# Rational Numbers - Class 8 - Mathematics

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## Examples - Rational Numbers | R.D. Sharma | Mathematics | Class 8

Add $\frac{3}{5}$ and $\frac{13}{5}$.

To add $\frac{3}{5}$ and $\frac{13}{5}$, we simply add their numerators because they have the same denominator:

$$ \frac{3}{5} + \frac{13}{5} = \frac{3 + 13}{5} = \frac{16}{5} $$

So, the sum is $\frac{16}{5}$.

Add $\frac{7}{9}$ and $\frac{-12}{9}$.

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Sign up nowAdd $\frac{-5}{9}$ and $\frac{-17}{9}$.

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Sign up nowAdd $\frac{4}{-11}$ and $\frac{7}{11}$.

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Sign up nowAdd $\frac{5}{12}$ and $\frac{3}{8}$.

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Add $\frac{7}{9}$ and $4$.

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Add $\frac{3}{8}$ and $\frac{-5}{12}$.

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Simplify: $\frac{8}{-15}+\frac{4}{-3}$.

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Simplify: $\frac{7}{-26}+\frac{16}{39}$.

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Verify: $\left(\frac{a}{b}+\frac{c}{d}\right)+\frac{e}{f}=\frac{a}{b}+\left(\frac{c}{d}+\frac{e}{f}\right)$ for $\frac{a}{b}=\frac{-2}{3}, \frac{c}{d}=\frac{5}{7}$ and $\frac{e}{f}=\frac{-1}{6}$

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Using commutativity and associativity of addition of rational numbers, express each of the following as a rational number:

(i) $\frac{3}{5}+\frac{-7}{6}+\frac{2}{5}+\frac{-5}{6}$

(ii) $\frac{4}{3}+\frac{-4}{5}+\frac{-2}{3}+\frac{7}{5}-2$

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Re-arrange suitably and find the sum in each of the following:

(i) $\frac{5}{3}+\frac{11}{2}+\frac{-9}{4}+\frac{-8}{3}+\frac{-7}{2}$

(ii) $\frac{-4}{7}+\frac{7}{6}+\frac{2}{7}+3+\frac{-11}{6}$

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Subtract $\frac{3}{4}$ from $\frac{5}{6}$.

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Subtract $\frac{-3}{8}$ from $\frac{-5}{7}$.

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Subtract $\frac{-3}{5}$ from $\frac{9}{10}$.

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The sum of two rational numbers is $\frac{-3}{5}$. If one of the number is $\frac{-9}{20}$, find the other.

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What number should be added to $\frac{-5}{8}$ so as to get $\frac{5}{9}$ ?

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What should be subtracted from $\frac{-3}{4}$ so as to get $\frac{5}{6}$ ?

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

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Find: $\frac{-7}{4}+\frac{5}{3}+\frac{-5}{6}+\frac{1}{3}+\frac{-1}{2}$

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Evaluate: $\frac{-12}{5}+\frac{-7}{20}+\frac{3}{14}+\frac{1}{7}+\frac{-1}{10}$

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Find: $\frac{3}{4}+\left(\frac{-3}{5}\right)+\left(\frac{-2}{3}\right)+\frac{5}{8}+\left(\frac{-4}{15}\right)$

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Evaluate: $\frac{6}{7}-2+\frac{-7}{9}+\frac{19}{21}$

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Multiply:

(i) $\frac{3}{4}$ by $\frac{5}{7}$

(ii) $\frac{3}{7}$ by $\left(\frac{-4}{5}\right)$

(iii) $\left(\frac{-5}{9}\right($ by 4

(iv) $\left(\frac{-36}{7}\right)$ by $\left(-\frac{28}{9}\right)$

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Simplify:

(i) $\frac{-8}{7} \times \frac{14}{5}$

(ii) $\frac{13}{6} \times \frac{-18}{91}$

(iii) $\frac{-5}{9} \times \frac{72}{-125}$

(iv) $\frac{-22}{9}+\frac{-51}{-88}$

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Simplify:

(i) $\left(\frac{-16}{5} \times \frac{20}{8}\right)-\left(\frac{15}{5} \times \frac{-35}{3}\right)$

(ii) $\left(\frac{-3}{2} \times \frac{4}{5}\right)+\left(\frac{9}{5} \times \frac{-10}{3}\right)-\left(\frac{1}{2} \times \frac{3}{4}\right($

(iii) $\left(\frac{-7}{18} \times \frac{15}{-7}\right)-\left(1 \times \frac{1}{4}\right)+\left(\frac{1}{2} \times \frac{1}{4}\right)$

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Write the reciprocal of each of the following rational numbers:

(i) 7

(ii) -11

(iii) $\frac{2}{5}$

(iv) $\frac{-7}{15}$

(v) $\frac{5}{-12}$

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Find the reciprocal of

(i) $\frac{2}{5} \times \frac{4}{9}$

(ii) $\frac{-3}{8} \times \frac{-7}{13}$

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Divide:

(i) $\frac{3}{5}$ by $\frac{4}{25}$

(ii) $\frac{-8}{9}$ by $\frac{4}{3}$

(iii) $\frac{-16}{21}$ by $\frac{-4}{3}$

(iv) $\frac{-8}{13}$ by $\frac{3}{-26}$

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The product of two rational numbers is $\frac{-28}{81}$. If one of the number is $\frac{14}{27}$, find the other.

