Selina Concise Mathematics Class 7 ICSE Solutions Chapter 1 Integers

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 1 Integers

Selina Publishers Concise Mathematics Class 7 ICSE Solutions Chapter 1 Integers

Integers Exercise 1A – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.

Evaluate:

1. 427 x 8 + 2 x 427
2. 394 x 12 + 394 x (-2)
3. 558 x 27 + 3 x 558

Solution:

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

Solution:

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)

Solution:

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

Solution:

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:

1. 4 x 6 x 8
2. 4 x 6 x (-8)
3. 4 x (-6) x 8
4. (-4) x 6 x 8
5. 4 x (-6) x (-8)
6. (-4) x (-6) x 8
7. (-4) x 6 x (- 8)
8. (-4) x (-6) x (-8)

Solution:

1. 4 x 6 x 8 = 192
2. 4 x 6 x (-8) = -192
        (It have one negative factor)
3. 4 x (-6) x 8 = -192
        (It have one negative factor)
4. (-4 )x 6 x 8 = -192
        (It have one negative factor)
5. 4 x (-6) x (-8) = 192
        (It have two negative factors)
6. (-4) x (-6) x 8 = 192
        (It have two negative factors)
7. (-4) x 6 x (-8) = 192
        (It have two negative factors)
8. (-4) x (-6) x (-8) = -192
        (It have three negative factors)

Question 6.
Evaluate:

1. 2 x 4 x 6 x 8
2. 2 x (-4) x 6 x 8
3. (-2) x 4 x (-6) x 8
4. (-2) x (-4) X 6 x (-8)
5. (-2) x (-4) x (-6) x (-8)

Solution:

1. 2 x 4 x 6 x 8 = 384
2. 2 x (-4) x 6 x 8 = -384
        (Number of negative integer in the product is odd)
3. (-2) x 4 x (-6) x 8 = 384
        (Number of negative integer in the product is even)
4. (-2) x (-4) x 6 x (-8) = -384
        (Number of negative integer in the product is odd)
5. (-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:

1. -47
2. 63
3. -1
4. 0

Solution:

1. -1 x 47 = -47
        Hence, integer is 47
2. -1 x -63 = 63
         Hence, integer is -63
3. -1 x 1 = -1
         Hence, integer is 1
4. -1 x 0 = 0
         Hence, integer is 0

Question 8.
Eighteen integers are multiplied together. What will be the sign of their product, if:

1. 15 of them are negative and 3 are positive?
2. 12 of them are negative and 6 are positive?
3. 9 of them are positive and the remaining are negative?
4. all are negative?

Solution:

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

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

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

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

Question 9.

Find which is greater?

1. (8 + 10) x 15 or 8 + 10 x 15
2. 12 x (6 – 8) or 12 x 6 – 8
3. {(-3) – 4} x (-5) or (-3) – 4 x (-5)

Solution:

1. (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
2. 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)
3. {(-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 :

1. product of two integers can be zero.
2. product of 120 negative integers and 121 positive integers is negative.
3. a x (b + c) = a x b + c
4. (b – c) x a=b – c x a

Solution:

1. False.
2. 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.

3. False.

      Correct :a x (b + c) ≠ a x b + c
       ab + ac ≠ ab + c
4. False.
      Correct: (b – c) x a ≠ b – c x a
       ab – ac ≠ b – ca

 Integers Exercise 1B – Selina Concise Mathematics Class 7 ICSE Solutions

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

Solution :



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)

Solution:





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)

Solution:

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

Solution:


Question 5.
State, true or false :
1. 0 ÷ 32 = 0
2. 0 ÷ (-9) = 0
3. (-37) ÷ 0 = 0
4. 0 ÷ 0 = 0

Solution:

1. True.
2. True.
3. False.
Correct: It is not meaningful (defined)
4. 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

Solution:

Integers Exercise 1C – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Evaluate:

18-(20- 15 ÷ 3)

Solution:
18-(20- 15 ÷ 3)
= 18 – 

= 18 – (20 – 5)
= 18 – 20 + 5
= 18 + 5 – 20
= 23 – 20
= 3

Question 2.
-15+ 24÷ (15-13)
Solution:
-15+ 24÷ (15- 13)
= -15 + 24 ÷ 2
= -15 + 12
= -3

Question 3.
35 – [15 + {14-(13 +  )}]

35- [15 + {14-(13 +   )}]

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

Question 11.
88-{5-(-48)+ (-16)}
Solution:
88- {5-(-48)+ (-16)}
=88 – 

= 88 – {5-3}
= 88 – 2
= 86

Question 12.
9 x (8- ) – 2 (2 +   )

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

Integers Exercise 1D – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
The sum of two integers is -15. If one of them is 9, find the other.
Solution:
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.
Solution:
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.
Solution:
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.
Solution:
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.
Solution:

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 ofx are -3 and -15.

