{2}$

$= \dfrac{-29}{2}$

Evaluate :
$2\dfrac{14}{25} \times \dfrac{-5}{16}$

$2\dfrac{14}{25} \times \dfrac{-5}{16}$
It can be written as

$= \dfrac{2 \times 25 + 14}{25} \times \dfrac{-5}{16}$
By further calculation

$= \dfrac{64}{25} \times \dfrac{-5}{16}$

$= \dfrac{64\times -5}{25 \times 16}$

On further simplification

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

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

Multiply:
$3\dfrac{3}{5}$ and $-10$

$3\dfrac{3}{5}$ and

$-10$
It can be written as

$= \dfrac{3 \times 5 + 3}{5} \times -10$
By further calculation

$= \dfrac{18}{5} \times -10$
So we get

$= 18 \times -2$

$= -36$

Evaluate :
$\dfrac{-8}{9} \times \dfrac{-3}{16}$

$\dfrac{-8}{9} \times \dfrac{-3}{16}$
It can be written as

$= \dfrac{-8 \times -3}{9 \times 16}$

By further calculation

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

$= \dfrac{1}{6}$

Multiply:
$5 \dfrac{1}{8}$ and $-16$

$5 \dfrac{1}{8}$ and

$-16$
It can be written as

$= \dfrac{41}{8} \times (-16)$
By further calculation

$= 41 \times -2$

$= -82$

Evaluate :
$\dfrac{5}{-27} \times \dfrac{-9}{20}$

$\dfrac{5}{-27} \times \dfrac{-9}{20}$
It can be written as

$= \dfrac{5 \times -9}{-27 \times 20}$

By further calculation

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

$= \dfrac{1}{12}$

The sum of two rational numbers is $-5$ . If one of them is $\dfrac{-52}{25}$ , find the other.

It is given that
Sum of two rational numbers

$= -5$
One rational numbers

$= \dfrac{-52}{52}$
Other rational numbers

$=x$
From the question

$\dfrac{-52}{25}+x=-5$

$x= -5 - \left (\dfrac{-52}{25} \right )$
Here LCM is

$25$

$= \dfrac{-5 \times 25}{1 \times 25} + \dfrac{52 \times 1}{25 \times 1}$

By further calculation

$= \dfrac{-125 + 52}{25}$

$= \dfrac{-73}{25}$

What rational number should be added to $\dfrac{-3}{16}$ to get $\dfrac{11}{24}$ ?

It is given that
Sum of two rational numbers

$= \dfrac{11}{24}$
One rational number

$= \dfrac{-3}{16}$

Other number

$=x$

$\dfrac{-3}{4}+x=\dfrac{11}{24}$

$x = \dfrac{11}{24} - \left ( \dfrac{-3}{16} \right )$
It can be written as

$= \dfrac{11}{24} + \dfrac{3}{16}$

Here LCM of

$16$ and

$24$ is

$48$

$= \dfrac{11 \times 2}{24 \times 2} + \dfrac{3 \times 3}{16 \times 3}$

By further calculation

$\dfrac{22 + 9}{48}$

$= \dfrac{31}{48}$
1739490_1826140_ans_aff0af65fc074cb0b40b02fee386bb45.png

What rational number should be added to $\dfrac{-3}{5}$ to get $2$ ?

So the required rational number is

$x$
It can be written as

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

$x=2+\dfrac{3}{5}$
LCM of

$1$ and

$5$ is

$5$

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

By further calculation

$= \dfrac{10 + 3}{5}$
So we get

$= \dfrac{13}{5}$

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

The product of two numbers is $14$ . If one of the numbers is $\dfrac{-8}{7}$ , find the other.

