# 7th Class Mathematics Rational Numbers Operation on Rational Numbers

## Operation on Rational Numbers

Category : 7th Class

### Operation on Rational Numbers

In this topic we will study about addition, subtraction, multiplication and division of rational numbers.

Step 1: Write the rational number in the standard form.

Step 2: Make the denominator same by taking the LCM of denominators.

Step 3: Write all the rational number with the same denominator.

Step 5: Write the numerator getting after addition on the denominator.

Step 6: Reduce the rational number to lowest term.

Add $2\frac{3}{5},\frac{-15}{13},\frac{-13}{-15},-4\frac{3}{5}$

Solution:

Step 1: The standard form of given rational numbers are

$\frac{13}{5},\frac{-15}{13},\frac{13}{15},\frac{-23}{5}$

Step 2: LCM of 5, 13, 15, 5 is 195

Step3: $\frac{13}{5}=\frac{13\times 39}{5\times 39}=\frac{507}{195}$

$\frac{-15}{13}=\frac{-15\times 15}{13\times 15}=\frac{-225}{195}$

$\frac{13}{15}=\frac{13\times 13}{15\times 13}=\frac{169}{195}$

$\frac{-23}{5}=\frac{23\times 39}{5\times 39}=\frac{-897}{195}$

Step 4: The sum of numerators

$=507+\left( -225 \right)+169+\left( -897 \right)$$=-446$

Step 5: $\frac{-446}{195}$

Step 6: $-2\frac{56}{195}$

Therefore

$2\frac{3}{5}+\frac{-15}{13}+\frac{-13}{-15}+\left( -4\frac{3}{5} \right)=-2\frac{56}{195}$

Subtraction of Two Rational Numbers

Step 1:  Write the rational numbers in standard form.

Step 2:   Make the denominator same by taking LCM.

Step 3:   Subtract numerators.

Step 4:   Write the result with denominator and reduce it to lowest terms.

If the sum of two rational numbers is $\frac{-5}{6}$ and if one of the number is $-\frac{9}{20}$ then find the other rational number.

Solution:

Sum of Rational number $=\frac{-5}{16}$

One number$=\frac{-9}{20}$, Then the other $=\frac{-5}{16}-\left( \frac{-9}{20} \right)$

$=\frac{-25-\left( -36 \right)}{80}=\frac{-25+36}{80}=\frac{11}{80}$

Multiplication of Rational Numbers

Product of rational numbers $=\frac{Product\text{ }of\,the\,numerators}{Product\text{ }of\text{ }the\text{ }denominators}$ and reduce it to the lowest terms.

Reciprocal of Rational Number

For non - zero rational numbers, a rational number is said to be reciprocal of other if the product is 1.

Find the product of  $\frac{13}{6}\times \frac{-18}{91}\times \frac{-5}{9}\times \frac{72}{-125}$

Solution:

$\frac{13}{6}\times \frac{-81}{91}\times \frac{-5}{9}\times \frac{72}{-125}=\frac{13\times \left( -18 \right)\times \left( -5 \right)\times 72}{6\times 91\times 9\left( -125 \right)}$

$=\frac{-3\times 8}{7\times 25}=\frac{-24}{175}$

Find the reciprocal of $\frac{4}{9}\times \frac{-7}{13}\times \frac{-3}{8}$

Solution:

$=\frac{4}{9}\times \frac{-7}{13}\times \frac{-3}{8}=\frac{{{\bcancel{4}}^{1}}}{{{\bcancel{9}}_{3}}}\times \frac{-7}{13}\times \frac{-{{\bcancel{3}}^{1}}}{{{\bcancel{8}}_{2}}}=\frac{7}{78}$

The reciprocal of $\left( \frac{7}{78} \right)$ is $\left( \frac{78}{7} \right)$ because $\frac{7}{78}\times \frac{78}{7}=1$

Division of Rational Numbers

If a and b are two rational numbers in such a way that $b\ne 0,$ then $a\div b$ is same as the product of a and reciprocal of b.

The product of two rational numbers is $\frac{-23}{88}$ If one of the number is$\frac{5}{22}$ then find the other rational number.