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By what number should we multiply $\frac{3}{-14}$, so that the product may be $\frac{5}{12}$.

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Write any three rational numbers between -2 and 0.

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Find four rational numbers between $\frac{2}{3}$ and $\frac{4}{5}$.

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Find five rational numbers between $\frac{-3}{2}$ and $\frac{5}{3}$.

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Find a rational number between $-2$ and $6$.

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Find a rational number between $\frac{-2}{3}$ and $\frac{1}{4}$.

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Find three rational numbers between $-2$ and $5$ .

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## Exercise 1.1 - Rational Numbers | R.D. Sharma | Mathematics | Class 8

Add the following rational numbers:

(i) $\frac{-5}{7}$ and $\frac{3}{7}$

(ii) $\frac{-15}{4}$ and $\frac{7}{4}$

(iii) $\frac{-8}{11}$ and $\frac{-4}{11}$

(iv) $\frac{6}{13}$ and $\frac{-9}{13}$

Here are the results of adding the given rational numbers:

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

Decimal approximation: $-0.285714$

(ii) $\frac{-15}{4} + \frac{7}{4} = -2$

(iii) $\frac{-8}{11} + \frac{-4}{11} = \frac{-12}{11}$

Decimal approximation: $-1.09$

Mixed fraction form: $-1 \frac{1}{11}$

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

Decimal approximation: $-0.230769$

These computations provide exact answers along with visual aids to better understand the results on a number line.

Add the following rational numbers with steps

(i) $\frac{3}{4}$ and $\frac{-5}{8}$

(ii) $\frac{5}{-9}$ and $\frac{7}{3}$

(iii) -3 and $\frac{3}{5}$

(iv) $\frac{-7}{27}$ and $\frac{11}{18}$

(v) $\frac{31}{-4}$ and $\frac{-5}{8}$

(vi) $\frac{5}{36}$ and $\frac{-7}{12}$

(vii) $\frac{-5}{16}$ and $\frac{7}{24}$

(viii) $\frac{7}{-18}$ and $\frac{8}{27}$

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Simplify: with steps

(i) $\frac{8}{9}+\frac{-11}{6}$

(ii) $3+\frac{5}{-7}$

(iii) $\frac{1}{-12}+\frac{2}{-15}$

(iv) $\frac{-8}{19}+\frac{-4}{57}$

(v) $\frac{7}{9}+\frac{3}{-4}$

(vi) $\frac{5}{26}+\frac{11}{-39}$

(vii) $\frac{-16}{9}+\frac{-5}{12}$

(viii) $\frac{-13}{8}+\frac{5}{36}$

(ix) $0+\frac{-3}{5}$

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

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Add and express the sum as a mixed fraction: with steps

(i) $\frac{-12}{5}$ and $\frac{43}{10}$

(ii) $\frac{24}{7}$ and $\frac{-11}{4}$

(iii) $\frac{-31}{6}$ and $\frac{-27}{8}$

(iv) $\frac{101}{6}$ and $\frac{7}{8}$

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## Exercise 1.2 - Rational Numbers | R.D. Sharma | Mathematics | Class 8

Verify commutativity of addition of rational numbers for each of the following pairs of rational numbers:

(i) $\frac{-11}{5}$ and $\frac{4}{7}$

(ii) $\frac{4}{9}$ and $\frac{7}{-12}$

(iii) $\frac{-3}{5}$ and $\frac{-2}{-15}$

(iv) $\frac{2}{-7}$ and $\frac{12}{-35}$

(v) 4 and $\frac{-3}{5}$

(vi) -4 and $\frac{4}{-7}$

(I) The results demonstrate the commutativity of addition for the given pairs of rational numbers:

- Adding $\frac{-11}{5}$ and $\frac{4}{7}$ gives $\frac{-11}{5} + \frac{4}{7} = \frac{-57}{35}$.

- Likewise, adding $\frac{4}{7}$ and $\frac{-11}{5}$ yields $\frac{4}{7} + \frac{-11}{5} = \frac{-57}{35}$.

Hence, in both sequences, the result is the same, confirming commutativity.

(ii) - Adding $\frac{4}{9}$ and $\frac{7}{-12}$ gives $\frac{4}{9} + \frac{7}{-12} = \frac{-5}{36}$.

- Similarly, adding $\frac{7}{-12}$ and $\frac{4}{9}$ results in $\frac{7}{-12} + \frac{4}{9} = \frac{-5}{36}$.

Thus, in both orders, the sum remains the same, verifying the commutativity for these pairs of rational numbers.

(iii) - Adding $\frac{-3}{5}$ and $\frac{-2}{-15}$ computes to $\frac{-3}{5} + \frac{-2}{-15} = \frac{-7}{15}$.

