Integers ICSE Class-7th Concise Selina Maths Solutions Chapter-1. We provide step by step Solutions of Exercise / lesson-1 Integer for ICSE Class-7 Concise Selina Mathematics. Our Solutions contain all type Questions with Exe-1 A, Exe-1 B, Exe-1 C and  Exe-1 D to develop skill and confidence. Visit official Website for detail information about ICSE Board Class-7.

## Integers ICSE Class-7th Concise Selina Maths Solutions Chapter-1

–: Select Topics :–

Exercise -1 A

Exercise -1 B

Exercise -1 C

Exercise -1 D

### Exercise -1 AIntegers for ICSE Class-7th Concise Selina Maths

#### Question 1 :-

Evaluate:

1. 427 x 8 + 2 x 427

2. 394 x 12 + 394 x (-2)

3. 558 x 27 + 3 x 558

(1) 427 x 8 + 2 x 427 = 427 x (8 + 2) (Distributive property)

= 427 x 10
= 4270

(2) 394 x 12 + 394 x (-2) = 394 x (12-2) (Distributive property)
= 394 x 10
= 3940

(3) 558 x 27 + 3 x 558 = 558 x (27 + 3) (Distributive property)
= 558 x 30
= 16740

#### Question 2 :-

Evaluate:

1. 673 x 9 + 673

2. 1925 x 101 – 1925

1. 673 x 9 + 673 = 673 x (9 + 1) (Distributive property) = 673 x 10 = 6730

2. 1925 x 101 – 1925 = 1925 x (101 – 1) (Distributive property) = 1925 x 100 = 192500

#### Question 3 :-

Verify:

1. 37 x {8 +(-3)} = 37 x 8 + 37 x – (3)

2. (-82) x {(-4) + 19} = (-82) x (-4) + (-82) x 19

3. {7 – (-7)} x 7 = 7 x 7 – (-7) x 7

4. {(-15) – 8} x -6 = (-15) x (-6) – 8 x (-6)

(1)

37 x {8 + (-3)} = 37 x 8 + 37 x – (3)
L.H.S. = 37 x {8 + (-3)}
= 37 x {8-3}
= 37 x {5}
= 37 x 5
= 185
R.H.S. = 37 x 8 + 37 – 3
= 37 x (8 – 3)
= 37 x 5
= 185
Hence, L.H.S. = R.H.S.

(2)

(-82) x {(-4) + 19} = (-82) x (-4) + (-82) x 19
L.H.S. = (-82) x {(_4) + 19}
= (-82) x {-4 + 19}
= (-82)x {15}
= -82 x 15
=-1230
R.H.S. = (-82) x (-4) + (-82) x 19
= -82 x (-4 + 19)
= -82 x 15
=-1230
Hence, L.H.S. = R.H.S.

(3)

7 – (-7)}. x 7 = 7 x 7 – (-1) x 7
L.H.S. = {7 – (-7)} x 7
= {7 + 7} x 7
= {14} x 7
= 14 x 7
= 98
R.H.S. = 7 x 7 – (-7) x 7
=7 x 7+7 x 7 =
7 x (7 + 7)
= 7 x (14)
= 98
Hence, L.H.S. = R.H.S.

(4)

(-15) – 8} x -6 = (-15) x (-6) – 8 x (-6)
L.H.S. = {(-15)-8} x-6
= {-15-8} x-6
= {-23} x-6
= -23 x- 6
= 138
R.H.S. = (-15) x (-6) – 8 x (-6)
= -6 x (-15-8)
= -6 x -23
= 138
Hence, L.H.S. = R.H.S.

#### Question 4 :-

Evaluate:

1. 15 x 8

2. 15 x (-8)

3. (-15) x 8

4. (-15) x -8

(1) 15 x 8= 120

(2) 15 x (-8) = -120

(3) (-15) x 8 = -120

(4) (-15) x -8 = 120
(Since the number of negative integers in the product is even)

#### Question 5 :-

Evaluate:

(i) 4 x 6 x 8

(ii) 4 x 6 x (-8)

(iii) 4 x (-6) x 8

(iv) (-4) x 6 x 8

(v) 4 x (-6) x (-8)

(vi) (-4) x (-6) x 8

(vii) (-4) x 6 x (- 8)

(viii) (-4) x (-6) x (-8)

