# GSEB Solutions Class 7 Maths Chapter 9 Rational Numbers Ex 9.2

Gujarat Board GSEB Textbook Solutions Class 7 Maths Chapter 9 Rational Numbers Ex 9.2 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 7 Maths Chapter 9 Rational Numbers Ex 9.2

Question 1.
Find the sum:
(i) $$\frac { 5 }{ 4 }$$ + ( $$\frac { -11 }{ 4 }$$ )
(ii) $$\frac { 5 }{ 3 }$$ + $$\frac { 3 }{ 5 }$$
(iii) $$\frac { -9 }{ 10 }$$ + $$\frac { 22 }{ 15 }$$
(iv) $$\frac { -3 }{ -11 }$$ and $$\frac { 5 }{ 9 }$$
(v) $$\frac { -8 }{ 19 }$$ + $$\frac { (-2) }{ 57 }$$
(vi) $$\frac { -2 }{ 3 }$$ + 0
(vii) – 2$$\frac { 1 }{ 3 }$$ + 4$$\frac { 3 }{ 5 }$$
Solution:
(i) $$\frac { 5 }{ 4 }$$ + ( $$\frac { -11 }{ 4 }$$ )
We have $$\frac { 5 }{ 4 }$$ + ( $$\frac { -11 }{ 4 }$$ ) = $$\frac { 5+(-11) }{ 4 }$$ = $$\frac { -6 }{ 4 }$$
= $$\frac { -3 }{ 2 }$$ or -1$$\frac { 1 }{ 2 }$$

(ii) $$\frac { 5 }{ 3 }$$ + $$\frac { 3 }{ 5 }$$
∵ LCM of 3 and 5 is 15.

(iv) $$\frac { -3 }{ -11 }$$ and $$\frac { 5 }{ 9 }$$
∵ LCM of 11 and 9 is 99.
∴ $$\frac { -3 }{ -11 }$$ = $$\frac { (-3)×9 }{ (-11)×9 }$$ = $$\frac { -27 }{ -99 }$$ = $$\frac { 27 }{ 99 }$$
and $$\frac { 5 }{ 9 }$$ = $$\frac { 5×11 }{ 9×11 }$$ = $$\frac { 55 }{ 99 }$$

(vi) $$\frac { -2 }{ 3 }$$ + 0
We have $$\frac { -2 }{ 3 }$$ + 0 = $$\frac { -2 }{ 3 }$$ + $$\frac { 0 }{ 3 }$$ = $$\frac { -2+0 }{ 3 }$$ = $$\frac { -2 }{ 3 }$$
Thus, $$\frac { -2 }{ 3 }$$ + 0 = $$\frac { -2 }{ 3 }$$

(vii) – 2$$\frac { 1 }{ 3 }$$ + 4$$\frac { 3 }{ 5 }$$

Question 2.
Find:
(i) $$\frac { 7 }{ 24 }$$ – $$\frac { 17 }{ 36 }$$
(ii) $$\frac { 5 }{ 63 }$$ – ( $$\frac { -6 }{ 21 }$$ )
(iii) $$\frac { -6 }{ 13 }$$ – $$\frac { -7 }{ 15 }$$
(iv) $$\frac { -3 }{ 8 }$$ – $$\frac { 7 }{ 11 }$$
(v) -2$$\frac { 1 }{ 9 }$$ – 6
Solution:
(i) $$\frac { 7 }{ 24 }$$ – $$\frac { 17 }{ 36 }$$
∵ LCM of 24 and 36 is 72.

(ii) $$\frac { 5 }{ 63 }$$ – ( $$\frac { -6 }{ 21 }$$ )
∵ LCM of 63 and 21 is 63.

Thus, $$\frac { 5 }{ 63 }$$ – ( $$\frac { -6 }{ 21 }$$ = $$\frac { -13 }{ 72 }$$

(iii) $$\frac { -6 }{ 13 }$$ – $$\frac { -7 }{ 15 }$$
∵ LCM of 13 and 15 is 195.

(v) -2$$\frac { 1 }{ 9 }$$ – 6

Question 3.
Find the product:
(i) $$\frac { 9 }{ 2 }$$ x ( $$\frac { -7 }{ 4 }$$ )
(ii) $$\frac { 3 }{ 10 }$$ x (-9)
(iii) $$\frac { -6 }{ 5 }$$ x $$\frac { 9 }{ 11 }$$
(iv) $$\frac { 3 }{ 7 }$$ x $$\frac { -2 }{ 5 }$$
(v) $$\frac { 3 }{ 11 }$$ x $$\frac { 2 }{ 5 }$$
(vi) $$\frac { 3 }{ -5 }$$ x $$\frac { -5 }{ 3 }$$
Solution:

Question 4.
Draw the number line and represent the following rational numbers on it:
(i) (- 4) ÷ $$\frac { 2 }{ 3 }$$
(ii) $$\frac { -3 }{ 5 }$$ ÷ 2
(iii) $$\frac { -4 }{ 5 }$$ ÷ (-3)
(iv) $$\frac { -1 }{ 8 }$$ ÷ $$\frac { 3 }{ 4 }$$
(v) $$\frac { -2 }{ 13 }$$ ÷ $$\frac { 1 }{ 7 }$$
(vi) $$\frac { -7 }{ 12 }$$ ÷ ( $$\frac { -2 }{ 13 }$$ )
(vii) $$\frac { 3 }{ 13 }$$ ÷ ( $$\frac { -4 }{ 65 }$$ )
Solution:
(i) (- 4) ÷ $$\frac { 2 }{ 3 }$$

Thus, (- 4) ÷ $$\frac { 2 }{ 3 }$$ = – 6

(ii) $$\frac { -3 }{ 5 }$$ ÷ 2
∵ The reciprocal of 2 is $$\frac { 1 }{ 2 }$$
∴ $$\frac { -3 }{ 5 }$$ ÷ 2 = $$\frac { -3 }{ 5 }$$ x $$\frac { 1 }{ 2 }$$
= $$\frac { (-3)×1 }{ 5×2 }$$ = $$\frac { -3 }{ 10 }$$
Thus, $$\frac { -3 }{ 5 }$$ ÷ 2 = $$\frac { -3 }{ 10 }$$

(iii) $$\frac { -4 }{ 5 }$$ ÷ (-3)

(iv) $$\frac { -1 }{ 8 }$$ ÷ $$\frac { 3 }{ 4 }$$

(vi) $$\frac { -7 }{ 12 }$$ ÷ ( $$\frac { -2 }{ 13 }$$ )
∵ The reciprocal of $$\frac { -2 }{ 13 }$$ is ( $$\frac { 13 }{ 2 }$$
∴ $$\frac { -7 }{ 12 }$$ ÷ ( $$\frac { -2 }{ 13 }$$ ) = $$\frac { -7 }{ 12 }$$ x ( $$\frac { 13 }{ 2 }$$
= $$\frac { (-7)×(-13) }{ 12×2 }$$
= $$\frac { 91 }{ 24 }$$ or 3$$\frac { 19 }{ 24 }$$
Thus,
$$\frac { -7 }{ 12 }$$ ÷ ( $$\frac { -2 }{ 13 }$$ )
= 3$$\frac { 19 }{ 24 }$$
= $$\frac { 91 }{ 24 }$$ or 3$$\frac { 19 }{ 24 }$$
Thus,
$$\frac { -7 }{ 12 }$$ ÷ ( $$\frac { -2 }{ 13 }$$ )
= 3$$\frac { 19 }{ 24 }$$

(vii) $$\frac { 3 }{ 13 }$$ ÷ ( $$\frac { -4 }{ 65 }$$ )