- Adding in the reverse order, $\frac{-2}{-15}$ and $\frac{-3}{5}$, also results in $\frac{-2}{-15} + \frac{-3}{5} = \frac{-7}{15}$.

Thus, the sum is the same in both cases, further confirming the commutativity of addition for these rational numbers.

(iv) - Adding $\frac{2}{-7}$ and $\frac{12}{-35}$ yields $\frac{2}{-7} + \frac{12}{-35} = \frac{-22}{35}$.

- Similarly, the reverse order, $\frac{12}{-35}$ and $\frac{2}{-7}$, results in $\frac{12}{-35} + \frac{2}{-7} = \frac{-22}{35}$.

In both sequences, the outcome remains the same, confirming the commutativity for these pairs as well.

(v) - Adding $4$ and $\frac{-3}{5}$ gives $4 + \frac{-3}{5} = \frac{17}{5}$.

- Adding in the reverse order, $\frac{-3}{5}$ and $4$, results in $\frac{-3}{5} + 4 = \frac{17}{5}$.

Thus, for this pair too, the sum remains consistent in both cases, demonstrating the commutativity of addition for these numbers.

(vi) - Adding $-4$ and $\frac{4}{-7}$ yields $-4 + \frac{4}{-7} = \frac{-32}{7}$.

- Similarly, adding $\frac{4}{-7}$ and $-4$ results in $\frac{4}{-7} + -4 = \frac{-32}{7}$.

In both sequences, we receive the same result, confirming the commutativity of addition for these numbers as well.

Verify associativity of addition of rational numbers i.e., $(x+y)+z=x+(y+z)$, when:

(i) $x=\frac{1}{2}, y=\frac{2}{3}, z=-\frac{1}{5}$

(ii) $x=\frac{-2}{5}, y=\frac{4}{3}, z=\frac{-7}{10}$

(iii) $x=\frac{-7}{11}, y=\frac{2}{-5}, z=\frac{-3}{22}$

(iv) $x=-2, y=\frac{3}{5}, z=\frac{-4}{3}$

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Write the additive inverse of each of the following rational numbers:

(i) $\frac{-2}{17}$

(ii) $\frac{3}{-11}$

(iii) $\frac{-17}{5}$

(iv) $\frac{-11}{-25}$

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Write the negative (additive inverse) of each of the following:

(i) $\frac{-2}{5}$

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

(iii) $\frac{-16}{13}$

(iv) $\frac{-5}{1}$

(v) 0

(vi) 1

(vii) -1

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Using commutativity and associativity of addition of rational numbers, express each of the following as a rational number:

(i) $\frac{2}{5}+\frac{7}{3}+\frac{-4}{5}+\frac{-1}{3}$

(ii) $\frac{3}{7}+\frac{-4}{9}+\frac{-11}{7}+\frac{7}{9}$

(iii) $\frac{2}{5}+\frac{8}{3}+\frac{-11}{15}+\frac{4}{5}+\frac{-2}{3}$

(iv) $\frac{4}{7}+0+\frac{-8}{9}+\frac{-13}{7}+\frac{17}{21}$

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Re-arrange suitably and find the sum in each of the following:

(i) $\frac{11}{12}+\frac{-17}{3}+\frac{11}{2}+\frac{-25}{2}$

(ii) $\frac{-6}{7}+\frac{-5}{6}+\frac{-4}{9}+\frac{-15}{7}$

(iii) $\frac{3}{5}+\frac{7}{3}+\frac{9}{5}+\frac{-13}{15}+\frac{-7}{3}$

(iv) $\frac{4}{13}+\frac{-5}{8}+\frac{-8}{13}+\frac{9}{13}$

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

(vi) $\frac{1}{8}+\frac{5}{12}+\frac{2}{7}+\frac{7}{12}+\frac{9}{7}+\frac{-5}{16}$

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## Exercise 1.3 - Rational Numbers | R.D. Sharma | Mathematics | Class 8

Subtract the first rational number from the second in each of the following:

(i) $\frac{3}{8}, \frac{5}{8}$

(ii) $\frac{-7}{9}, \frac{4}{9}$

(iii) $\frac{-2}{11}, \frac{-9}{11}$

(iv) $\frac{11}{13}, \frac{-4}{13}$

(v) $\frac{1}{4}, \frac{-3}{8}$

(vi) $\frac{-2}{3}, \frac{5}{6}$

(vii) $\frac{-6}{7}, \frac{-13}{14}$

(viii) $\frac{-8}{33}, \frac{-7}{22}$

Here are the results of subtracting the first rational number from the second for each pair:

(i) To subtract the first rational number from the second, $\frac{5}{8} - \frac{3}{8}$:

Given the same denominator, we subtract the numerators directly: $5 - 3 = 2$.

So, $\frac{5}{8} - \frac{3}{8} = \frac{2}{8}$. Simplifying $\frac{2}{8}$ gives $\frac{1}{4}$ (or $0.25$).