It is given that
Product of two numbers

$= 14$
One of the number

$= \dfrac{-8}{7}$
Other number

$= x$

$x\times \dfrac{-8}{7}=14$

$x=14 \div \dfrac{-8}{7}$
We cam write it as

$= 14 \times \dfrac{-7}{8}$

$= \dfrac{-98}{8}$

$= \dfrac{-49}{4}$

Multiply:
$\dfrac{27}{28}$ and $-14$

$\dfrac{27}{28}$ and

$-14$
It can be written as

$= \dfrac{27}{28} \times -14$
By further calculation

$= \dfrac{27 \times -1}{2}$

$= \dfrac{-27}{2}$

What rational number should be subtracted from $\dfrac{-5}{12}$ to get $\dfrac{5}{24}$ ?

Required rational number

$=x$

$= \dfrac{-5}{12} - x=\dfrac{5}{24}$

$x=\dfrac{-5}{12}-\dfrac{5}{24}$

Here the LCM of

$12$ and

$24$ is

$24$

$= \dfrac{-5 \times 2}{12 \times 2} - \dfrac{5 \times 1}{24 \times 1}$

By further calculation

$= \dfrac{-10 - 5}{24}$

So we get

$= \dfrac{-15}{24}$

$= \dfrac{-5}{8}$
1739479_1826142_ans_9015e4be34ed4a43ba9da0bdbf7be288.png

Evaluate :
$-1 + \dfrac{2}{-3} + \dfrac{5}{6}$

$-1 + \dfrac{2}{-3} + \dfrac{5}{6}$
Here LCM of

$3$ and

$6$ is

$6$
So we get

$= \dfrac{-1 \times 6}{1 \times 6} - \dfrac{2 \times 2}{3 \times 2} + \dfrac{5 \times 1}{6 \times 1}$
By further calculation

$= \dfrac{-6 -4 + 5}{6}$
We get

$= \dfrac{-10 + 5}{6}$

$= \dfrac{-5}{6}$
1747179_1826138_ans_e34e906fc1bd4c5fa4214d0798c2a006.png

Evaluate :
$\dfrac{-2}{3} - \left (\dfrac{-5}{7} \right)$

$\dfrac{-2}{3} - \left (\dfrac{-5}{7} \right)$
It can be written as

$= \dfrac{-2}{3} + \dfrac{5}{7}$
Here LCM of

$3$ and

$7$ is

$21$
So we get

$= \dfrac{-2 \times 7}{3 \times 7} + \dfrac{5 \times 3}{7 \times 3}$
By further calculation

$= \dfrac{-14 + 15}{21}$

$= \dfrac{1}{21}$
1747138_1826135_ans_907f3cdb42a84939bcf7041878792748.png

By what number should $\dfrac{-3}{8}$ be multiplied so that the product is $\dfrac{-9}{16}$ ?

Number

$= \dfrac{-3}{8} \div \dfrac{-9}{16}$
We can write it as

$= \dfrac{-3}{8} \times \dfrac{16}{-9}$
By further calculation

$= \dfrac{2}{3}$

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

What rational number should be subtracted from $\dfrac{5}{8}$ to get $\dfrac{8}{5}$ ?

Rational number

$=x$

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

$x=\dfrac{5}{8}-\dfrac{8}{5}$
Here LCM of

$8$ and

$5$ is

$40$

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

By further calculation

$= \dfrac{25 - 64}{40}$

$= \dfrac{-39}{40}$
1739460_1826143_ans_4edb829f7e9942048a82e464cc7d47dd.png

The product of two rational numbers is $24$ . If one of them is $\dfrac{-36}{11}$ , find the other.

It is given that
Product of two rational numbers

$= 24$
One rational number

$=x$

$= \dfrac{-36}{11} \times x=24$
Other rational number

$= 24 \div \left ( \dfrac{-36}{11} \right )$
It can be written as

$= 24 \times \dfrac{-11}{36}$
By further calculation

$= 2 \times \dfrac{-11}{3}$

$= \dfrac{-22}{3}$

By what rational number should we multiply $\dfrac{20}{-9}$ , so that the product may be $\dfrac{-5}{9}$ ?