(i) 4 x 6 x 8 = 192

(ii) 4 x 6 x (-8) = -192
(It have one negative factor)

(iii) 4 x (-6) x 8 = -192
(It have one negative factor)

(iv) (-4 )x 6 x 8 = -192
(It have one negative factor)

(v) 4 x (-6) x (-8) = 192
(It have two negative factors)

(vi) (-4) x (-6) x 8 = 192
(It have two negative factors)

(vii) (-4) x 6 x (-8) = 192
(It have two negative factors)

(viii) (-4) x (-6) x (-8) = -192
(It have three negative factors)

#### Question 6 :-

Evaluate:

(i) 2 x 4 x 6 x 8

(ii) 2 x (-4) x 6 x 8

(iii) (-2) x 4 x (-6) x 8

(iv) (-2) x (-4) X 6 x (-8)

(v) (-2) x (-4) x (-6) x (-8)

(i) 2 x 4 x 6 x 8 = 384

(ii) 2 x (-4) x 6 x 8 = -384
(Number of negative integer in the product is odd)

(iii) (-2) x 4 x (-6) x 8 = 384
(Number of negative integer in the product is even)

(iv) (-2) x (-4) x 6 x (-8) = -384
(Number of negative integer in the product is odd)

(v) (-2) x (-4) x (-6) x (-8) = 384
(Number of negative integer in the product is even)

#### Question 7 :-

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

(i) -47

(ii) 63

(iii) -1

(iv) 0

(i) -1 x 47 = -47
Hence, integer is 47

(ii) -1 x -63 = 63
Hence, integer is -63

(ii) -1 x 1 = -1
Hence, integer is 1

(iv) -1 x 0 = 0
Hence, integer is 0

#### Question 8 :-

Eighteen integers are multiplied together. What will be the sign of their product, if:

(i) 15 of them are negative and 3 are positive?

(ii) 12 of them are negative and 6 are positive?

(iii) 9 of them are positive and the remaining are negative?

(iv) all are negative?

(i) Since out of eighteen integers, 15 of them are negative, which is odd number. Hence, sign of product will be negative (-).

(ii) Since out of eighteen integers 12 of them are negative, which is even number. Hence sign of product will be positive (+).

(iii) Since out of eighteen integers 9 of them are negative, which is odd number. Hence, sign of product will be negative (-).

(iv) Since all are negative, which is even number. Hence sign of product will be positive (+).

#### Question 9 :-

Find which is greater?

(i) (8 + 10) x 15 or 8 + 10 x 15

(ii) 12 x (6 – 8) or 12 x 6 – 8

(iii) {(-3) – 4} x (-5) or (-3) – 4 x (-5)

(i) (8 + 10) x 15 or 8 + 10 x 15
(8 + 10) x 15 = 18 x 15 = 270
8 + 10 x 15 = 8 + 150 = 158
∴(8 + 10) x 15 > 8 + 10 x 15

(ii) 12 x (6 – 8) or 12 x 6 – 8
12 x (6 – 8) = 12 (-2) = -24
12 x 6 – 8 = 72 – 8 = 64
∴12 x 6 – 8 > 12 x (6-8)

(iii) {(-3) – 4} x (-5) or (-3) – 4 x (-5)
{(-3) – 4} x (-5) = {-3 – 4} x (-5) = -7 x -5 = 35
(-3) – 4 x (-5) = -7 x (-5) = 35
∴{(-3) – 4} x (-5) = (-3) – 4 x (-5)

#### Question 10 :-

State, true or false :

(i) product of two integers can be zero.

(ii) product of 120 negative integers and 121 positive integers is negative.

(iii) a x (b + c) = a x b + c

(iv) (b – c) x a=b – c x a

(i) False.

(ii) False.
Correct : Since 120 integers are even numbers, hence product will be positive and for 121 integers are positive in numbers, hence product will be positive.

