## NCERT Solutions For Class 6 Maths Integers Exercise 6.2

Free download NCERT Solutions for Class 6 Maths Chapter 6 Ex 6.2 Integers PDF for CBSE 2020 Exams.

You can also Download __NCERT Solutions Class 6 Maths__ to help you to revise complete Syllabus and score more marks in your examinations.

- Integers Class 6 Ex 6.1
- Integers Class 6 Ex 6.2
- Integers Class 6 Ex 6.3
- Integers Class 6 Extra Questions

**NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.2**

**Exercise 6.2**

Question 1.

Using the number line write the integer which is:

(a) 3 more than 5

(b) 5 more than -5

(c) 6 less than 2

Solution:

(a) 3 more than 5

Moving right 3 steps from 5, we reach at 8. Hence, 3 more than 5 = 8.

(b) 5 more than -5

Moving right 5 steps from -5 we reach at 0. Hence, 5 more than -5 = 0

(c) 6 less than 2

Moving left 6 steps from 2, we reach at -4. Hence, 6 less than 2 = -4

(d) 3 less than -2

Moving left 3 steps from -2, we reach at -5.

Question 2.

Use number line and add the following integers:

(a) 9 + (-6)

(b) 5 + (-11)

(c) (-1) + (-7)

(d) (- 5) + 10

(e) (-1) + (-2) + (-3)

Solution:

(a) 9 + (- 6)

Question 3.

Add without using number line:

(а) 11 + (-7)

(b) (-13) + (+18)

(c) (-10) + (+19)

(d) (-250) + (+150)

(e) (-380) + (-270)

(f) (-217) + (-100).

Solution:

(a) 11 + (-7) = 4 + (+7) + (-7)

[∵ (+7) + (-7) = 0]

= 4 + 0 = 4

Hence, 11 + (-7) = 4.

(b) (-13) + (+18) = (-13) + (+13) + (+5)

[∵ (-13) + (+13) = 0]

= 0 + (+ 5) = 5

Hence, (- 13) + (+ 18) = 5.

(c) (-10) + (+19) = (-10) + (+10) + (+9)

[∵ (-10) + (10) = 0] = 0 + (+9) = 9

Hence, (-10) + (19) = 9.

(d) (-250) + (+150) = (-100) + (-150) + (+150)

= (-100) + 0 = -100 [∵(-150) + (+150) = 0]

Hence, (-250) + (+150) = – 100.

(e) (-380) + (-270) = – [380 + 270] = (-650)

Hence, (-380) + (-270) = (-650).

(f) (-217) + (-100) = – [217 + 100] = -317

Question 4.

Find the sum of:

(a) 137 and -354

(b) -52 and 52 .

(d) -312, 39 and 192

(d) -50, -200 and 300

Solution:

(a) 137 and -354

(137) + ( -354) = (137) + (-137) + (-217) [∵ (137) + (-137) = 0]

= 0 + (-217) = (-217)

(b) -52 and 52

(-52) + (+52) = 0 [ ∵ (-a) + (+ a) = 0]

(c) -312, 39 and 192

(-312) + (+ 39) + (+192)

= (-231) + (- 81) + (+39) + (+192)

= (-231) + (-81) + (+ 231)

= (-231) + (+ 231) + (-81)

[∵ (-a) + (a) = 0]

= 0 + (- 81) = – 81

(d) – 50, -200 and 300

(-50) + (-200) + (+ 300)

= (-50) + (-200) + (+200) + (+100)

= (-50) + 0 + (+100)[∵ (-a) + (+ a) = 0]

= (-50) + (+100)

= (-50) + (+50) + (+ 50)

= 0 + (+ 50) = 50 [ ∵ (-a) + (+ a) = 0]

Question 5.

Find the sum of:

(a) (- 7) + (-9) + 4 + 16

(b) (37) + (-2) + (-65) + (-18)

Solution:

(a) (- 7) + (-9) + 4 + 16

= (-7) + (-9) + 4 + (+ 7) + (+ 9)

= (-7) + (+7) + (-9) + (+9) + 4

= 0 + 0 + 4 = 4 [∵ (-a) + (a) = 0]

(b) (37) + (-2) + (-65) + (-8)

= (+37) + (-75)

= (+37) + (-37) + (-38)

= 0 + (-38) = (-38) [∵ (-a) + (+a) = 0]