Integers Class 7 Exercise 1.1 |
NCERT Answers for Class 7 Maths Chapter 1 Integers Exercise 1.2
Page 9
Ex 1.2 Class 7 Maths Question 1. Write down a pair of integers whose:
(a) sum is –7
(b) difference is –10
(c) sum is 0
Answer
(a)
= – 4 + (-3)
= – 4 – 3 … [∵ (+ × – = -)]
= – 7
(b)
= -25 – (-15)
= – 25 + 15 … [∵ (- × – = +)]
= -10
(c)
= 4 + (-4)
= 4 – 4
= 0
Ex 1.2 Class 7 Maths Question 2.
(a) Write a pair of negative integers whose difference gives 8.
(b) Write a negative integer and a positive integer whose sum is –5.
(c) Write a negative integer and a positive integer whose difference is –3.
Answer
(a) Write a pair of negative integers whose difference gives 8
= (-5) – (- 13)
= -5 + 13 … [∵ (- × – = +)]
= 8
(b) Write a negative integer and a positive integer whose sum is –5
= -25 + 20
= -5
(c) Write a negative integer and a positive integer whose difference is – 3.
= – 6 – (-3)
= – 6 + 3 … [∵ (- × – = +)]
= – 3
Ex 1.2 Class 7 Maths Question 3. In a quiz, team A scored – 40, 10, 0 and team B scored 10, 0, – 40 in three successiverounds. Which team scored more? Can we say that we can add integers inany order?
Answer
From the question, it is given that
Score of team A = -40, 10, 0
Total score obtained by team A = – 40 + 10 + 0
= – 30
Score of team B = 10, 0, -40
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.
Ex 1.2 Class 7 Maths 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] + …………
Answer
(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
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
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
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
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
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)]
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
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] +…………
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
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]
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