Here are the results for each subtraction:

(ii) Subtracting the first rational number from the second, $\frac{-7}{9}, \frac{4}{9}$ gives:

$\frac{4}{9} - \left(\frac{-7}{9}\right) = \frac{11}{9} \text{ (or approximately } 1.2222\text{)}.$

(iii) Subtracting the first rational number from the second, $\frac{-2}{11}, \frac{-9}{11}$ results in:

$\frac{-9}{11} - \left(\frac{-2}{11}\right) = \frac{-7}{11} \text{ (or approximately } -0.6364\text{)}.$

(iv) Subtracting the first rational number from the second, $\frac{11}{13}, \frac{-4}{13}$ yields:

$\frac{-4}{13} - \frac{11}{13} = \frac{-15}{13} \text{ (or approximately } -1.1538\text{)}.$

(v) Subtracting the first rational number from the second, $\frac{-3}{8} - \frac{1}{4}$:

Convert $\frac{1}{4}$ into eighths to simplify: $\frac{1 \times 2}{4 \times 2} = \frac{2}{8}$.

Now, subtract: $\frac{-3}{8} - \frac{2}{8} = \frac{-3 - 2}{8} = \frac{-5}{8}$.

Hence, the result is $-\frac{5}{8}$ (or $-0.625$).

(vi) Subtracting the first rational number from the second, $\frac{-2}{3}$ and $\frac{5}{6}$ results in:

$\frac{5}{6} - \left(\frac{-2}{3}\right) = \frac{3}{2} \text{ (or } 1.5\text{)}.$

(vii) Subtracting the first rational number from the second, $\frac{-6}{7}, \frac{-13}{14}$ gives:

$\frac{-13}{14} - \left(\frac{-6}{7}\right) = \frac{-1}{14} \text{ (or approximately } -0.0714\text{)}.$

(viii) Subtracting the first rational number from the second, $\frac{-8}{33}, \frac{-7}{22}$ results in:

$\frac{-7}{22} - \left(\frac{-8}{33}\right) = \frac{-5}{66} \text{ (or approximately } -0.0758\text{)}.$

Evaluate each of the following:

(i) $\frac{2}{3}-\frac{3}{5}$

(ii) $-\frac{4}{7}-\frac{2}{-3}$

(iii) $\frac{4}{7}-\frac{-5}{-7}$

(iv) $-2-\frac{5}{9}$

(v) $\frac{-3}{-8}-\frac{-2}{7}$

(vi) $\frac{-4}{13}-\frac{-5}{26}$

(vii) $\frac{-5}{14}-\frac{-2}{7}$

(viii) $\frac{13}{15}-\frac{12}{25}$

(ix) $\frac{-6}{13}-\frac{-7}{13}$

(x) $\frac{7}{24}-\frac{19}{36}$

(xi) $\frac{5}{63}-\frac{-8}{21}$

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The sum of the two numbers is $\frac{5}{9}$. If one of the numbers is $\frac{1}{3}$, find the other.

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The sum of two numbers is $\frac{-1}{3}$. If one of the numbers is $\frac{-12}{3}$, find the other.

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The sum of two numbers is $\frac{-4}{3}$. If one of the numbers is -5 , find the other.

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The sum of two rational numbers is -8 . If one of the numbers is $\frac{-15}{7}$, find the other.

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What should be added to $\frac{-7}{8}$ so as to get $\frac{5}{9}$ ?

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What number should be added to $\frac{-5}{11}$ so as to get $\frac{26}{33}$ ?

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What number should be added to $\frac{-5}{7}$ to get $\frac{-2}{3}$ ?

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

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What number should be subtracted from $\frac{3}{7}$ to get $\frac{5}{4}$ ?

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What should be added to $\left(\frac{2}{3}+\frac{3}{5}\right)$ to get $\frac{-2}{15}$ ?

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What should be added to $\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{5}\right)$ to get 3 ?

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What should be subtracted from $\left(\frac{3}{4}-\frac{2}{3}\right)$ to get $\frac{-1}{6}$ ?

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Fill in the blanks:

(i) $\frac{-4}{13}-\frac{-3}{26}=\ldots$

(ii) $\frac{-9}{14}+\ldots=-1$

(iii) $\frac{-7}{9}+\ldots=3$

(iv) $\ldots+\frac{15}{23}=4$

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## Exercise 1.4 - Rational Numbers | R.D. Sharma | Mathematics | Class 8

Simplify each of the following and write as a rational number of the the form $\frac{p}{q}$ :

(i) $\frac{3}{4}+\frac{5}{6}+\frac{-7}{8}$

(ii) $\frac{2}{3}+\frac{-5}{6}+\frac{-7}{9}$

(iii) $\frac{-11}{2}+\frac{7}{6}+\frac{-5}{8}$

(iv) $\frac{-4}{5}+\frac{-7}{10}+\frac{-8}{15}$

(v) $\frac{-9}{10}+\frac{22}{15}+\frac{13}{-20}$

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

To simplify $\frac{3}{4} + \frac{5}{6} + \frac{-7}{8}$ and express it as a rational number of the form $\frac{p}{q}$, follow these steps:

1. **Find a common denominator** for the fractions $\frac{3}{4}$, $\frac{5}{6}$, and $\frac{-7}{8}$. The denominators are $4$, $6$, and $8$.