(iii) False

Correct: a x (b + c) ≠ a x b + c

ab + ac ≠ ab + c

(iv) False

Correct: (b – c) x a ≠ b – c x a

ab – ac ≠ b – ca

### Exercise- 1 B of Exe- Integers ICSE Class-7th Concise

#### Question 1 :-

Divide:
(i) 117 by 9
(ii) (-117) by 9
(iii) 117 by (-9)
(iv) (-117) by (-9)
(v) 225 by (-15)
(vi) (-552) ÷ 24
(vii) (-798) by (-21)
(viii) (-910) ÷ – 26

#### Question 2 :-

Evaluate:
(i) (-234) ÷ 13
(ii) 234 ÷ (-13)
(iii) (-234) ÷ (-13)
(iv) 374 ÷ (-17)
(v) (-374) ÷ 17
(vi) (-374) ÷ (-17)
(vii) (-728) ÷ 14
(viii) 272 ÷ (-17)

#### Question 3 :-

Find the quotient in each of the following divisions:
(i) 299 ÷ 23
(ii) 299 ÷ (-23)
(iii) (-384) ÷ 16
(iv) (-572) ÷ (-22)
(v) 408 ÷ (-17)

#### Question 4 :-

Divide:
(i) 204 by 17
(ii) 152 by-19
(iii) 0 by 35
(iv) 0 by (-82)
(v) 5490 by 10
(vi) 762800 by 100

#### Question 5 :-

State, true or false :

(i) 0 ÷ 32 = 0

(ii) 0 ÷ (-9) = 0

(iii) (-37) ÷ 0 = 0

(iv) 0 ÷ 0 = 0

(i) True.

(ii) True.

(iii) False.
Correct: It is not meaningful (defined)

(iv) False.
Correct: It is not defined.

#### Question 6 :-

Evaluate:
(i) 42 ÷ 7 + 4
(ii) 12+18 ÷ 3
(iii) 19 – 20 ÷ 4
(iv) 16 – 5 x 3+4
(v) 6 – 8 – (-6) ÷ 2
(vi) 13 -12 ÷ 4 x 2
(vii) 16 + 8 ÷ 4- 2 x 3
(viii) 16 ÷ 8 + 4 – 2 x 3
(ix) 16 – 8 + 4 ÷ 2 x 3
(x) (-4) + (-12) ÷ (-6)
(xi) (-18) + 6 ÷ 3 + 5
(xii) (-20) x (-1) + 14 – 7

### Exercise – 1 C Solutions of Integers for ICSE Class-7th Concise Selina

Evaluate:
18-(20- 15 ÷ 3)
18-(20- 15 ÷ 3)
= 18 – (20-15/3)
= 18 – (20 – 5)
= 18 – 20 + 5
= 18 + 5 – 20
= 23 – 20
= 3

#### Question 2 :-

-15+ 24÷ (15-13)
-15+ 24÷ (15- 13)
= -15 + 24 ÷ 2
= -15 + 12
= -3

#### Question 3 :-

35 – [15 + {14-(13 + ……)}]
35- [15 + {14-(13 + +2-1+3)}] = 35-[15+ 14-(13+4)] = 35 — [15 + 14 – (13 + 4}] = 35-{15 + 14-17] = 35-15-14+ 17
= 35 + 17-15-14
= 52 – 29
= 23

#### Question 4 :-

27- [13 + {4-(8 + 4 – …….)}]
27- [13 + {4-(8 + 4 – …….)}] = 27-[13 +{4-(8+ 4-4)}] = 27-[13 + {4-8}] = 27 – [13 + (-4)] = 21 –  = 27-9
= 18

#### Question 5 :-

32 – [43-{51 -(20 – ………)}]
32 – [43 – {51 – (20 – ………….)}] = 32-[43 – {51 -(20- 11)}] = 32-[43-{51 -9}] = 32-[43 -42] = 32-1
=31

#### Question 6 :-

46-[26-{14-(15-4÷ 2 x 2)}]
46 – [26 – {14 – (15 – 4 ÷ 2 x 2)}] = 46-[26- {14-(15-2 x 2)}] = 46-[26- {14-(15 -4)}] = 46-[26- {14- 11}] = 46 – [26 – 3] = 46 – 23
= 23

#### Question 7 :-

45 – [38 – {60 ÷ 3 – (6 – 9 ÷ 3) ÷ 3}]
45 – [38 – {60 ÷ 3 – (6 – 9 ÷ 3) ÷ 3}] = 45-[38- {60 ÷ 3-(6-3)÷ 3}] = 45-[38 -{20-3 ÷ 3}] = 45-[38- {20-1}] = 45-[38- 19] = 45-19
= 26

