7th Class Mathematics Rational Numbers Operation on Rational Numbers

     Operation on Rational Numbers                

 

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

 

     Addition 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 4: Add the numerators                

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                                

 

Answer: (b)                

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                  

 

Answer: (c)                

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                  

 

Answer: (c)                

 

 

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                  

 

Answer: (d)    

 

 

 

• 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.