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.                

 

*     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.  

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