NCERT Solutions Class 7 Maths Chapter 9 Rational Numbers Ex 9.2

NCERT Solutions Class 7 Mathematics 
Chapter – 9 (Rational Numbers)

The NCERT Solutions in English Language for Class 7 Mathematics Chapter – 9 Rational Numbers Exercise 9.2 has been provided here to help the students in solving the questions from this exercise. 

Chapter : 9 Rational Numbers

Exercise – 9.2 

1. Find the sum:

(i) \mathbf{\frac{5}{4}}+ \mathbf{\left ( -\frac{11}{4} \right )}         (ii) \mathbf{\frac{5}{3}} + \mathbf{\frac{3}{5}}                        (iii) \mathbf{\frac{-9}{10}} + \mathbf{\frac{22}{15}}

(iv) \mathbf{\frac{-3}{-11}}+\mathbf{\frac{5}{9}}          (v) \mathbf{\frac{-8}{19}}+\mathbf{\frac{(-2)}{57}}                (vi) \mathbf{\frac{-2}{3} + 0}

(vii) \mathbf{-2\frac{1}{3} + 4\frac{3}{5}}

Solution –

(i) \mathbf{\frac{5}{4}}+ \mathbf{\left ( -\frac{11}{4} \right )} 

= \frac{5}{4} – \frac{11}{4}

= \frac{5-11}{4} [∵ denominator is same in both the rational numbers]

= \frac{-6}{4}

= \frac{-3}{2}                 ———–[∵ Divide both numerator and denominator by 3]

(ii) \mathbf{\frac{5}{3}} + \mathbf{\frac{3}{5}}

LCM of 3 and 5 is 15

= \frac{5\times 5}{3\times 5} + \frac{2\times 3}{5\times 3}

= \frac{25}{15} + \frac{9}{15}

= \frac{34}{15}

(iii) \mathbf{\frac{-9}{10}} + \mathbf{\frac{22}{15}}

LCM of 10 and 15 is 30

= \frac{-9\times 3}{10\times 3} +\frac{22\times 2}{15\times 2}

= \frac{-27}{30} + \frac{44}{30}

= \frac{17}{30}

(iv) \mathbf{\frac{-3}{-11}}+\mathbf{\frac{5}{9}}

= \frac{3}{11} + \frac{5}{9}

LCM of 11 and 9 is 99

= \frac{3\times 9}{11\times 9} + \frac{5\times 11}{9\times 11}

= \frac{27}{99}+\frac{55}{99}

= \frac{82}{99}

(v) \mathbf{\frac{-8}{19}}+\mathbf{\frac{(-2)}{57}}

= \frac{-8}{19}-\frac{2}{57}

LCM of 19 and 57 is 57

= \frac{-8\times 3}{19\times 3} - \frac{2}{57}

= \frac{-24}{57} - \frac{2}{57}

= \frac{-24-2}{57}

= \frac{-26}{57}

(vi) \mathbf{\frac{-2}{3} + 0}

= -\frac{2}{3} + 0

= -\frac{2}{3}

(vii) \mathbf{-2\frac{1}{3} + 4\frac{3}{5}}

= \frac{-7}{3} + \frac{23}{5}

LCM of 3 and 5 is 15

= \frac{-7\times 5}{3\times 5} + \frac{23\times 3}{5\times 3}

= \frac{-35}{15} + \frac{69}{15}

= \frac{-35+69}{15}

= \frac{34}{15}

2. Find

(i) \mathbf{\frac{7}{24}-\frac{17}{36}}                       (ii) \mathbf{\frac{5}{63}-\left ( \frac{-6}{21} \right )}                          (iii) \mathbf{\frac{-6}{13} - \left ( \frac{-7}{15} \right )}

(iv) \mathbf{\frac{-3}{8}-\frac{7}{11}}                    (v) \mathbf{-2\frac{1}{9} - 6}                        

