Mathematics- Class 7 Chapter 1 – Integers – NCERT Exercise 1.2 Solutions is provided below. Total Questions are in this CBSE NCERT Exercise, all are solved here.
Q1. Write down a pair of integers whose:
(a) sum is -7
Solution:
= – 5 + (-2)
= – 5 – 2 … (+ × – = -)
= – 7
(b) difference is – 10
Solution:
= -25 – (-15)
= – 25 + 15 … (- × – = +)
= -10
(c) sum is 0
Solution:
= 2 + (-2)
= 2 – 2
= 0
Q2.
(a) Write a pair of negative integers whose difference gives 8.
Solution:
= (-10) – (- 18)
= -10 + 18 (- × – = +)
= 8
(b) Write a negative integer and a positive integer whose sum is – 5.
Solution:
= -26 + 21
= -5
(c) Write a negative integer and a positive integer whose difference is – 3.
Solution:
= – 6 – (-3)
= – 6 + 3 (- × – = +)
= – 3
Q3. 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:
Given,
Score of team A = -40, 10, 0
Score of team B = 10, 0, -40
According to question:
Total score obtained by team A = – 40 + 10 + 0
= – 30
Total score obtained by team B = 10 + 0 + (-40)
= 10 + 0 – 40
= – 30
Thus, the score of the both A team and B team is same.
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) + (.…)
Solution:
Let the missing integer be x
Then,
L.H.S = R.H.S
(–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
Now substitute the x value in the blank place,
(–5) + (– 8) = (– 8) + (- 5)
(This equation is in the form of Commutative law of Addition)
(ii) –53 + …… = –53
Solution:
Let the missing integer be x
Then,
L.H.S = R.H.S
–53 + x = –53
By sending – 53 from LHS to RHS it becomes 53,
x = -53 + 53
x = 0
Now substitute the x value in the blank place,
–53 + 0 = –53
(This equation is in the form of Closure property of Addition)
(iii) 17 + ………… = 0
Solution:
Let the missing integer be x
Then,
L.H.S = R.H.S
17 + x = 0
By sending 17 from LHS to RHS it becomes -17,
x = 0 – 17
x = – 17
Now substitute the x value in the blank place,
17 + (-17) = 0
(This equation is in the form of Closure property of Addition)
17 – 17 = 0
(iv) [13 + (– 12)] + (…………) = 13 + [(–12) + (–7)]
Solution:
Let the missing integer be x
Then,
L.H.S = R.H.S
[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
Now substitute the x value in the blank place,
= [13 + (– 12)] + (-7) = 13 + [(–12) + (–7)]
(This equation is in the form of Associative property of Addition)
(v) (– 4) + [15 + (–3)] = [– 4 + 15] +…………
Solution:
Let the missing integer be x
Then,
L.H.S = R.H.S
(– 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
Now substitute the x value in the blank place,
= (– 4) + [15 + (–3)] = [– 4 + 15] + -3
(This equation is in the form of Associative property of Addition)