Question 6.
Evaluate:

1. (-1) x (-1) x (-1) x  ….60 times.
2. (-1) x (-1) x (-1) x (-1) x …. 75 times.

Solution:

1. 1 (because (-1) is multiplied even times.)
2. -1 (because (-1) is multiplied odd times.)

Question 7.
Evaluate:

1. (-2) x (-3) x (-4) x (-5) X (-6)
2. (-3) x (-6) x (-9) x (-12)
3. (-11) x (-15) + (-11) x (-25)
4. 10 x (-12) + 5 x (-12)

Solution:
1. (-2) x (-3) x (-4) x (-5) x (-6)
   ⇒ 6 x 20 x (-6) = 120 x (-6)
   = -720
2. (-3) x (-6) x (-9) x (-12)
   ⇒ 18 x 108
   = 1944
3. (-11) x (-15) + (-11) x (-25)
   ⇒ 165 + 275
   = 440
4. 10 x (-12) + 5 x (-12)
   ⇒ -120-60
   = -180

Question 8.

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

Solution:

1. x x (-1) = -36
  -lx = -36
   x = 

 x = 36
  ∵ x = 36
  ∴ It is a positive integer.
2. x x (-1) = 36
   -1x = 36
    x = 
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.
Solution:
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.
Solution:
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:

1. (-20) + (-8) ÷ (-2) x 3
2. (-5) – (-48) ÷ (-16) + (-2) x 6
3. 16 + 8 ÷ 4- 2 x 3
4. 16 ÷ 8 x 4 – 2 x 3
5. 27 – [5 + {28 – (29 – 7)}]
6. 48 – [18 – {16 – (5 – )}]
7. -8 – {-6 (9 – 11) + 18 = -3}
8. (24 ÷ – 12) – (3 x 8 ÷ 4 + 1)

Solution:

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).

1. (-20) + (-8) ÷ (-2) x 3

   ⇒ -20 + 4 x 3
   ⇒ -20+ 12
   =-8
2. (-5) – (-48) ÷ (-16) + (-2) x 6
   ⇒ (-5) – 3 + (-2) x 6
   ⇒ -5 – 3 – 12
   ⇒ -8- 12
   = -20
3. 16 + 8 ÷ 4 – 2 x 3
   ⇒ 16 + 2 – 2 x 3
   ⇒16 + 2 – 6
   ⇒ 18-6
   = 12
4. 16 ÷ 8 x 4 – 2 x 3
   ⇒ 2 x 4 – 2 x 3
   ⇒ 8 – 6
   = 2
5. 27 – [5 + {28 – (29 – 7)}]
   ⇒ 27 – [5 + {28 – 22}]
   ⇒ 27 – [5 + 6]
   ⇒ 27 — 11
   = 16
6. 48-[18-{16-(5 –  )}]
   ⇒ 48-[18-{16-(5-5)}]
   ⇒ 48-[18- {16-0)}]
   ⇒ 48-[18- 16]
   ⇒ 48 – 2
   = 46
7. -8 – {-6 (9 – 11) + 18 ÷ -3}
   ⇒ -8 – {-6 (-2) – 6}
   ⇒ -8- {12-6}
   ⇒ -8 – {6}
   ⇒ -8-6
   = -14
8. (24 ÷  – 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.

Solution:

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.
Solution:
(-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.
Solution:

The product of two integers = -180 One integer = 12

∴ Second integer = -180 – 12 = -15

Question 15.

1. A number changes from -20 to 30. What is the increase or decrease in the number?
2. A number changes from 40 to -30. What is the increase or decrease in the number?

Solution:
1. ∵A number changes from = -20 to 30
   ⇒ -20 – 30 = -50
   ∴-50, it will be increases.
2. ∵A number changes from = 40 to -30
   ⇒ 40 – (-30)
      40 + 30 = 70
   ∴70, it will be decreases