Solution –

(i) \mathbf{\frac{7}{24}-\frac{17}{36}}

LCM of 24 and 36 is 72

= \frac{7\times 3}{24\times 3} - \frac{17\times 2}{36\times 2}

= \frac{21}{72} - \frac{34}{72}

= \frac{21-34}{72}

= \frac{-13}{72}

(ii) \mathbf{\frac{5}{63}-\left ( \frac{-6}{21} \right )}

= \frac{5}{63}+\frac{6}{21}

LCM of 63 and 21 is 63

= \frac{5}{63} + \frac{6\times 3}{21\times 3}

= \frac{5}{63}+\frac{18}{63}

= \frac{5+18}{63}

= \frac{23}{63}

(iii) \mathbf{\frac{-6}{13} - \left ( \frac{-7}{15} \right )}

= \frac{-6}{13}+\frac{7}{15}

LCM of 13 and 15 is 195

= \frac{-6\times 15}{13\times 15} + \frac{7\times 13}{15\times 13}

= \frac{-90}{195} +\frac{91}{195}

= \frac{-90+91}{195}

= \frac{1}{195}

(iv) \mathbf{\frac{-3}{8}-\frac{7}{11}}

LCM of 8 and 11 is 88

= \frac{-3\times 11}{8\times 11}-\frac{7\times 8}{11\times 8}

= \frac{-33}{88}-\frac{56}{88}

= \frac{-33-56}{88}

= \frac{-89}{88}

(v) \mathbf{-2\frac{1}{9} - 6}

= \frac{-19}{9} -6

LCM of 9 and 1 is 9

= \frac{-19}{9} -\frac{6\times 9}{1\times 9}

= \frac{-19}{9} -\frac{54}{9}

= \frac{-19-54}{9}

= \frac{-73}{9}

3. Find the product:

(i) \mathbf{\frac{9}{2}\times \left (\frac{-7}{4} \right )}                      (ii) \mathbf{\frac{3}{10}\times (-9)}                    (iii) \mathbf{\frac{-6}{5}\times \frac{9}{11}}

(iv) \mathbf{\frac{3}{7}\times \left (\frac{-2}{5} \right )}                    (v) \mathbf{\frac{3}{11}\times \frac{2}{5}}                          (vi) \mathbf{\frac{3}{-5}\times \frac{-5}{3}}

Solution –

(i) \mathbf{\frac{9}{2}\times \left (\frac{-7}{4} \right )} 

= \frac{9\times -7}{2\times 4}

= \frac{-63}{8}

(ii) \mathbf{\frac{3}{10}\times (-9)}

= \frac{3}{10}\times \frac{-9}{1}

= \frac{3\times -9}{10\times 1}

= \frac{-27}{10}

(iii) \mathbf{\frac{-6}{5}\times \frac{9}{11}}

= \frac{-6\times 9}{5\times 11}

= \frac{-54}{55}

(iv) \mathbf{\frac{3}{7}\times \left (\frac{-2}{5} \right )}

= \frac{3\times -2}{7\times 5}

= \frac{-6}{35}

(v) \mathbf{\frac{3}{11}\times \frac{2}{5}}

= \frac{3\times 2}{11\times 5}

= \frac{6}{55}

(vi) \mathbf{\frac{3}{-5}\times \frac{-5}{3}}

= \frac{3\times -5}{-5\times 3}

On simplifying,

= \frac{-1}{-1}

= 1

4. Find the value of:

(i) \mathbf{(-4)\div \frac{2}{3}}                    (ii) \mathbf{\frac{-3}{5}\div 2}                          (iii) \mathbf{\frac{-4}{5}\div (-3)}

(iv) \mathbf{\frac{-1}{8}\div \frac{3}{4}}                    (v) \mathbf{\frac{-2}{13}\div \frac{1}{7}}                          (vi) \mathbf{\frac{-7}{12}\div \left (\frac{-2}{13} \right )}     

(vii) \mathbf{\frac{3}{13}\div \left (\frac{-4}{65} \right )} 

Solution –

(i) \mathbf{(-4)\div \frac{2}{3}}   

= \frac{-4}{1} \times\frac{3}{2}  —————- [∵ reciprocal of \frac{2}{3} is \frac{3}{2}]

= \frac{-4\times 3}{1\times 2}

= \frac{-2\times 3}{1\times 1}

= – 6

(ii) \mathbf{\frac{-3}{5}\div 2}

= \frac{-3}{5}\times \frac{1}{2}    —————- [∵ reciprocal of \frac{2}{1} is \frac{1}{2}]

= \frac{-3\times 1}{5\times 2}

= \frac{-3}{10}

(iii) \mathbf{\frac{-4}{5}\div (-3)}

= \frac{-4}{5}\times \frac{1}{-3}  ———— [∵ reciprocal of (-3) is \frac{1}{-3} ]

= \frac{-4\times 1}{5\times -3}

= \frac{-4}{-15}

= \frac{4}{15}

(iv) \mathbf{\frac{-1}{8}\div \frac{3}{4}}

= \frac{-1}{8}\times \frac{4}{3}  ————– [∵ reciprocal of \frac{3}{4} is \frac{4}{3} ]

= \frac{-1\times 4}{8\times 3}

= \frac{-1\times 1}{2\times 3}

= \frac{-1}{6}

(v) \mathbf{\frac{-2}{13}\div \frac{1}{7}}

= \frac{-2}{13}\times \frac{7}{1}  ————- [∵ reciprocal of \frac{1}{7} is \frac{7}{1} ]

= \frac{-2\times 7}{13\times 1}

= \frac{-14}{13}

(vi) \mathbf{\frac{-7}{12}\div \left (\frac{-2}{13} \right )}

= \frac{-7}{12}\times \frac{13}{-2}  ———— [∵ reciprocal of  \frac{-2}{13} is \frac{13}{-2} ]

= \frac{-7\times 13}{12\times -2}

= –\frac{-91}{-24}

= \frac{91}{24}

(vii) \mathbf{\frac{3}{13}\div \left (\frac{-4}{65} \right )}

= \frac{3}{13}\times \frac{65}{-4}  ———— [∵ reciprocal of \frac{-4}{65} is \frac{65}{-4}]

= \frac{3\times 65}{13\times -4}

= \frac{3\times 5}{1\times -4}

= \frac{15}{-4} 

 

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