Mathematics- Class 7 – Chapter 1 – Integers – NCERT Exercise 1.2 Solutions

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)

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