Solution:

Product of two number $=\frac{-23}{88}$

one number $=\frac{5}{22},$ So the other number $=\frac{-23}{88}\div \frac{5}{22}=\frac{-23}{88}\times \frac{22}{5}=\frac{-23}{20}$

If $p=\frac{-4}{5}\times \left( \frac{10}{9} \right)\times \left( \frac{3}{4} \right)$ and $Q=(-6)\times \frac{7}{5}\times \frac{2}{3}$then P + Q is equal to:

(a) $-2\frac{14}{15}$

(b) $-4\frac{14}{15}$

(c) $-3\frac{14}{15}$

(d) $-1\frac{14}{15}$

(e) None of these

Solution:

Here $P=\underset{1}{\overset{-1}{\mathop{\left( \frac{-\cancel{4}}{\cancel{5}} \right)}}}\,\times \underset{3}{\overset{-2}{\mathop{\left( \frac{-\cancel{10}}{\cancel{9}} \right)}}}\,\times \frac{\cancel{3}}{\cancel{4}}=\frac{2}{3}$and Q $=\cancel{\overset{2}{\mathop{6}}\,}\times \frac{7}{5}\times \frac{2}{\cancel{3}}=\frac{-28}{5}$

$\Rightarrow P+Q=\frac{2}{3}+\left( \frac{-28}{5} \right)=\frac{10+\left( -84 \right)}{15}=\frac{-74}{15}=-4\frac{14}{15}$

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

(a) $\frac{42}{286}$

(b) $\frac{109}{286}$

(c) $\frac{-21}{286}$

(d) $\frac{21}{286}$

(e) None of these

Explanation

$\left( \frac{\cancel{5}}{13}\times \frac{\overset{2}{\mathop{\cancel{6}}}\,}{\underset{\begin{smallmatrix} \cancel{3} \\ 1 \end{smallmatrix}}{\mathop{\cancel{15}}}\,} \right)\div \left( \underset{\cancel{3}}{\overset{\cancel{3}}{\mathop{\frac{\cancel{9}}{\cancel{12}}}}}\,\times \frac{\cancel{4}}{\cancel{3}} \right)-\left( \frac{\cancel{3}}{11}\times \frac{5}{\underset{2}{\mathop{\cancel{6}}}\,} \right)$

$=\frac{2}{23}\div 1-\frac{5}{22}\Rightarrow \frac{2}{13}-\frac{5}{22}=\frac{44-65}{286}=\frac{-21}{286}$

Find the value of $\frac{-1}{2}+\frac{1}{3}+\frac{-5}{6}+\frac{5}{3}+\frac{-7}{4}.$

(a) $\frac{23}{12}$

(b) $\frac{15}{12}$

(c) $\frac{-13}{12}$

(d) $\frac{-11}{12}$

(e) None of these

The value of ${{(17\times 12)}^{-1}}$Is equal to :

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

(b) $17\times {{\left( \frac{1}{12} \right)}^{-1}}$

(c) ${{\left( \frac{1}{17} \right)}^{-1}}\times {{12}^{-1}}$

(d) $\left( \frac{1}{17} \right)\times {{12}^{-1}}$

(e) None of these

• A number which is in the form of $\frac{a}{b},$ where a and b are integers and $b\ne 0.$
• Every positive rational number is greater than the negative rational number.
• For two rational numbers which are in the standard form$\frac{A}{B}$ and $\frac{C}{D}$ (I) if AD > BC Then $\frac{A}{B}>\frac{C}{D}$ (II) If AD < BC , Then $\frac{A}{B}<\frac{C}{D}$

• A continued fraction is expression of a number as the sum of an integer and a quotient, the denominator of which is the sum of an integer and a quotient, and so on. In general, $y={{a}_{0}}+\frac{{{b}_{0}}}{{{a}_{1}}+\frac{{{b}_{1}}}{{{a}_{2}}+\frac{{{b}_{2}}}{{{a}_{3}}+.....}}}$,where ${{a}_{o}},{{a}_{1}},{{a}_{2}},...$and ${{b}_{o}},{{b}_{1}},{{b}_{2}},...$are all integers.
• In a simple continued fraction, all the b; are equal to 1 and all the a. are positive integers.

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