Here the required rational number

$=x$

$=\dfrac{20}{-9}\times x=\dfrac{-5}{9}$

$x = \dfrac{-5}{9} \div \dfrac{20}{-9}$
By firther calculation

$= \dfrac{-5}{9} \times \dfrac{-9}{20}$

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

$\dfrac{1}{3}+\dfrac{4}{5}$ of $\dfrac{7}{5}\div 1\dfrac{17}{25}$ .

1322967_1584842_ans_65d7db03d0b549bb8fd6d4c4ef2a06e4.jpg

+	$-\dfrac{1}{9}$	$\dfrac{4}{11}$	$\dfrac{-5}{6}$
$\dfrac{2}{3}$
$-\dfrac{5}{4}$		$-\dfrac{39}{44}$
$-\dfrac{1}{3}$

Material	Recycled
Paper	$\dfrac{5}{11}$
Aluminium Cans	$\dfrac{5}{8}$
Glass	$\dfrac{2}{5}$
Scap	$\dfrac{3}{4}$

Rational Numbers - Class 7 Maths - Extra Questions

Does the following pairs represent the same rational number:−1620\dfrac {-16}{20} and 20−25\dfrac {20}{-25}

Does the following pairs represent the same rational number:−2−3\dfrac {-2}{-3} and 23\dfrac {2}{3}

Which of the following pairs represent the same rational numbers?8−5\dfrac {8}{-5} and −2415\dfrac {-24}{15}

Multiply:−−611-\dfrac{-6}{11} by −5536.\dfrac{-55}{36}. If the answer is ab\dfrac ab, then the value of (a+b)(a+b) is

Multiply:−−825-\dfrac{-8}{25} by −516\dfrac{-5}{16}

Simplify:−1936×16\dfrac{-19}{36}\times 16

Simplify:−916×−64−27\dfrac{-9}{16}\times \dfrac{-64}{-27}

Simplify:76×−328\dfrac{7}{6}\times \dfrac{-3}{28}

Simplify:−139×27−26\dfrac{-13}{9}\times \dfrac{27}{-26}

Simplify and express the result as a rational number in standard form, and find the value of aa−507×143=−ab\dfrac{-50}{7}\times \dfrac{14}{3}=\dfrac{-a}{b}

Simplify:−59×72−5\dfrac{-5}{9}\times \dfrac{72}{-5}

Simplify:(135×83)−(−52×113)\left(\dfrac{13}{5}\times \dfrac{8}{3}\right)-\left(\dfrac{-5}{2}\times \dfrac{11}{3}\right)

Simplify:(137×1126)−(−43×56)\left(\dfrac{13}{7}\times \dfrac{11}{26}\right)-\left(\dfrac{-4}{3}\times \dfrac{5}{6}\right)

Simplify:(12×14)+(12×6)\left(\dfrac{1}{2}\times \dfrac{1}{4}\right)+\left(\dfrac{1}{2}\times 6\right)

Simplify:(−5×215)−(6×29)\left(-5\times \dfrac{2}{15}\right)-\left(6\times \dfrac{2}{9}\right)

Simplify:(−43×12−5)+(37×2115)\left(\dfrac{-4}{3}\times \dfrac{12}{-5}\right)+\left(\dfrac{3}{7}\times \dfrac{21}{15}\right)

Simplify:(85×−32)+(−310×1116)\left(\dfrac{8}{5}\times \dfrac{-3}{2}\right)+\left(\dfrac{-3}{10}\times \dfrac{11}{16}\right)

Simplify:(258×25)−(35×−109)\left(\dfrac{25}{8}\times \dfrac{2}{5}\right)-\left(\dfrac{3}{5}\times \dfrac{-10}{9}\right).If the answer will be ab\dfrac ab, then the value of a+ba+b will be

Simplify:(−94×53)+(132×56)\left(\dfrac{-9}{4}\times \dfrac{5}{3}\right)+\left(\dfrac{13}{2}\times \dfrac{5}{6}\right)

If (32×16)+(53×72)−(138×43)=474×a\left(\dfrac{3}{2}\times \dfrac{1}{6}\right)+\left(\dfrac{5}{3}\times \dfrac{7}{2}\right)-\left(\dfrac{13}{8}\times \dfrac{4}{3}\right)=\dfrac{47}{4 \times a}, then what is the value of a?a?

