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

Integers Class- 7th RS Aggarwal Exe-1C 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 C to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-7 Mathematics.

## Integers Class- 7th RS Aggarwal Exe-1C 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 C Academic Session 2023 – 2024

### Exercise – 1 C

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

#### Question: 1. Find the product :

(i) 18 × 4

18 × 4

= 72

(ii) (-25) × 6

(-25) × 6

= -125

(iii) (-30) × 7

(-30) × 7

= -210

(iv) 8 × (-15)

8 × (-15)

= -120

(v) 20 ×  (-10)

20 ×  (-10)

= -200

(vi) (-12) × (-15)

(-12) × (-15)

= 180

(vii) (-8) × (-13)

(-8) × (-13)

= 104

(viii) (-20) × (-1)

(-20) × (-1)

= 20

(ix) (-9) × 0

(-9) × 0

= 0

(x) 0 × (-11)

0 × (-11)

= 0

#### Question: 2. Find the product :

(i) {(-9) × 8} × (-5)

Answer: Given: {(-9) × 8} × (-5)

{(-72)} × (-5)

= 360

(ii) {(-10) × (-5)} × 6

Answer: Given: {(-10) × (-5)} × 6

{(-50)} × 6

= -300

(iii) {(-12) × (-15)} × (-2)

Answer: Given: {(-12) × (-15)} × (-2)

180 × (-2)

= -360

(iv) (-8) × {(-5) × (-3)}

Answer: Given: (-8) × {(-5) × (-3)}

(-8) × {15}

= -120

(v) (-11) × {(-8) × 5}

Answer: Given: (-11) × {(-8) × 5}

(-11) × {-40}

= 440

#### Question: 3. Verify the following :

(i) (-14) × (-8) = (-8) × (-14)

Answer: Given: (-14) × (-8) = (-8) × (-14)

112 = 112

∴ L.H.S = R.H.S

(ii) {(-7) × 5} × (-6) = (-7) × {5 × (-6)}

Answer: Given: {(-7) × 5} × (-6) = (-7) × {5 × (-6)}

{-35} × (-6) = (-7) × {-30}

210 = 210

∴ L.H.S = R.H.S

(iii) (-10) × {(-7) + (-9)} = {(-10) × (-7)} + {(-10) × (-9)}

Answer: Given: (-10) × {(-7) + (-9)} = {(-10) × (-7)} + {(-10) × (-9)}

(-10) × {-7 -9} = {70} + {90}

(-10) × (-16) = 70 + 90

160 = 160

∴ L.H.S = R.H.S

#### Question: 4. Find the quotient :

(i) 28 ÷ (-7)

28 ÷ (-7)

= -4

(ii) (-65) ÷ 13

(-65) ÷ 13

= -5

(iii) (-66) ÷ (-6)

(-66) ÷ (-6)

= 11

(iv) (-9) ÷ (-1)

(-9) ÷ (-1)

= 9

(v) 0 ÷ (-2)

0 ÷ (-2)

= 0

(vi) (-12) ÷ (-12)

(-12) ÷ (-12)

= 1

#### Question: 5. Write all even integers between (i) (-4) and 11   (ii) (-13) and (-7)

All even integer between -4 and 11 will be –

-2, 0, 2, 4, 6, 8, 10.

#### (ii) (-13) and (-7)

All even integer between (-13) and (-7) will be –

-12, -10, -8.

#### Question: 6. Write all odd integers between  (i) (-1) and 7      (ii) (-20) and (-14)

All odd integer between (-1) and 7 will be –

1, 3, 5.

#### (ii) (-20) and (-14)

All odd integer between (-20) and (-14) will be –

-19, -17, -15.

#### Question: 7. Write five consecutive even integers succeeding -21.

Answer: Solution: five consecutive even integers succeeding -21 will be –

-20, -18, -16, -14, -12.

#### Question: 8. Write five consecutive odd integers preceding -36.

Answer: Solution: five consecutive odd integers preceding -36 will be –

-37, -39, -41, -43, -45.

#### Question: 9. The product of two integers is -120. If one number is 15, find the other.

[Hint: Other number = -120 ÷ 15]

one number = 15

other number = ?

product of both number = -120

∴ Let the other No. be x

So, According to question –

15 × x =-120

x = -120/15

x = -8

∴ The other number is -8.

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