#### Question 8 :-

17- [17 — {17 — (17 – ……..)}]
17- [17-{17-(17 –……..)}] = 17-[17-{17-(17-0)}] = 17 – [17 – {17 — 17}] = 17 — [17 — 0] = 17-17
= 0

#### Question 9 :-

2550 – [510 – {270 – (90 – ………..)}]
2550- [510-{270-(90-……..)}] = 2550 – [510 – {270 – (90 – 87)}] = 2550 -[510- {270 -3}] = 2550-[510-267] = 2550 – 243
= 2307

#### Question 10 :-

30+ [{-2 x (25-……….)}]
30+ [{-2 x (25-……….)}] = 30 + [{-2 x (25 – 10)}] = 30 + [{-2 x 15}] = 30 + [-30] = 30-30
= 0

#### Question 11 :-

88-{5-(-48)+ (-16)}
88- {5-(-48)+ (-16)}
=88 – {5-(-48/-16)
= 88 – {5-3}
= 88 – 2
= 86

#### Question 12 :-

9 x (8-…….) – 2 (2 + ……)
9 x (8-…..) -2(2 + ……..)
= 9 x (8 – 5) – 2(2 + 6)
= 9 x 3 – 2 x 8
= 27- 16
= 11

#### Question 13 :-

2 – [3 – {6 – (5 – …)}]
2 – [3 – {6 – (5 – …….)}] ⇒ 2 – [3 – {6 – (5 – 1)}] ⇒ 2 – [3 – {6 – 4}] ⇒2 – (3 – 2)
⇒2-1 = 1

### Exercise – 1 DIntegers  Concise  Maths Selina Publishers ICSE Class-7

#### Question 1 :-

The sum of two integers is -15. If one of them is 9, find the other.
Sum of two integers = -15
One integer = 9
∴ Second integer = -15 – 9
= -(15 + 9)
= -24

#### Question 2 :-

The difference between an integer and -6 is -5. Find the values of x.
The difference between an integer
= x-(-6) = -5
∴ Value of
⇒ x – (-6) = -5
⇒ x + 6 = -5
x = -5 – 6
x = -11

#### Question 3 :-

The sum of two integers is 28. If one integer is -45, find the other.
Sum of two integers = 28
One integer = -45
∴ Second integer = 28 – (-45)
= 28 + 45
= 73

#### Question 4 :-

The sum of two integers is -56. If one integer is -42, find the other.
Sum of two integers = -56
One integer = -42
∴Second integer = -56 – (-42)
= -56+ 42
=-14

#### Question 5 :-

The difference between an integer x and (-9) is 6. Find all possible values ofx.
The difference between an integer x – (-9) = 6 or -9 – x = 6
∴ Value of x
⇒ x – (-9) = 6 or ⇒ -9 – x = 6
⇒ x + 9 = 6 or Answer-x = 6 + 9
⇒ x = 6 – 9 or ⇒ -x = 15
⇒x = -3 or ⇒ x = -15
Hence, possible values of x are -3 and -15.

#### Question 6 :-

Evaluate:

(i) (-1) x (-1) x (-1) x  ….60 times.

(ii) (-1) x (-1) x (-1) x (-1) x …. 75 times.

(i) 1 (because (-1) is multiplied even times.)

(ii) -1 (because (-1) is multiplied odd times.)

#### Question 7 :-

Evaluate:

(i) (-2) x (-3) x (-4) x (-5) X (-6)

(i) (-3) x (-6) x (-9) x (-12)

(i) (-11) x (-15) + (-11) x (-25)

(i) 10 x (-12) + 5 x (-12)

(i) (-2) x (-3) x (-4) x (-5) x (-6)
⇒ 6 x 20 x (-6) = 120 x (-6)
= -720

(ii) (-3) x (-6) x (-9) x (-12)
⇒ 18 x 108
= 1944

(iii) (-11) x (-15) + (-11) x (-25)
⇒ 165 + 275
= 440

(iv) 10 x (-12) + 5 x (-12)
⇒ -120-60
= -180

#### Question 8 :-

(i) If x x (-1) = -36, is x positive or negative?

(ii) If x x (-1) = 36, is x positive or negative?

(i) x x (-1) = -36
-lx = -36
x = $\frac { -36 }{ -1 }$
x = 36
∵ x = 36
∴ It is a positive integer.