The product of two rational numbers is −89\dfrac{-8}{9}. If one of the numbers is −415\dfrac{-4}{15} then find the other.

Fill in the blank:The product of a rational number and its reciprocal is _____.1

Fill in the blank:The number 00 is ____ the reciprocal of any number.

Fill in the blank:The reciprocal of a negative rational number is _____.

Fill in the blank:Reciprocal of 1a , a≠0\dfrac{1}{a}\ ,\ a\neq 0 is .........

Fill in the blank:−45×(57+−89)=(−45×........)+−45×−89\dfrac{-4}{5}\times \left(\dfrac{5}{7}+\dfrac{-8}{9}\right)=\left(\dfrac{-4}{5}\times ........\right)+\dfrac{-4}{5}\times \dfrac{-8}{9}

Fill in the blank:Zero has _____ reciprocal.

Divide:11 by 12\dfrac{1}{2}

Fill in the blank:The reciprocal of a positive rational number is _____.

Simplify the following: √24/8+√54/9\sqrt{24}/8 + \sqrt{54}/9

Compare :−112 -1 \dfrac{1}{2} and 00

Compare :−72 \dfrac{-7}{2} and 52 \dfrac{5}{2}

Compare :−3 -3 and 234 2\dfrac{3}{4}

Compare :14 \dfrac{1}{4} and 00

Compare :00 and 34 \dfrac{3}{4}

Write the following rational numbers in ascending order:13\cfrac{1}{3}, −29\cfrac{-2}{9}, −43\cfrac{-4}{3}

Which of the two rational numbers in each of the following pairs of rational numbers is greater?−411,311\dfrac{-4}{11}, \dfrac{3}{11}

Which of the two rational numbers in each of the following pairs of rational numbers is greater?−38,0\dfrac{-3}{8}, 0

Which of the two rational numbers in each of the following pairs of rational numbers is smaller?−43,8−7\dfrac{-4}{3}, \dfrac{8}{-7}

Which of the two rational numbers in each of the following pairs of rational numbers is smaller?16−5,3\dfrac{16}{-5}, 3

Which of the two rational numbers in each of the following pairs of rational numbers is smaller?−125,−3\dfrac{-12}{5}, -3

Which of the two rational numbers in each of the following pairs of rational numbers is smaller?−6−13,713\dfrac{-6}{-13}, \dfrac{7}{13}

Which of the two rational numbers in each of the following pairs of rational numbers is greater?5−8,−712\dfrac{5}{-8}, \dfrac{-7}{12}

Add the following rational numbers:−811\dfrac{-8}{11} and −411\dfrac{-4}{11}

Add the following rational numbers:−57\dfrac{-5}{7} and 37\dfrac{3}{7}

Add the following rational numbers:613\dfrac{6}{13} and −913\dfrac{-9}{13}

Add the following rational numbers:−727\dfrac{-7}{27} and 1118\dfrac{11}{18}

Subtract the first rational number from the second in each of the following:38,58\dfrac{3}{8}, \dfrac{5}{8}

Fill in the blanks:−167+...=...+−167=−167\dfrac {-16}{7} + ... = ... + \dfrac {-16}{7} = \dfrac {-16}{7}.

Which of the two rational numbers is greater in each of the following pairs?79\cfrac79 or −59\cfrac{-5}9

Simplify:65×37−15×37\dfrac{6}{5}\times\dfrac{3}{7}-\dfrac{1}{5}\times\dfrac{3}{7}

Solve513−−826\dfrac{5}{13}-\dfrac{-8}{26}

Multiply:−24 -24 and 516 \dfrac{5}{16}

Fill in the following blanks:The number _______ and _____ are their own reciprocals.

