Home » R.D. Sharma Solutions » R.D. Sharma Class 7 Solutions » Chapter 1 Integers » Integers Exercise 1.1

Question: 1Determine each of the following products:

(i) 12 × 7

(ii) (-15) × 8

(iii) (- 25) × (- 9)

(iv) (125) × (- 8)

Solution:(i) We have,

12 × 7 = 84 [The product of two integers of like signs is equal to the product of their absolutevalue]

(ii) We have,

(- 15) × 8 [The product of two integers of opposite

= (- 15 × 8) signs is equal to the additive inverse of the

= –120 [product of their absolute values]

(iii) We have,

(-25) × (-9)

= + (25 × 9)

= 225

(iv) We have,

(125) × (- 8)

= - (125 × 8)

= –1000

Question: 2Find each of the following products:

(i) 3 × (- 8) × 5

(ii) 9 × (- 3) × (- 6)

(iii) (- 2) × 36 × (- 5)

(iv) (- 2) × (- 4) × (- 6) × (- 8)

Solution:(i) We have,

3 × (- 8) × 5

= – (3 × 8) × 5

= (- 24) × 5

= – (24 × 5)

= – 120

(ii) We have,

9 × (-3) × (- 6)

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= - (9 × 3) × (- 6)

= (- 27) × (- 6)

= + (27 × 6)

= 162

(iii) We have,

(-2) × 36 × (- 5)

= - (2 × 36) × (- 5)

= (- 72) × (- 5)

= (72 × 5)

= 360

(iv) We have,

(- 2) × (- 4) × (- 6) × (- 8)

= (2 × 4) × (6 × 8)

= (8 × 48)

= 384

Question: 3Find the value of:

(i) 1487 × 327 + (- 487) × 327

(ii) 28945 × 99 - (- 28945)

Solution:(i) We have,

1487 × 327 + (- 487) × 327

= 486249 – 159249

= 327000

(ii) We have,

28945 × 99 - (- 28945)

= 2865555 – 28945

= 2894500

Question: 4Complete the following multiplication table:

Second Number

X - 4 -3 - 2 -1 0 1 2 3 4

- 4

FirstNumber

- 2

-1

0

1

2

3

4

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Is the multiplication table symmetrical about the diagonal joining the upper left corner tothe lower right corner?

Solution:Second number

X - 4 - 3 - 2 - 1 0 1 2 3 4

- 4 16 12 8 4 0 -4 -8 -12 -16

First number12 9 6 3 0 -3 -6 -9 -12

-3

-2 8 6 4 2 0 -2 -4 -6 -8

-1 4 3 2 1 0 -1 -2 -3 -4

0 0 0 0 0 0 0 0 0 0

1 -4 -3 -2 -1 0 1 2 3 4

2 -8 -6 -4 -2 0 2 4 6 8

3 -12 -9 -6 -3 0 3 6 9 12

4 -16 -12 -8 -4 0 4 8 12 16

Question: 5Determine the integer whose product with ‘-1’ is

(i) 58

(ii) 0

(iii) – 225

Solution:(i) 58 x (–1) = – (58 x 1)

= – 58

(ii) 0 x (–1) = 0

(iii) (–225) x (–1) = + (225 x 1)

= 225

Question: 6What will be the sign of the product if we multiply together

(i) 8 negative integers and 1 positive integer?

(ii) 21 negative integers and 3 positive integers?

(iii) 199 negative integers and 10 positive integers?

Solution:(i) Positive ∵ [- ve × - ve = + ve]

(ii) Negative ∵ [- ve × + ve = - ve]

(iii) Negative

Question: 7State which is greater:

(i) (8 + 9) × 10 and 8 + 9 × 10

(ii) (8 - 9) × 10 and 8 – 9 × 10

(iii) ((-2) - 5) × - 6 and (-2) - 5 × (- 6)

Solution:(i) (8 + 9) × 10 = 17 × 10

= 170

8 + 9 × 10 = 8 + 90 = 98

(8 + 9) × 10 > 8 + 9 × 10

(ii) (8 - 9) × 10 = - 1 × 10

= - 10

8 – 9 × 10 = 8 – 90 = - 82

(8 - 9) × 10 > 8 – 9 × 10

(iii) ((-2) – 5) × – 6 = (- 7) × (- 6)

= (7 x 6)

= 42

(– 2) – 5 x (– 6) = – 2 + (5 x 6)

= 30 – 2

= 28

Therefore, ((-2) – 5×(- 6)) > (- 2) – 5 × (- 6)

Question: 8(i) If a× (-1) = - 30, is the integer a positive or negative?

(ii) If a × (-1) = 30, is the integer a positive or negative?

Solution:(i) When multiplied by ‘a’ negative integer, a gives a negative integer. Hence, ‘a’ should be

a positive integer.

a = 30

(ii) When multiplied by ‘a’ negative integer, a gives a positive integer. Hence, ‘a’ should be

a negative integer.

a = – 30

Question: 9 Verify the following:

(i) 19 × (7 + (-3)) = 19 × 7 + 19 × (-3)

(ii) (-23)[(-5)+ (+19)] = (-23) × (- 5) + (- 23) × (+19)

Solution:(i) L.H.S = 19 × (7+ (-3))

= 19 × (7-3)

= 19 × 4

= 76

R.H.S = 19 × 7 + 19 × (-3)

= 133 – 57

= 76

Therefore, L.H.S = R.H.S

(ii) L.H.S = (-23)[(-5) + (+19)]

= (-23)[-5 + 19]

= (-23)[14]

= – 322

R.H.S = (-23) × (-5) + (-23) × (+19)

= 115 – 437

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= –322

Therefore, L.H.S = R.H.S

Question: 10Which of the following statements are true?

