Integers ICSE Class-7th Concise Selina Maths Solutions

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 CISCE 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 A Integers 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

Answer :-

(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

Answer :-

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)

Answer :-

(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

Answer :-

(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)

Answer :-

(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)

Answer :-

(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

Answer :-

(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?

Answer :-

(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)

Answer :-

(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

Answer :-

(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

Answer :-

Ans 1 Exercise- 1 B Integers ICSE Class-6th

Ans 1 Exercise- 1 B Integers ICSE Class-6th

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)

Answer :-

Ans 2 Exercise- 1 B Integers ICSE Class-6th

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)

Answer :-

Ans 3 Exercise- 1 B Integers ICSE Class-6th

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

Answer

Ans 4 Exercise- 1 B Integers ICSE Class-6th

Question 5 :-

State, true or false :

(i) 0 ÷ 32 = 0

(ii) 0 ÷ (-9) = 0

(iii) (-37) ÷ 0 = 0

(iv) 0 ÷ 0 = 0

Answer :-

(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

Answer :-

 

Ans 6 Exercise- 1 B Integers ICSE Class-6th


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

Question 1 :-

Evaluate:
18-(20- 15 ÷ 3)
Answer :-
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)
Answer
-15+ 24÷ (15- 13)
= -15 + 24 ÷ 2
= -15 + 12
= -3

Question 3 :-

35 – [15 + {14-(13 + ……)}]
Answer :-
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 – …….)}]
Answer :-
27- [13 + {4-(8 + 4 – …….)}]
= 27-[13 +{4-(8+ 4-4)}]
= 27-[13 + {4-8}]
= 27 – [13 + (-4)]
= 21 – [9]
= 27-9
= 18

Question 5 :-

32 – [43-{51 -(20 – ………)}]
Answer :-
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)}]
Answer :-
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}]
Answer :-
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 – ……..)}]
Answer :-
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 – ………..)}]
Answer :-
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-……….)}]
Answer :-
30+ [{-2 x (25-……….)}]
= 30 + [{-2 x (25 – 10)}]
= 30 + [{-2 x 15}]
= 30 + [-30]
= 30-30
= 0

Question 11 :-

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

Question 12 :-

9 x (8-…….) – 2 (2 + ……)
Answer :-
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 – …)}]
Answer :-
2 – [3 – {6 – (5 – …….)}]
⇒ 2 – [3 – {6 – (5 – 1)}]
⇒ 2 – [3 – {6 – 4}]
⇒2 – (3 – 2)
⇒2-1 = 1


Exercise – 1 D Integers  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.
Answer :-
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.
Answer :-
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.
Answer :-
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.
Answer :-
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.
Answer :-
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.

Answer :-

(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)

Answer :-

(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?

Answer :-

(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.
Answer :-
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.
Answer :-
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)

Answer :-
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.
Answer :-

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.
Answer :-
(-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.
Answer :-

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.

Answer :-

(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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