Simplify:1+−451+\dfrac{-4}{5}

Simplify:0+−350+\dfrac{-3}{5}

Simplify:−1621×145\dfrac{-16}{21}\times \dfrac{14}{5}

Multiply:−8−9-\dfrac{8}{-9} by −7−16\dfrac{-7}{-16}

Multiply:−67-\dfrac{6}{7} by −4936\dfrac{-49}{36}

Multiply:−−89-\dfrac{-8}{9} by 364\dfrac{3}{64}

solve213+114+256+37122\frac{1}{3} + 1\frac{1}{4} + 2\frac{5}{6} + 3\frac{7}{{12}}

Find the rational number that should be subtracted from −57\dfrac{-5}{7} to get −1-1?

Subtract :0 0 from −38 \dfrac{-3}{8}

Which of the two rational numbers in each of the following pairs of rational numbers is greater?52,0\dfrac{5}{2}, 0

Which of the two rational numbers in each of the following pairs of rational numbers is greater?59,−3−8\dfrac{5}{9}, \dfrac{-3}{-8}

Which of the two rational numbers in each of the following pairs of rational numbers is greater?−58,3−4\dfrac{-5}{8}, \dfrac{3}{-4}

Which of the two rational numbers in each of the following pairs of rational numbers is greater?−712,5−12\dfrac{-7}{12}, \dfrac{5}{-12}

Which of the two rational numbers in each of the following pairs of rational numbers is greater?4−9,−3−7\dfrac{4}{-9}, \dfrac{-3}{-7}

Fill in the blanks:−9+−218=(...)+(−9)-9 + \dfrac {-21}{8} = (...) + (-9).

Give four rational numbers equivalent to 49\dfrac {4}{9}

Arrange 58,316,−14\displaystyle \frac{5}{8}, \frac{3}{16}, -\frac{1}{4} and 1732\displaystyle \frac{17}{32} in descending order of their magnitudes.Identify the largest rational number.

Multiply 613\dfrac{6}{13} by the reciprocal of −716\dfrac{-7}{16}.

Does the following pair represents the same rational number:−721\dfrac {-7}{21} and 39\dfrac {3}{9}

The sum of two rational numbers is −45\dfrac{-4}{5}.If one of them is −1120\dfrac{-11}{20}, find the other

Rewrite the following rational numbers in the simplest form:−86\cfrac{-8}{6}

Rewrite the following rational numbers in the simplest form:−4472\cfrac{-44}{72}

Rewrite the following rational number in the simplest form:2545\cfrac{25}{45}

Simplify:−138+536\dfrac{-13}{8}+\dfrac{5}{36}

Add the following rational numbers:613\dfrac{6}{13} and −913\dfrac{-9}{13}.

Does the following pairs represent the same rational number:
$\dfrac {-16}{20}$ and $\dfrac {20}{-25}$

Does the following pairs represent the same rational number:
$\dfrac {-2}{-3}$ and $\dfrac {2}{3}$

Which of the following pairs represent the same rational numbers?
$\dfrac {8}{-5}$ and $\dfrac {-24}{15}$

Multiply:
$-\dfrac{-6}{11}$ by $\dfrac{-55}{36}.$
If the answer is $\dfrac ab$ , then the value of $(a+b)$ is

Multiply:
$-\dfrac{-8}{25}$ by $\dfrac{-5}{16}$

Simplify:
$\dfrac{-19}{36}\times 16$

Simplify:
$\dfrac{-9}{16}\times \dfrac{-64}{-27}$

Simplify:
$\dfrac{7}{6}\times \dfrac{-3}{28}$

Simplify:
$\dfrac{-13}{9}\times \dfrac{27}{-26}$

Simplify and express the result as a rational number in standard form, and find the value of $a$
$\dfrac{-50}{7}\times \dfrac{14}{3}=\dfrac{-a}{b}$

Simplify:
$\dfrac{-5}{9}\times \dfrac{72}{-5}$

Simplify:
$\left(\dfrac{13}{5}\times \dfrac{8}{3}\right)-\left(\dfrac{-5}{2}\times \dfrac{11}{3}\right)$