- Least Common Multiple (LCM) of $4$, $6$, and $8$ is $24$.

2. **Convert each fraction** to have the common denominator of $24$:

- $\frac{3}{4}$ becomes $\frac{3 \times 6}{4 \times 6} = \frac{18}{24}$,

- $\frac{5}{6}$ becomes $\frac{5 \times 4}{6 \times 4} = \frac{20}{24}$,

- $\frac{-7}{8}$ becomes $\frac{-7 \times 3}{8 \times 3} = \frac{-21}{24}$.

3. **Add the fractions** now that they have the same denominator:

- $\frac{18}{24} + \frac{20}{24} + \frac{-21}{24} = \frac{18 + 20 - 21}{24} = \frac{17}{24}$.

Thus, the simplified form of $\frac{3}{4} + \frac{5}{6} + \frac{-7}{8}$ is $\frac{17}{24}$.

Here are the simplified forms for each expression as rational numbers:

(ii) $\frac{2}{3} + \frac{-5}{6} + \frac{-7}{9}$ simplifies to $\frac{-17}{18}$ (or approximately $-0.9444$).

(iii) $\frac{-11}{2} + \frac{7}{6} + \frac{-5}{8}$ simplifies to $\frac{-119}{24}$ (or approximately $-4.9583$).

(iv) $\frac{-4}{5} + \frac{-7}{10} + \frac{-8}{15}$ simplifies to $\frac{-61}{30}$ (or approximately $-2.0333$).

(v) $\frac{-9}{10} + \frac{22}{15} + \frac{13}{-20}$ simplifies to $\frac{-1}{12}$ (or approximately $-0.08333$).

(vi) $\frac{5}{3} + \frac{3}{-2} + \frac{-7}{3} + 3$ simplifies to $\frac{5}{6}$ (or approximately $0.8333$).

Express each of the following as a rational number of the form $\frac{p}{q}$ :

(i) $\frac{-8}{3}+\frac{-1}{4}+\frac{-11}{6}+\frac{3}{8}-3$

(ii) $\frac{6}{7}+1+\frac{-7}{9}+\frac{19}{21}+\frac{-12}{7}$

(iii) $\frac{15}{2}+\frac{9}{8}+\frac{-11}{3}+6+\frac{-7}{6}$

(iv) $\frac{-7}{4}+0+\frac{-9}{5}+\frac{19}{10}+\frac{11}{14}$

(v) $\frac{-7}{4}+\frac{5}{3}+\frac{-1}{2}+\frac{-5}{6}+2$

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Simplify:

(i) $\frac{-3}{2}+\frac{5}{4}-\frac{7}{4}$

(ii) $\frac{5}{3}-\frac{7}{6}+\frac{-2}{3}$

(iii) $\frac{5}{4}-\frac{7}{6}-\frac{-2}{3}$

(iv) $\frac{-2}{5}-\frac{-3}{10}-\frac{-4}{7}$

(v) $\frac{5}{6}+\frac{-2}{5}-\frac{-2}{15}$

(vi) $\frac{3}{8}-\frac{-2}{9}+\frac{-5}{36}$

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## Exercise 1.5 - Rational Numbers | R.D. Sharma | Mathematics | Class 8

Multiply:

(i) $\frac{7}{11}$ by $\frac{5}{4}$

(ii) $\frac{5}{7}$ by $\frac{-3}{4}$

(iii) $\frac{-2}{9}$ by $\frac{5}{11}$

(iv) $\frac{-3}{17}$ by $\frac{-5}{-4}$

(v) $\frac{9}{-7}$ by $\frac{36}{-11}$

(vi) $\frac{-11}{13}$ by $\frac{-21}{7}$

(vii) $-\frac{3}{5}$ by $-\frac{4}{7}$ (viii) $-\frac{15}{11}$ by 7

To multiply $\frac{7}{11}$ by $\frac{5}{4}$, follow these steps:

1. **Multiply the numerators**: Multiply the numerators of the two fractions together.

- $7 \times 5 = 35$.

2. **Multiply the denominators**: Multiply the denominators of the two fractions together.

- $11 \times 4 = 44$.

3. **Combine the results**: Place the product of the numerators over the product of the denominators to form the result.

- $\frac{35}{44}$.

Thus, the product of $\frac{7}{11}$ and $\frac{5}{4}$ is $\frac{35}{44}$.

