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)