NCERT Solutions Class 7 Maths Chapter 1 Integers Ex 1.2

NCERT Solutions Class 7 Maths
Chapter – 1 (Integers)

The NCERT Solutions in English Language for Class 7 Mathematics Chapter – 1 Integers Exercise 1.2 has been provided here to help the students in solving the questions from this exercise.

Chapter 1: Integers

Exercise – 1.2

1. Write down a pair of integers whose:
(a) sum is -7
(b) difference is -10
(c) sum is 0.
Solution –
(a) sum is -7
Let us take a pair of integers -3 and -4.
∴ (-3) + (-4)
= -3 – 4 [∵ (+ × – = -)]
= – 7

(b) difference is – 10
Let us take a pair of integers -12 and -2
∴ (-12) – (-2)
= -12 + 2 [∵ (- × – = +)]
= -10

(c) sum is 0
Let us take a pair of integers -3 and 3
∴ (-3) + (3)
= -3 + 3
= 0

2.
(a) Write a pair of negative integers whose difference gives 8.
(b) Write a negative integer and positive integer whose sum is -5.
(c) Write a negative integer and a positive integer whose difference is -3.

Solution –
(a) Write a pair of negative integers whose difference gives 8.
Let us have -2 and -10
∴ Difference = (-2) – (-10)
= -2 + 10
= 8

(b) Write a negative integer and positive integer whose sum is -5.
Let us have -7 and 2
∴ (-7) + (2)
= -7 + 2
= -5

(c) Write a negative integer and a positive integer whose difference is -3.
Let us have -2 and 1
∴ (-2) – (1) = – 2 – 1 = -3
= (-5) – (- 13)
= -5 + 13 [∵ (- × – = +)]
= 8

3. In a quiz, team A scored – 40, 10, 0 and team B scored 10, 0, – 40 in three successive rounds. Which team scored more? Can we say that we can add integers in any order?
Solution –
Total score of team A = (-40) + (10) + (0)
= – 40 + 10 + 0
= – 30
Total score of team B = 10 + 0 + (-40)
= 10 + 0 – 40
= – 30
∴ The scores of both the teams are same i.e. -30.
Yes, we can say that we can add integers in any order.

4. Fill in the blanks to make the following statements true:
(i) (–5) + (– 8) = (– 8) + (____)
(ii) -53 + ____ = -53
(iii) 17 + ____ = 0
(iv) [13 + (-12)] + (____) = 13 + [(-12) + (-7)]
(v) (-4) + [15 + (-3)] = [-4 + 15] + ____

Solution –
(i) (–5) + (– 8) = (– 8) + (____)
Let us assume the missing integer be x,
Then,
= (–5) + (– 8) = (– 8) + (x)
= – 5 – 8 = – 8 + x
= – 13 = – 8 + x
By sending – 8 from RHS to LHS it becomes 8,
= – 13 + 8 = x
= x = – 5
(–5) + (– 8) = (– 8) + (– 5) [Commutative law of Addition]

(ii) – 53 + ____ = –53
Let us assume the missing integer be x,
Then,
= –53 + x = –53
By sending – 53 from LHS to RHS it becomes 53,
= x = -53 + 53
= x = 0
= –53 + 0 = –53 [Additive Identity]

(iii) 17 + ____ = 0
Let us assume the missing integer be x,
Then,
= 17 + x = 0
By sending 17 from LHS to RHS it becomes -17,
= x = 0 – 17
= x = – 17
= 17 + (-17) = 0 [Additive inverse]

(iv) [13 + (– 12)] + (____) = 13 + [(–12) + (–7)]
Let us assume the missing integer be x,
Then,
= [13 + (– 12)] + (x) = 13 + [(–12) + (–7)]
= [13 – 12] + (x) = 13 + [–12 –7]
= [1] + (x) = 13 + [-19]
= 1 + (x) = 13 – 19
= 1 + (x) = -6
By sending 1 from LHS to RHS it becomes -1,
= x = -6 – 1
= x = -7
= [13 + (– 12)] + (– 7) = 13 + [(–12) + (–7)]
[Associative law of addition]

(v) (– 4) + [15 + (–3)] = [– 4 + 15] + ____
Let us assume the missing integer be x,
Then,
= (– 4) + [15 + (–3)] = [– 4 + 15] + x
= (– 4) + [15 – 3)] = [– 4 + 15] + x
= (-4) + [12] = [11] + x
= 8 = 11 + x
By sending 11 from RHS to LHS it becomes -11,
= 8 – 11 = x
= x = -3
= (– 4) + [15 + (–3)] = [– 4 + 15] + -3
[Associative law of addition]

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