(i) The product of a positive and a negative integer is negative.

(ii) The product of three negative integers is a negative integer.

(iii) Of the two integers, if one is negative, then their product must be positive.

(iv) For all non-zero integers a and b, a × b is always greater than

either a or b.

(v) The product of a negative and a positive integer may be zero.

(vi) There does not exist an integer b such that for a >1, a × b = b × a = b. <

Solution:(i) True

(ii) True

(iii) False

(iv) False

(v) False

(vi) True

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Home » R.D. Sharma Solutions » R.D. Sharma Class 7 Solutions » Chapter 1 Integers » Integers Exercise 1.2

Question: 1Divide:

(i) 102 by 17

(ii) –85 by 5

(iii) –161 by –23

(iv) 76 by –19

(v) 17654 by –17654

(vi) (–729) by (–27)

(vii) 21590 by – 10

(viii) 0 by –135

Solution:(i) We have,

(ii) We have,

(iii) We have,

(iv) We have,

(v) We have,

(vi) We have,

(vii) We have,

(viii) We have,

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Question: 2 Fill in the blanks:

(i) 296 ÷ …. = - 148

(ii) – 88 ÷ …. = 11

(iii) 84 ÷ …. = 12

(v) …. ÷ 156 = - 2

(vi) …. ÷ 567 = - 1

Solution:(i) We have,

(ii) We have,

(iii) 84/12 = 7

(iv) …. ÷ -5 = 25

25 × (- 5) = - 125

(v) …. ÷ 156 = - 2

- (156 × 2)

= – 312

(vi) x/567 = - 1

⇒ x = - (567 × 1)

= – 567

∴ - (567/567) = -1

Question: 3Which of the following statements are true?

(i) 0 ÷ 4 = 0

(ii) 0 ÷ (-7) = 0

(iii) -15 ÷ 0 = 0

(iv) 0 ÷ 0 = 0

(v) (- 8) ÷ (- 1) = - 8

(vi) – 8 ÷ (- 2) = 4<

Solution:(i) True

(ii) True

(iii) False

(iv) False

(v) False

(vi) True

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Home » R.D. Sharma Solutions » R.D. Sharma Class 7 Solutions » Chapter 1 Integers » Integers Exercise 1.3

Question: 1Find the value of 36 ÷ 6 + 3.

Solution:36 ÷ 6 + 3 = 6 + 3

= 9

Question: 2Find the value of 24 + 15 ÷ 3.

Solution:24 + 15 ÷ 3 = 24 + 5

= 29

Question: 3Find the value of 120 – 20 ÷ 4.

Solution:120 – 20 ÷ 4 = 120 - 5

= 115

Question: 4Find the value of 32 - (3 × 5) + 4.

Solution:32 – (3 x 5) + 4 = 32 – 15 + 4

= 17 + 4

= 21

Question: 5Find the value of 3 - (5 – 6 ÷ 3).

Solution:3 - (5 – 6 ÷ 3) = 3 - (5 - 2)

= 3 – 3

= 0

Question: 6Find the value of 21 – 12 ÷ 3 × 2.

Solution:

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21 – 12 ÷ 3 × 2 = 21 – 123 × 2

= 21 – 4 × 2

= 21 – 8

= 13

Question: 7Find the value of 16 + 8 ÷ 4 – 2 × 3.

Solution:16 + 8 ÷ 4 – 2 × 3

= 16 + 2 – 6

= 18 – 6

= 12

∴ 16 + 8 ÷ 4 – 2 × 3 = 12

Question: 8Find the value of 28 – 5 × 6 + 2.

Solution:28 – 5 × 6 + 2 = 28 - (5 × 6) + 2

= 28 – 30 + 2

= 30 – 30

= 0

Question: 9Find the value of (-20) × (-1) + (-28) ÷ 7.

Solution:(-20) × (-1) + (-28) ÷ 7 = 20 + ∣-28∣ ∣7∣

= 20 - 287

= 20 – 4

= 16

Question: 10Find the value of (-2) + (-8) ÷ (- 4).

Solution:(-2) + (-8) ÷ (- 4) = - 2 + ∣- 8∣ ∣- 4∣

= – 2 + 2

= 0

Question: 11Find the value of (- 15) + 4 ÷ (5 - 3).

Solution:-15 + 4 ÷ (5 - 3) = - 15 + 4 ÷ 2

= – 15 + 2

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= – 13

-15 + 4 ÷ (5 - 3) = - 13

Question: 12Find the value of (- 40) × (-1) + (-28) ÷ 7.

Solution:(- 40) × (-1) + (-28) ÷ 7 = 40 + (- 4)

= 40 – 4

= 36

Question: 13Find the value of (-3) + (-8) ÷ (-4) - 2× (-2).

Solution:(-3) + (-8) ÷ (-4) – 2 × (-2) = (-3) + (-8)(-4) - 2×(-2)

= – 3 + 2 + 4

= 6 – 3

= 3

Question: 14Find the value of (-3) × (-4) ÷ (-2) + (-1).

Solution:(-3) × (-4) ÷ (-2) + (-1) = 12 ÷ (-2) + (-1)

= – 6 – 1

= – 7

∴ (-3) × (-4) ÷ (-2) + (-1) = - 7

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