Simplify:
$\left(\dfrac{13}{7}\times \dfrac{11}{26}\right)-\left(\dfrac{-4}{3}\times \dfrac{5}{6}\right)$

Simplify:
$\left(\dfrac{1}{2}\times \dfrac{1}{4}\right)+\left(\dfrac{1}{2}\times 6\right)$

Simplify:
$\left(-5\times \dfrac{2}{15}\right)-\left(6\times \dfrac{2}{9}\right)$

Simplify:
$\left(\dfrac{-4}{3}\times \dfrac{12}{-5}\right)+\left(\dfrac{3}{7}\times \dfrac{21}{15}\right)$

Simplify:
$\left(\dfrac{8}{5}\times \dfrac{-3}{2}\right)+\left(\dfrac{-3}{10}\times \dfrac{11}{16}\right)$

Simplify:
$\left(\dfrac{25}{8}\times \dfrac{2}{5}\right)-\left(\dfrac{3}{5}\times \dfrac{-10}{9}\right)$ .

If the answer will be $\dfrac ab$ , then the value of $a+b$ will be

Simplify:
$\left(\dfrac{-9}{4}\times \dfrac{5}{3}\right)+\left(\dfrac{13}{2}\times \dfrac{5}{6}\right)$

If $\left(\dfrac{3}{2}\times \dfrac{1}{6}\right)+\left(\dfrac{5}{3}\times \dfrac{7}{2}\right)-\left(\dfrac{13}{8}\times \dfrac{4}{3}\right)=\dfrac{47}{4 \times a}$ , then what is the value of $a?$

The product of two rational numbers is $\dfrac{-8}{9}$ . If one of the numbers is $\dfrac{-4}{15}$ then find the other.

Fill in the blank:
The product of a rational number and its reciprocal is _____.1

Fill in the blank:
The number $0$ is ____ the reciprocal of any number.

Fill in the blank:
The reciprocal of a negative rational number is _____.

Fill in the blank:
Reciprocal of $\dfrac{1}{a}\ ,\ a\neq 0$ is .........

Fill in the blank:
$\dfrac{-4}{5}\times \left(\dfrac{5}{7}+\dfrac{-8}{9}\right)=\left(\dfrac{-4}{5}\times ........\right)+\dfrac{-4}{5}\times \dfrac{-8}{9}$

Fill in the blank:
Zero has _____ reciprocal.

Divide:
$1$ by $\dfrac{1}{2}$

Fill in the blank:
The reciprocal of a positive rational number is _____.

Simplify the following:
$\sqrt{24}/8 + \sqrt{54}/9$

Compare :
$-1 \dfrac{1}{2}$ and $0$

Compare :
$\dfrac{-7}{2}$ and $\dfrac{5}{2}$

Compare :
$-3$ and $2\dfrac{3}{4}$

Compare :
$\dfrac{1}{4}$ and $0$

Compare :
$0$ and $\dfrac{3}{4}$

Write the following rational numbers in ascending order:
$\cfrac{1}{3}$ , $\cfrac{-2}{9}$ , $\cfrac{-4}{3}$

Which of the two rational numbers in each of the following pairs of rational numbers is greater?
$\dfrac{-4}{11}, \dfrac{3}{11}$

Which of the two rational numbers in each of the following pairs of rational numbers is greater?
$\dfrac{-3}{8}, 0$

Which of the two rational numbers in each of the following pairs of rational numbers is smaller?
$\dfrac{-4}{3}, \dfrac{8}{-7}$

Which of the two rational numbers in each of the following pairs of rational numbers is smaller?
$\dfrac{16}{-5}, 3$

Which of the two rational numbers in each of the following pairs of rational numbers is smaller?
$\dfrac{-12}{5}, -3$

Which of the two rational numbers in each of the following pairs of rational numbers is smaller?
$\dfrac{-6}{-13}, \dfrac{7}{13}$

Which of the two rational numbers in each of the following pairs of rational numbers is greater?
$\dfrac{5}{-8}, \dfrac{-7}{12}$