(ii) $\frac{5}{7} \times \left(-\frac{3}{4}\right) = -\frac{15}{28}$

(iii) $\frac{-2}{9} \times \frac{5}{11} = -\frac{10}{99}$

(iv) $\frac{-3}{17} \times \frac{-5}{-4} = -\frac{15}{68}$

(v) $\frac{9}{-7} \times \frac{36}{-11} = \frac{324}{77}$

(vi) $\frac{-11}{13} \times \frac{-21}{7} = \frac{33}{13}$

(vii) $-\frac{3}{5} \times -\frac{4}{7} = \frac{12}{35}$

(viii) $-\frac{15}{11} \times 7 = -\frac{105}{11}$

Multiply:

(i) $\frac{-5}{17}$ by $\frac{51}{-60}$

(ii) $\frac{-6}{11}$ by $\frac{-55}{36}$

(iii) $\frac{-8}{25}$ by $\frac{-5}{16}$

(iv) $\frac{6}{7}$ by $\frac{-49}{36}$

(v) $\frac{8}{-9}$ by $\frac{-7}{-16}$

(vi) $\frac{-8}{9}$ by $\frac{3}{64}$

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Simplify each of the following and express the result as a rational number in standard form:

(i) $\frac{-16}{21} \times \frac{14}{5}$

(ii) $\frac{7}{6} \times \frac{-3}{28}$

(iii) $\frac{-19}{36} \times 16$

(iv) $\frac{-13}{9} \times \frac{27}{-26}$

(v) $\frac{-9}{16} \times \frac{-64}{-27}$

(vi) $\frac{-50}{7} \times \frac{14}{3}$

(vii) $\frac{-11}{9} \times \frac{-81}{-88}$

(viii) $\frac{-5}{9} \times \frac{72}{-25}$

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Simplify:

(i) $\left(\frac{25}{8} \times \frac{2}{5}\right)-\left(\frac{3}{5} \times \frac{-10}{9}\right)$

(ii) $\left(\frac{1}{2} \times \frac{1}{4}\right)+\left(\frac{1}{2} \times 6\right)$

(iii) $\left(-5 \times \frac{2}{15}\right)-\left(-6 \times \frac{2}{9}\right)$

(iv) $\left(\frac{-9}{4} \times \frac{5}{3}\right)+\left(\frac{13}{2} \times \frac{5}{6}\right)$

(v) $\left(\frac{-4}{3} \times \frac{12}{-5}\right)+\left(\frac{3}{7} \times \frac{21}{15}\right)$

(vi) $\left(\frac{13}{5} \times \frac{8}{3}\right)-\left(\frac{-5}{2} \times \frac{11}{3}\right)$

(vii) $\left(\frac{13}{7} \times \frac{11}{26}\right)-\left(\frac{-4}{3} \times \frac{5}{6}\right)$

(viii) $\left(\frac{8}{5} \times \frac{-3}{2}\right)+\left(\frac{-3}{10} \times \frac{11}{16}\right)$

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Simplify:

(i) $\left(\frac{3}{2} \times \frac{1}{6}\right)+\left(\frac{5}{3} \times \frac{7}{2}\right)-\left(\frac{13}{8} \times \frac{4}{3}\right)$

(ii) $\left(\frac{1}{4} \times \frac{2}{7}\right)-\left(\frac{5}{14} \times \frac{-2}{3}\right)+\left(\frac{3}{7} \times \frac{9}{2}\right)$

(iii) $\left(\frac{13}{9} \times \frac{-15}{2}\right)+\left(\frac{7}{3} \times \frac{8}{5}\right)+\left(\frac{3}{5} \times \frac{1}{2}\right)$

(iv) $\left(\frac{3}{11} \times \frac{5}{6}\right)-\left(\frac{9}{12} \times \frac{4}{3}\right)+\left(\frac{5}{13} \times \frac{6}{15}\right)$

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## Exercise 1.6 - Rational Numbers | R.D. Sharma | Mathematics | Class 8

Verify the property: $x \times y=y \times x$ by taking:

(i) $x=-\frac{1}{3}, y=\frac{2}{7}$

(ii) $x=\frac{-3}{5}, y=\frac{-11}{13}$

(iii) $x=2, y=\frac{7}{-8}$ (iv) $x=0, y=\frac{-15}{8}$

(i) By taking $x=-\frac{1}{3}$ and $y=\frac{2}{7}$ and applying the commutative property of multiplication $x \times y = y \times x$, we have:

- Multiplying $x$ by $y$: $-\frac{1}{3} \times \frac{2}{7} = -\frac{2}{21}$.

- Multiplying $y$ by $x$: $\frac{2}{7} \times -\frac{1}{3} = -\frac{2}{21}$.

Hence, in both cases, the result is the same: $-\frac{2}{21}$, verifying the commutative property of multiplication for these values.

(ii) For $x = \frac{-3}{5}, y = \frac{-11}{13}$, both $x \times y$ and $y \times x$ yield $\frac{33}{65}$.

(iii) For $x = 2, y = \frac{7}{-8}$, both $x \times y$ and $y \times x$ yield $-\frac{7}{4}$.

(iv) For $x = 0, y = \frac{-15}{8}$, both $x \times y$ and $y \times x$ yield $0$.

This demonstrates the commutative property of multiplication, indicating that the order in which two numbers are multiplied does not change the product.

