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

Q.1. Multiply:
(i) 16 by 9
(ii) 18 by − 6
(iii) 36 by − 11
(iv) − 28 by 14
(v) − 53 by 18
(vi) − 35 by 0
(vii) 0 by − 23
(viii) − 16 by − 12
(ix) − 105 by − 8
(x) − 36 by − 50
(xi) − 28 by − 1
(xii) 25 by − 11
Ans. (i) 16 ×× 9 = 144
(ii) 18 ×× (−6) = −(18×6) = -(18×6) = −108
(iii) 36 ×× (−11) = − (36×11) = - (36×11) = −396
(iv)  (−28) ××14 = −(28×14) = -(28×14) = −392
(v) (−53) ×× 18 = −(53×18) = -(53×18) = −954
(vi) (−35) ×× 0 = 0
(vii) 0 ×× (−23) = 0
(viii) (−16) ×× (−12) = 192
(ix) (−105) ×× (−8) = 840
(x) (−36) ×× (−50) = 1800
(xi) (−28) ×× (−1) = 28
(xii)  25 ×× (−11) = − (25×11) = - (25×11) = −275

Q.2. Find each of the following products:
(i) 3 × 4 × (−5)
(ii) 2 × (−5) × (−6)
(iii) (−5) × (−8) × (−3)
(iv) (−6) × 6 × (−10)
(v) 7 × (−8) × 3
(vi) (−7) × (−3) × 4
Ans. (i) 3 × 4 × (−5) = (12) × (−5) = −60
(ii) 2 × (−5) × (−6) = (−10) × (−6) = 60
(iii) (−5) × (−8) × (−3) = (−5) × (24) = −120
(iv)  (−6) × 6 × (−10) = 6 × (60) = 360
(v)  7 × (−8) × 3 = 21 × (−8) = −168
(vi)  (−7) × (−3) × 4 = 21 × 4 = 84

Q.3. Find each of the following products:
(i) (−4) × (−5) × (−8) × (−10)
(ii) (−6) × (−5) × (−7) × (−2) × (−3)
(iii) (−60) × (−10) × (−5) × (−1)
(iv) (−30) × (−20) × (−5)
(v) (−3) × (−3) × (−3) × ...6 times
(vi) (−5) × (−5) × (−5) × ...5 times
(vii) (−1) × (−1) × (−1) × ...200 times
(viii) (−1) × (−1) × (−1) × ...171 times
Ans. (i)  Since the number of negative integers in the product is even, the product will be positive.
(4) × (5) × (8) × (10) = 1600
(ii) Since the number of negative integers in the product is odd, the product will be negative.
−(6) × (5) × (7) × (2) × (3) = −1260
(iii) Since the number of negative integers in the product is even, the product will be positive.
(60) × (10) × (5) × (1) = 3000
(iv) Since the number of negative integers in the product is odd, the product will be negative.
−(30) × (20) × (5) = −3000
(v) Since the number of negative integers in the product is even, the product will be positive.
(−3)6(-3)6 = 729
(vi) Since the number of negative integers in the product is odd, the product will be negative.
(−5)5(-5)5 = −3125
(vii) Since the number of negative integers in the product is even, the product will be positive.
(−1)200(-1)200= 1
(viii) Since the number of negative integers in the product is odd, the product will be negative.
(−1)171(-1)171 = −1

Q.4. What will be the sign of the product, if we multiply 90 negative integers and 9 positive integers?
Ans. Multiplying 90 negative integers will yield a positive sign as the number of integers is even.
Multiplying any two or more positive integers always gives a positive integer.
The product of both(the above two cases) the positive and negative integers is also positive.
Therefore, the final product will have a positive sign.

Q.5. What will be the sign of the product, if we multiply 103 negative integers and 65 positive integers?
Ans. Multiplying 103 negative integers will yield a negative integer, whereas 65 positive integers will give a positive integer.
The product of a negative integer and a positive integer is a negative integer.

Q.6. Simplify:
(i) (−8) × 9 + (−8) × 7
(ii) 9 × (−13) + 9 × (−7)
(iii) 20 × (−16) + 20 × 14
(iv) (−16) × (−15) + (−16) × (−5)
(v) (−11) × (−15) + (−11) × (−25)
(vi) 10 × (−12) + 5 × (−12)
(vii) (−16) × (−8) + (−4) × (−8)
(viii) (−26) × 72 + (−26) × 28
Ans. (i) (−8) ×× (9 + 7)   [using the distributive law]
= (−8) ×× 16 = −128
(ii)  9 ×× (−13 + (−7))  [using the distributive law]
= 9 ×× (−20) = −180
(iii)  20 ×× (−16 + 14)    [using the distributive law]
= 20 ×× (−2) = −40
(iv) (−16) ×× (−15 + (−5))  [using the distributive law]
= (−16) ×× (−20) = 320
(v) (−11) ×× (−15 +(−25))  [using the distributive law]
= (−11) ×× (−40)
= 440
(vi) (−12) ×× (10 + 5)   [using the distributive law]
= (−12) ×× 15 = −180
(vii) (−16 + (−4)) ×× (−8)  [using the distributive law]
= (−20) ×× (−8) = 160
(viii) (−26) ×× (72 + 28)    [using the distributive law]
= (−26) ××100 = −2600

Q.7. Fill in the blanks:
(i) (−6) × (......) = 6
(ii) (−18) × (......) = (−18)
(iii) (−8) × (−9) = (−9) × (......)
(iv) 7 × (−3) = (−3) × (......)
(v) {(−5)×3} × (−6) = (......) × {3×(−6)}
(vi) (−5) × (......) = 0
Ans. (i) (−6) × (x) = 6
x = 6−6 = −66= −1x = 6-6 = -66= -1
Thus, x = (−1)
(ii) 1 [∵ Multiplicative identity]
(iii) (−8) [∵ Commutative law]
(iv) 7 [∵ Commutative law]
(v) (−5) [∵ Associative law]
(vi) 0 [∵ Property of zero]

Q.8. In a class test containing 10 questions, 5 marks are awarded for every correct answer and (−2) marks are awarded for every incorrect answer and 0 for each question not attempted.
(i) Ravi gets 4 correct and 6 incorrect answers. What is his score?
(ii) Reenu gets 5 correct and 5 incorrect answers. What is her score?
(iii) Heena gets 2 correct and 5 incorrect answers. What is her score?
Ans. We have 5 marks for correct answer and (−2) marks for an incorrect answer.
Now, we have the following:
(i) Ravi's score = 4 ×× 5 + 6 ×× (−2)
= 20 + (−12) =8
(ii) Reenu's score = 5 ×× 5 + 5 ×× (−2)
= 25 − 10 = 15
(iii) Heena's score = 2 ×× 5 + 5 ×× (−2)
= 10 − 10 = 0

Q.9. Which of the following statements are true and which are false?
(i) The product of a positive and a negative integer is negative.
(ii) The product of two negative integers is a negative integer.
(iii) The product of three negative integers is a negative integer.
(iv) Every integer when multiplied with −1 gives its multiplicative inverse.
(v) Multiplication on integers is commutative.
(vi) Multiplication on integers is associative.
(vii) Every nonzero integer has a multiplicative inverse as an integer.
Ans. (i) True.
(ii) False. Since the number of negative signs is even, the product will be a positive integer.
(iii) True. The number of negative signs is odd.
(iv) False. a ×× (−1) = −a, which is not the multiplicative inverse of a.
(v) True. a ×× b = b ×× a
(vi) True. (a ×× b) ×× c = a ×× (b ×× c)
(vii) False. Every non-zero integer a has a multiplicative inverse 1a1a, which is not an integer.