Add the following rational numbers:
$\dfrac{-8}{11}$ and $\dfrac{-4}{11}$

Add the following rational numbers:
$\dfrac{-5}{7}$ and $\dfrac{3}{7}$

Add the following rational numbers:
$\dfrac{6}{13}$ and $\dfrac{-9}{13}$

Add the following rational numbers:
$\dfrac{-7}{27}$ and $\dfrac{11}{18}$

Subtract the first rational number from the second in each of the following:
$\dfrac{3}{8}, \dfrac{5}{8}$

Fill in the blanks:
$\dfrac {-16}{7} + ... = ... + \dfrac {-16}{7} = \dfrac {-16}{7}$ .

Which of the two rational numbers is greater in each of the following pairs?
$\cfrac79$ or $\cfrac{-5}9$

Simplify:
$\dfrac{6}{5}\times\dfrac{3}{7}-\dfrac{1}{5}\times\dfrac{3}{7}$

Solve
$\dfrac{5}{13}-\dfrac{-8}{26}$

Multiply:
$-24$ and $\dfrac{5}{16}$

Fill in the following blanks:
The number ___ and _ are their own reciprocals.

Simplify:
$1+\dfrac{-4}{5}$

Simplify:
$0+\dfrac{-3}{5}$

Simplify:
$\dfrac{-16}{21}\times \dfrac{14}{5}$

Multiply:
$-\dfrac{8}{-9}$ by $\dfrac{-7}{-16}$

Multiply:
$-\dfrac{6}{7}$ by $\dfrac{-49}{36}$

Multiply:
$-\dfrac{-8}{9}$ by $\dfrac{3}{64}$

solve
$2\frac{1}{3} + 1\frac{1}{4} + 2\frac{5}{6} + 3\frac{7}{{12}}$

Find the rational number that should be subtracted from $\dfrac{-5}{7}$ to get $-1$ ?

Subtract :
$0$ from $\dfrac{-3}{8}$

Which of the two rational numbers in each of the following pairs of rational numbers is greater?
$\dfrac{5}{2}, 0$

Which of the two rational numbers in each of the following pairs of rational numbers is greater?
$\dfrac{5}{9}, \dfrac{-3}{-8}$

Which of the two rational numbers in each of the following pairs of rational numbers is greater?
$\dfrac{-5}{8}, \dfrac{3}{-4}$

Which of the two rational numbers in each of the following pairs of rational numbers is greater?
$\dfrac{-7}{12}, \dfrac{5}{-12}$

Which of the two rational numbers in each of the following pairs of rational numbers is greater?
$\dfrac{4}{-9}, \dfrac{-3}{-7}$

Fill in the blanks:
$-9 + \dfrac {-21}{8} = (...) + (-9)$ .

Give four rational numbers equivalent to $\dfrac {4}{9}$

Arrange $\displaystyle \frac{5}{8}, \frac{3}{16}, -\frac{1}{4}$ and $\displaystyle \frac{17}{32}$ in descending order of their magnitudes.Identify the largest rational number.

Multiply $\dfrac{6}{13}$ by the reciprocal of $\dfrac{-7}{16}$ .

Does the following pair represents the same rational number:
$\dfrac {-7}{21}$ and $\dfrac {3}{9}$

The sum of two rational numbers is $\dfrac{-4}{5}$ .If one of them is $\dfrac{-11}{20}$ , find the other

Rewrite the following rational numbers in the simplest form:
$\cfrac{-8}{6}$

Rewrite the following rational numbers in the simplest form:
$\cfrac{-44}{72}$

Rewrite the following rational number in the simplest form:
$\cfrac{25}{45}$

Simplify:
$\dfrac{-13}{8}+\dfrac{5}{36}$

Add the following rational numbers:
$\dfrac{6}{13}$ and $\dfrac{-9}{13}$ .

Simplify:
$\dfrac{-16}{9}+\dfrac{-5}{12}=\dfrac{-39.5 \times a}{36}$

What is the value of $'a'?$