# Integers Class- 7th RS Aggarwal Exe-1B Goyal Brothers ICSE Math Solution

Integers Class- 7th RS Aggarwal Exe-1B Goyal Brothers ICSE Math Solution . We provide step by step Solutions of lesson-1 Integers for ICSE Class-7 Foundation RS Aggarwal Mathematics of Goyal Brothers Prakashan . Our Solutions contain all type Questions of Exe-1 B to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-7 Mathematics.

## Integers Class- 7th RS Aggarwal Exe-1B Goyal Brothers ICSE Math Solution

 Board ICSE Publications Goyal brothers Prakashan Subject Maths Class 7th Chapter-1 Integers Writer RS Aggrawal Book Name Foundation Topics Solution of Exe-1 B Academic Session 2023 – 2024

### Exercise – 1 B

Integers Class- 7th RS Aggarwal Exe-1B Goyal Brothers ICSE Math Solution

#### Question: 1. Add the following integers :

(i) 64 and 36

= 64 + 36

= 100

(ii) 73 and -37

= 73 + (-37)

= 73 – 37

= 36

(iii) -26 and -45

= -26 + (-45)

= -26 – 45

= -71

(iv) -51 and 25

= -51 + 25

= -26

(v) 100 and -32

= 100 + (-32)

= 100 – 32

= 68

(vi) 0 and -21

= 0 + (-21)

= 0 – 21

= -21

(i) 23

(ii) -33

(iii) -1

(iv) -476

#### Question: 3. Evaluate :

(i) 6 – 24

= 6 – 24

= 18

(ii) 18 – (-8)

= 18 + 8

= 26

(iii) (-16) – (-5)

= (-16 + 5)

= -11

(iv) (-20) – 6

= -20 – 6

= -26

(v) (-1) – (-19)

= -1 + 19

= 18

(vi) 8 – (-23)

= 8 + 23

= 31

[Hint. a – (-b) = a + b]

#### Question: 4. Verify the following :

(i) (-14) + 9 = 9 + (-14)

Answer: Given: (-14) + 9 = 9 + (-14)

-14 + 9 = 9 – 14

-5 = -5

∴ L.H.S = R.H.S

(ii) (-8) + (-12) = (-12) + (-8)

Answer: Given: (-8) + (-12) = (-12) + (-8)

-8 – 12 = -12 – 8

-20 = -20

∴ L.H.S = R.H.S

(iii) (-6) + [(-8) + 12] = [(-6) + (-8)] + 12

Answer: Given:  (-6) + [(-8) + 12] = [(-6) + (-8)] + 12

-6 + [-8 + 12] = [-6 – 8] + 12

-6 + [4] = -14 + 12

-2 = -2

∴ L.H.S = R.H.S

(iv) [(-9) + (-7)] + (-14) = (-9) + [(-7) + (-14)]

Answer: Given: [(-9) + (-7)] + (-14) = (-9) + [(-7) + (-14)]

[-9 -7] – 14 = -9 + [-7 -14]

[-16] – 14 = -9 + [-21]

-16 – 14 = -9 – 21

-30 = -30

∴ L.H.S = R.H.S

#### Question: 5. Fill in the blanks :

(i) 8 + (-8) = 0

(ii) (-10) + 0 = (-10)

(iii) (-6) + (-8) = (-8) + (-6)

(iv) (-6) + (-8) = -14

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

(vi) (-9) + 8 = -1

#### Question: 6. Subtract the sum of -137 and -43 from the sum of -103 and 27.

Answer: According to the question: (-103 + 27) – [-137 + (-43)]

Given: (-103 + 27) – [-137 + (-43)]

= -76 – [-180]

= -76 + 180

= 104

#### Question: 7. Subtract -29 from -53 and add -16 to the result.

Answer: According to the question: -53 – (-29)

Given: -53 – (-29)

= -53 + 29

= -24

Now, We add (-16) with (-24), we get.

= -24 + (-16)

= -24 – 16

= -40

#### Question: 8. The sum of two integers is 43. If one of them is -27, find the other.

Given is,

one integer = -27

second integer = ?

sum of two integer = 43

Now,

second integer = -27 – (43)

= -70

second integer is -70.

#### Question: 9. Find the successor of :

(i) 30

= 30 + 1                                [ 1 more than integer]

= 31

(ii) -70

= -70 + 1                                [ 1 more than integer]

= -69

(iii) -206

= -206 + 1                                [ 1 more than integer]

= -205

(iv) -1

= -1 + 1                                [ 1 more than integer]

= 0

#### Question: 10. Find the predecessor of :

(i) 60

= 60 – 1                                [ 1 less than the integer]

= 59

(ii) -351

= -351 – 1                                [ 1 less than the integer]

= -352

(iii) 0

= 0 – 1                                [ 1 less than the integer]

= -1

(iv) -99

= -99 – 1                                [ 1 less than the integer]

= -100

— : end of Integers Class- 7th RS Aggarwal Exe-1B Goyal Brothers ICSE Math Solution :–