Verify the property: $x \times(y \times z)=(x \times y) \times z$ by taking:

(i) $x=\frac{-7}{3}, y=\frac{12}{5}, z=\frac{4}{9}$

(ii) $x=0, y=\frac{-3}{5}, z=\frac{-9}{4}$

(iii) $x=\frac{1}{2}, y=\frac{5}{-4}, z=\frac{-7}{5}$

(iv) $x=\frac{5}{7}, y=\frac{-12}{13}, z=\frac{-7}{18}$

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Verify the property: $x \times(y+z)=x \times y+x \times z$ by taking:

(i) $x=\frac{-3}{7}, y=\frac{12}{13}, z=\frac{-5}{6}$

(ii) $x=\frac{-12}{5}, y=\frac{-15}{4}, z=\frac{8}{3}$

(iii) $x=\frac{-8}{3}, y=\frac{5}{6}, z=\frac{-13}{12}$

(iv) $x=\frac{-3}{4}, y=\frac{-5}{2}, z=\frac{7}{6}$

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Use the distributivity of multiplication of rational numbers over their addition to simplify:

(i) $\frac{3}{5} \times\left(\frac{35}{24}+\frac{10}{1}\right)$

(ii) $\frac{-5}{4} \times\left(\frac{8}{5}+\frac{16}{5}\right)$

(iii) $\frac{2}{7} \times\left(\frac{7}{16}-\frac{21}{4}\right)$ (iv) $\frac{3}{4} \times\left(\frac{8}{9}-40\right)$

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Find the multiplicative inverse (reciprocal) of each of the following rational numbers:

(i) 9

(ii) -7

(iii) $\frac{12}{5}$

(iv) $\frac{-7}{9}$

(v) $\frac{-3}{-5}$

(vi) $\frac{2}{3} \times \frac{9}{4}$

(vii) $\frac{-5}{8} \times \frac{16}{15}$

(viii) $-2 \times \frac{-3}{5}$

(ix) -1

(x) $\frac{0}{3}$

(xi) 1

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Name the property of multiplication of rational numbers illustrated by the following statements:

(i) $\frac{-5}{16} \times \frac{8}{15}=\frac{8}{15} \times \frac{-5}{16}$

(ii) $\frac{-17}{5} \times 9=9 \times \frac{-17}{5}$

(iii) $\frac{7}{4} \times\left(\frac{-8}{3}+\frac{-13}{12}\right)=\frac{7}{4} \times \frac{-8}{3}+\frac{7}{4} \times \frac{-13}{12}$

(iv) $\frac{-5}{9} \times\left(\frac{4}{15} \times \frac{-9}{8}\right)=\left(\frac{-5}{9} \times \frac{4}{15}\right) \times \frac{-9}{8}$

(v) $\frac{13}{-17} \times 1=\frac{13}{-17}=1 \times \frac{13}{-17}$

(vi) $\frac{-11}{16} \times \frac{16}{-11}=1$

(vii) $\frac{2}{13} \times 0=0=0 \times \frac{2}{13}$

(viii) $\frac{-3}{2} \times \frac{5}{4}+\frac{-3}{2} \times \frac{-7}{6}=\frac{-3}{2} \times\left(\frac{5}{4}+\frac{-7}{6}\right)$

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Fill in the blanks:

(i) The product of two positive rational numbers is always

(ii) The product of a positive rational number and a negative rational number is always

(iii) The product of two negative rational numbers is always

(iv) The reciprocal of a positive rational number is

(v) The reciprocal of a negative rational number is

(vi) Zero has reciprocal.

(vii) The product of a rational number and its reciprocal is

(viii) The numbers and are their own reciprocals.

(ix) If $a$ is reciprocal of $b$, then the reciprocal of $b$ is

(x) The number 0 is the reciprocal of any number.

(xi) Reciprocal of $\frac{1}{a}, a \neq 0$ is

(xii) $(17 \times 12)^{-1}=17^{-1} \times$

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Fill in the blanks:

(i) $-4 \times \frac{7}{9}=\frac{7}{9} \times \ldots \ldots$

(ii) $\frac{5}{11} \times \frac{-3}{8}=\frac{-3}{8} \times \ldots \ldots$

(iii) $\frac{1}{2} \times\left(\frac{3}{4}+\frac{-5}{12}\right)=\frac{1}{2} \times \ldots \ldots+\ldots . \times \frac{-5}{12}$

(iv) $\frac{-4}{5} \times\left(\frac{5}{7}+\frac{-8}{9}\right)=\left(\frac{-4}{5} \times \ldots.\right)+\frac{-4}{5} \times \frac{-8}{9}$

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## Exercise 1.7 - Rational Numbers | R.D. Sharma | Mathematics | Class 8

Divide:

(i) 1 by $\frac{1}{2}$

(ii) 5 by $\frac{-5}{7}$

(iii) $\frac{-3}{4}$ by $\frac{9}{-16}$

(iv) $\frac{-7}{8}$ by $\frac{-21}{16}$

(v) $\frac{7}{-4}$ by $\frac{63}{64}$

(vi) 0 by $\frac{-7}{5}$

(vii), $\frac{-3}{4}$ by -6

(viii) $\frac{2}{3}$ by $\frac{-7}{12}$

(ix) -4 by $\frac{-3}{5}$

(x) $\frac{-3}{-13}$ by $\frac{-4}{65}$

(i) To divide $1$ by $\frac{1}{2}$, follow these steps:

1. **Identify the numbers**: The divisor is $\frac{1}{2}$, and the dividend is $1$.

