Integers Class 7 Ex. 1.1
New Ncert Solution for Class 7 Maths Chapter 1 Integers Free Solution.
Exercise 1.1
Question 1 :- Write-down a-pair of integers whose:
(a) sum is -7
(b) difference is -10
(c) sum is 0.
Solution 1 :-
(a) Let us, take a pair of integers -2 and -5.
So, (-2) + (-5) = -2 – 5 = -7
(b) Let us, take a pair of integers -14 and -4
So, (-14) – (-4) = -14 + 4 = -10
(c) Let us, take a pair of integers -4 and 4
So, (-4) + (4) = -4 + 4 = 0
Question 2 :- (a) Write a pair of negative-integers; whose difference, 8.
(b) Write a negative-integer and positive-integer; whose sum, -5.
(c) Write a negative-integer and a positive-integer; whose difference, -3.
Solution 2:-
(a) Let us, take -2 and -10 integer
So, the difference = (-2) – (-10) = -2 + 10 = 8
(b) Let us, take -6 and 1 integer
So, the sum = (-6) + (1) = -7 + 2 = -5
(c) Let us, take -1 and 2 integer
So, the difference = (-1) – (2) = – 2 – 1 = -3
Question 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 you say, that we can add integers in any-order?
Solution 3:-
Total score of team A = (-40) + (10) + (0) = -40 + 10 + 0 = -30;
Total score of team B = 10 + 0 + (-40) = 10 + 0 – 40 = -30;
So, the scores of both the teams are same which is = -30;
Yes, we can add the integers in any order due to commutative property.
Question 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 4:-
(i) (-5) + (-8) = (-8) + (-5). [Commutative law of additions]
(ii) -53 + 0 = -53. [Additive Identity][Adding 0 to any integer, it gives the same value]
(iii) 17 + (-17) = 0. [Additive inverse]
(iv) [13 + (-12)] + (-7) = 13 + [(-12) + (-7)]. [Associative law of addition]
(v) (-4) + [15 + (-3)] = [-4 + 15] + (-3). [Associative law of addition]
Chapter 1 Integer Class 7 Exercise. 1.1 for Free