(ii) x x (-1) = 36
-1x = 36
x = $\frac { 36 }{ -1 }$
x = -36
∵x = -36
∴It is a negative integer.

#### Question 9 :-

Write all the integers between -15 and 15, which are divisible by 2 and 3.
The integers between -15 and 15 are :
-12, -6, 0, 6 and 12
That are divisible by 2 and 3.

#### Question 10 :-

Write all the integers between -5 and 5, which are divisible by 2 or 3.
The integers between -5 and 5 are :
-4, -3, -2, 0, 0, 2, 3 and 4
That are divisible by 2 or 3.

#### Question 11 :-

Evaluate:

(i) (-20) + (-8) ÷ (-2) x 3

(ii) (-5) – (-48) ÷ (-16) + (-2) x 6

(iii) 16 + 8 ÷ 4- 2 x 3

(iv) 16 ÷ 8 x 4 – 2 x 3

(v) 27 – [5 + {28 – (29 – 7)}]

(vi) 48 – [18 – {16 – (5 – $\overline { 4 +1 }$)}]

(vii) -8 – {-6 (9 – 11) + 18 = -3}

(viii) (24 ÷ $\overline { 12 -9 }$ – 12) – (3 x 8 ÷ 4 + 1)

We know that, if these type of expressions that has more than one fundamental operations, we use the rule of DMAS i.e., First of all we perform D (division), then M (multiplication), then A (addition) and in the last S (subtraction).

(i) (-20) + (-8) ÷ (-2) x 3
⇒ -20 + 4 x 3
⇒ -20+ 12
=-8

(ii) (-5) – (-48) ÷ (-16) + (-2) x 6
⇒ (-5) – 3 + (-2) x 6
⇒ -5 – 3 – 12
⇒ -8- 12
= -20

(iii) 16 + 8 ÷ 4 – 2 x 3
⇒ 16 + 2 – 2 x 3
⇒16 + 2 – 6
⇒ 18-6
= 12

(iv) 16 ÷ 8 x 4 – 2 x 3
⇒ 2 x 4 – 2 x 3
⇒ 8 – 6
= 2

(v) 27 – [5 + {28 – (29 – 7)}]
⇒ 27 – [5 + {28 – 22}] ⇒ 27 – [5 + 6] ⇒ 27 — 11
= 16

(vi) 48-[18-{16-(5 – $\overline { 4 +1 }$)}]
⇒ 48-[18-{16-(5-5)}] ⇒ 48-[18- {16-0)}] ⇒ 48-[18- 16] ⇒ 48 – 2
= 46

(viii) -8 – {-6 (9 – 11) + 18 ÷ -3}
⇒ -8 – {-6 (-2) – 6}
⇒ -8- {12-6}
⇒ -8 – {6}
⇒ -8-6
= -14

(viii) (24 ÷ $\overline { 12 -9 }$ – 12) – (3 x 8 = 4 + 1)
⇒ (24 ÷ 3-12)-(3 x 2 + 1)
⇒ (8- 12)-(6+ 1)
⇒ —4 — 7
= —11

#### Question 12 :-

Find the result of subtracting the sum of all integers between 20 and 30 from the sum of all integers from 20 to 30.

Required number

= (Sum of all integers between 20 and 30 – Integers between 20 and 30)

(20 + 21 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29 + 30) – (21 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29 )

⇒ 20 + 30 = 50

∴ Required number = 50

#### Question 13 :-

Add the product of (-13) and (-17) to the quotient of (-187) and 11.
(-13) x (-17)+ (-187- 11)
⇒ (-13) x (-17) + (-17)
⇒ 221 – 17 = 204

#### Question 14 :-

The product of two integers is-180. If one of them is 12, find the other.

The product of two integers = -180 One integer = 12
∴ Second integer = -180 – 12 = -15

#### Question 15 :-

What is the increase or decrease in the number?

(i) A number changes from -20 to 30.

(ii) A number changes from 40 to – 30.

(i)

∵ A number changes from = – 20 to 30

⇒ – 20 – 30 = – 50

∴ – 50, it will be increases.

(ii)

∵A number changes from = 40 to – 30

⇒ 40 – (- 30)

40 + 30 = 70

∴70, it will be decreases.

— End of Integers Solutions :–

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