2. **Reciprocal of the divisor**: Find the multiplicative inverse (reciprocal) of the divisor. The reciprocal of $\frac{1}{2}$ is $2$ because $\frac{1}{2} \times 2 = 1$.

3. **Multiply the dividend by the reciprocal of the divisor**: Multiply $1$ (the dividend) by $2$ (the reciprocal of $\frac{1}{2}$).

- $1 \times 2 = 2$.

Thus, dividing $1$ by $\frac{1}{2}$ results in $2$.

(ii) $5$ by $\frac{-5}{7}$ results in $-7$.

(iii) $\frac{-3}{4}$ by $\frac{9}{-16}$ results in $\frac{4}{3}$ (or approximately $1.3333$).

(iv) $\frac{-7}{8}$ by $\frac{-21}{16}$ results in $\frac{2}{3}$ (or approximately $0.6667$).

(v) $\frac{7}{-4}$ by $\frac{63}{64}$ results in $-\frac{16}{9}$ (or approximately $-1.7778$).

(vi) $0$ by $\frac{-7}{5}$ results in $0$ (since multiplying or dividing zero by any rational number yields zero).

(vii) $\frac{-3}{4}$ by $-6$ results in $\frac{1}{8}$ (or $0.125$).

(viii) $\frac{2}{3}$ by $\frac{-7}{12}$ results in $-\frac{8}{7}$ (or approximately $-1.1429$).

(ix) $-4$ by $\frac{-3}{5}$ results in $\frac{20}{3}$ (or approximately $6.6667$).

(x) $\frac{-3}{-13}$ by $\frac{-4}{65}$ results in $-\frac{15}{4}$ (or $-3.75$).

Find the value and express as a rational number in standard form:

(i) $\frac{2}{5} \div \frac{26}{15}$

(ii) $\frac{10}{3}+\frac{-35}{12}$

(iii) $-6 \div\left(\frac{-8}{17}\right)$

(iv) $\frac{-40}{99} \div(-20)$

(v) $\frac{-22}{27} \div \frac{-110}{18}$

(vi) $\frac{-36}{125}+\frac{-3}{75}$

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The product of two rational numbers is 15 . If one of the numbers is -10 , find the other.

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

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

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By what number should we multiply $\frac{-15}{28}$ so that the product may be $\frac{-5}{7}$ ?

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By what number should we multiply $\frac{-8}{13}$ so that the product may be 24 ?

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By what number should $\frac{-3}{4}$ be multiplied in order to produce $\frac{2}{3}$ ?

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Find $(x+y) \div(x-y)$, if

(i) $x=\frac{2}{3}, y=\frac{3}{2}$

(ii) $x=\frac{2}{5}, y=\frac{1}{2}$

(iii) $x=\frac{5}{4}, y=\frac{-1}{3}$

(iv) $x=\frac{2}{7}, y=\frac{4}{3}$

(v) $x=\frac{1}{4}, y=\frac{3}{2}$

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The cost of $7 \frac{2}{3}$ metres of rope is Rs $12 \frac{3}{4}$. Find its cost per metre.

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The cost of $2 \frac{1}{3}$ metres of cloth is Rs $75 \frac{1}{4}$. Find the cost of cloth per metre.

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By what number should $\frac{-33}{16}$ be divided to get $\frac{-11}{4}$ ?

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Divide the sum of $\frac{-13}{5}$ and $\frac{12}{7}$ by the product of $\frac{-31}{7}$ and $\frac{-1}{2}$.

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Divide the sum of $\frac{65}{12}$ and $\frac{12}{7}$ by their difference. is required for each trouser?

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## Exercise 1.8 - Rational Numbers | R.D. Sharma | Mathematics | Class 8

Find a rational number between -3 and 1 .

To find a rational number between -3 and 1, we can use the method of taking the average of the two numbers. The average of -3 and 1 is given by:

$$ \frac{-3 + 1}{2} = \frac{-2}{2} = -1 $$

Therefore, $-1$ is a rational number between -3 and 1.

Find any five rational numbers less than 2.

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Find two rational numbers between $\frac{-2}{9}$ and $\frac{5}{9}$.

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Find two rational numbers between $\frac{1}{5}$ and $\frac{1}{2}$.

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Find ten rational numbers between $\frac{1}{4}$ and $\frac{1}{2}$.

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Find ten rational numbers between $\frac{-2}{5}$ and $\frac{1}{2}$.

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Find ten rational numbers between $\frac{3}{5}$ and $\frac{3}{4}$.