RBSE Solutions for Class 7 Maths Chapter 9 Rational Numbers Ex 9.2

Rajasthan Board RBSE Solutions for Class 7 Maths Chapter 9 Rational Numbers Ex 9.2 Textbook Exercise Questions and Answers.

RBSE Class 7 Maths Solutions Chapter 9 Rational Numbers Ex 9.2

Question 1.
Find the sum
(i) \(\frac{5}{4}+\left(\frac{-11}{4}\right)\)
Answer:
\(\frac{5}{4}+\left(\frac{-11}{4}\right)=\frac{5}{4}+\left(\frac{-11}{4}\right)\)
= \(\frac{5+(-11)}{4}=\frac{-6}{4}=\frac{-3}{2}\) or -1 \(\frac{1}{2}\)

(ii) \(\frac{5}{3}+\frac{3}{5}\)
Answer:
LCM of 3 and 5 = 15

(iii) \(\frac{-9}{10}+\frac{22}{15}\)
Answer:
LCM of 10 and 15 = 30

(iv) \(\frac{-3}{-11}+\frac{5}{9}\)
Answer:
LCM of 11 and 9 = 99

(v) \(\frac{-8}{19}+\frac{(-2)}{57}\)
Answer:
LCM of 19 and 57 = 57

(vi) \(\frac{-2}{3}\) + 0
Answer:
0 can be written as \(\frac{0}{3}\)
∴ \(\frac{-2}{3}\) + 0 = \(\frac{0}{3}+\frac{-2}{3}=\frac{-2+0}{3}=\frac{-2}{3}\)

(vii) -2\(\frac{1}{3}\) + 4 \(\frac{3}{4}\)
Answer:
-2 \(\frac{1}{3}=\frac{-7}{3}\) and 4 \(\frac{3}{5}=\frac{23}{5}\)
LCM of 3 and 5 = 15

Question 2. 
Find : 
(i) \(\frac{7}{24}-\frac{17}{36}\)
Answer:
LCM of 24 and 36 = 72

(ii) \(\frac{5}{63}-\left(\frac{-6}{21}\right)\)
Answer:
LCM of 63 and 21 = 63

(iii) \(\frac{-6}{13}-\left(\frac{-7}{15}\right)\)
Answer:
LCM of 13 and 15 = 195

(iv) \(\frac{-3}{8}-\frac{7}{11}\)
Answer:
LCM of 8 and 11 = 88

(v) -2\(\frac{1}{9}\) - 6
Answer:
-2 \(\frac{1}{9}=\frac{-19}{9}\) and 6 = \(\frac{6}{1}\)
LCM of 9 and 1 = 9

Question 3.
Find the product:
(i) \(\frac{9}{2} \times\left(\frac{-7}{4}\right)\)
Answer:
\(\frac{9}{2} \times \frac{-7}{4}=\frac{9 \times(-7)}{2 \times 4}=\frac{-63}{8} = -7 \frac{7}{8}\)

(ii) \(\frac{3}{10}\) × (-9)
Answer:
\(\frac{3}{10}\)× (-9) = \(\frac{3}{10} \times \frac{(-9)}{1}=\frac{-27}{10} = -2 \frac{7}{10}\)

(iii) \(\frac{-6}{5} \times \frac{9}{11}\)
Answer:
\(\frac{-6}{5} \times \frac{9}{11}=\frac{-6 \times 9}{5 \times 11}=\frac{-54}{55}\)

(iv) \(\frac{3}{7} \times\left(\frac{-2}{5}\right)\)
Answer:
\(\frac{3}{7} \times\left(\frac{-2}{5}\right)=\frac{3 \times(-2)}{7 \times 5}=\frac{-6}{35}\)

(v) \(\frac{3}{11} \times \frac{2}{5}\)
Answer:
\(\frac{3}{11} \times \frac{2}{5}=\frac{3 \times 2}{11 \times 5}=\frac{6}{55}\)

(vi) \(\frac{3}{-5} \times \frac{-5}{3}\)
Answer:
\(\frac{3}{-5} \times \frac{-5}{3}=\frac{3 \times-5}{-5 \times 3}=\frac{-15}{-15}\) = 1

Question 4.
Find the value of:
(i) (-4) ÷ \(\frac{2}{3}\)
Answer:
(-4) ÷ \(\frac{2}{3}\) = (-4) × \(\frac{3}{2} \times \frac{(-4)}{1} \times \frac{3}{2}\)
= \(\frac{(-4) \times 3}{1 \times 2}=\frac{-12}{2}\) = -6

(ii) \(\frac{-3}{5}\) ÷ 2
Answer:
\(\frac{-3}{5}\) ÷ 2 =\(\frac{(-3)}{5} \times \frac{1}{2}=\frac{(-3) \times 1}{5 \times 2}=\frac{-3}{10}\)

(iii) \(\frac{-4}{5}\) ÷ (-3) 
Answer:
\(\frac{-4}{5}\) ÷ (-3)  = \(\frac{(-4)}{5} \times \frac{1}{(-3)}\)
= \(\frac{(-4) \times 1}{5 \times(-3)}=\frac{-4}{-15}=\frac{4}{15}\)

(iv) \(\frac{-1}{8} ÷ \frac{3}{4}\)
Answer:
\(\frac{-1}{8} \div \frac{3}{4}=\frac{(-1)}{8} \times \frac{4}{3}\)
= \(\frac{(-1) \times 4}{8 \times 3}=\frac{-4}{24}=\frac{-4 \div 4}{24 \div 4}=\frac{-1}{6}\)

(v) \(\frac{-2}{13} ÷ \frac{1}{7}\)
\(\frac{-2}{13} \div \frac{1}{7}=\frac{-12}{13} \times \frac{7}{1}\)
= \(\frac{(-2) \times 7}{13 \times 1}=\frac{-14}{13} = -1 \frac{1}{13}\)

(vi) \(\frac{-7}{12} ÷ \left(\frac{-2}{13}\right)\)
Answer:
\(\frac{-7}{12} \div\left(\frac{-2}{13}\right)=\frac{-7}{12} \times \frac{13}{-2}\)
= \(\frac{(-7) \times 13}{12 \times(-2)}=\frac{-91}{-24}=\frac{91}{24} = 3 \frac{19}{24}\)

(vii) \(\frac{3}{13} \div\left(\frac{-4}{65}\right)\)
Answer:
\(\frac{3}{13} \div\left(\frac{-4}{65}\right)=\frac{3}{13} \times\left(\frac{65}{-4}\right)\)
= \(\frac{3 \times 65}{13 \times(-4)}=\frac{195}{-52}=\frac{195 \div 13}{-52 \div 13}=\frac{15}{-4}\)
= \(\frac{-15}{4} = -3 \frac{3}{4}\)