RS Aggarwal Solutions: Integers (Exercise 1C) | Mathematics (Maths) Class 7 (Old NCERT) PDF Download

Q.1. Divide:
(i) 65 by −13
(ii) −84 by 12
(iii) −76 by 19
(iv) −132 by 12
(v) −150 by 25
(vi) −72 by −18
(vii) −105 by −21
(viii) −36 by −1
(ix) 0 by −31
(x) −63 by 63
(xi) −23 by −23
(xii) −8 by 1
Ans. (i) 65 ÷ (−13) = 65/−13 = −5
(ii) (−84) ÷ 12 = −84/12 =  −7
(iii) (−76) ÷ 19 = −76/19 = −4
(iv) (−132) ÷ 12 = −132/12 = −11
(v) (−150) ÷ 25 = −150/25 = −6
(vi) (−72) ÷ (−18) = −72/−18 = 4
(vii)  (−105) ÷ (−21) = −105/−21 = 5
(viii) (−36) ÷ (−1) = −36/−1 = 36
(ix) 0 ÷ (−31) = 0/-31  = 0
(x)  (−63) ÷ 63 = −63/63 = −1
(xi)  (−23) ÷ (−23) = −23/−23 = 1
(xii) (−8) ÷ 1 =  −8/1 = -8

Q.2. Fill in the blanks
(i) 72 ÷ (......) = −4
(ii) −36 ÷ (......) = −4
(iii) (......) ÷ (−4) = 24
(iv) (......) ÷ 25 = 0
(v) (......) ÷ (−1) = 36
(vi) (......) ÷ 1 = −37
(vii) 39 ÷ (......) = −1
(viii) 1 ÷ (......) = −1
(ix) −1 ÷ (......) = −1
Ans. (i) 72 ÷ (x) = −4
⇒72/x = −4
⇒x = 72/−4 = −18
(ii) −36 ÷ (x) = −4
⇒−36/x = −4
⇒x = -36/-4 = 9
(iii) (x) ÷ (−4) = 24
⇒x/−4 = 24
⇒x = 24×(−4) = −96
(iv) (x) ÷ 25 = 0
⇒x/25 = 0
⇒x = 25×0 = 0
(v) (x) ÷ (−1) = 36
⇒x/−1 = 36
⇒x = 36×(−1) = −36
(vi) (x) ÷ 1 = −37
⇒x1= −37
⇒x = −37×1 = −37
⇒x/1= -37
⇒x = -37×1 = -37
(vii) 39 ÷ (x) = −1
⇒39/x = −1
⇒x = −1×39 = −39
(viii) 1 ÷ (x) = −1
⇒1/x= −1
⇒x = −1×1 = −1
(ix) −1 ÷ (x) = −1
⇒−1/x = −1
⇒x = −1/−1= 1

Q.3. Write (T) for true and (F) for false for each of the following statements.
(i) 0 ÷ (−4) = 0
(ii) (−6) ÷ 0 = 0
(iii) (−5) ÷ (−1) = −5
(iv) (−8) ÷ 1 = −8
(v) (−1) ÷ (−1) = −1
(vi) (−9) ÷ (−1) = 9
Ans. (i) True (T). Dividing zero by any integer gives zero.
(ii) False (F). Division by zero gives an indefinite number.
(iii) False (F). −5/−1 = 5
(iv)  True (T). −8/1 = -8
(v)  False (F). −1/−1 = 1
(vi) True (T). −9/−1 = 9