Class 7: Maths Chapter 1 solutions. Complete Class 7 Maths Chapter 1 Notes.
Contents
Selina Class 7 ICSE Solutions Mathematics : Chapter 1- Integers
Selina 7th Maths Chapter 1, Class 7 Maths Chapter 1 solutions
Exercise 1A page: 6
1. Evaluate:
(i) 427 × 8 + 2 × 427
(ii) 394 × 12 + 394 × (-2)
(iii) 558 × 27 + 3 × 558
Solution:
(i) 427 × 8 + 2 × 427
Using Distributive property
= 427 × (8 + 2)
By further calculation
= 427 × 10
= 4270
(ii) 394 × 12 + 394 × (-2)
Using Distributive property
= 394 × (12 – 2)
By further calculation
= 394 × 10
= 3940
(iii) 558 × 27 + 3 × 558
Using Distributive property
= 558 × (27 + 3)
By further calculation
= 558 × 30
= 16740
2. Evaluate:
(i) 673 × 9 + 673
(ii) 1925 × 101 – 1925
Solution:
(i) 673 × 9 + 673
Using Distributive property
= 673 × (9 + 1)
By further calculation
= 673 × 10
= 6730
(ii) 1925 × 101 – 1925
Using Distributive property
= 1925 × (101 – 1)
By further calculation
= 1925 × 100
= 192500
3. Verify:
(i) 37 × {8 + (-3)} = 37 × 8 + 37 × (-3)
(ii) (-82) × {(- 4) + 19} = (-82) × (- 4) + (-82) × 19
(iii) {7 – (-7)} × 7 = 7 × 7 – (-7) × 7
(iv) {(-15) – 8} × -6 = (-15) × (-6) – 8 × (-6)
Solution:
(i) 37 × {8 + (-3)} = 37 × 8 + 37 × (-3)
Consider LHS = 37 × {8 + (-3)}
It can be written as
= 37 × {8 – 3}
By further calculation
= 37 × {5}
= 185
Similarly RHS = 37 × 8 + 37 × (-3)
It can be written as
= 37 × (8 – 3)
By further calculation
= 37 × 5
= 185
Therefore, LHS = RHS.
(ii) (-82) × {(- 4) + 19} = (-82) × (- 4) + (-82) × 19
Consider LHS = (-82) × {(- 4) + 19}
It can be written as
= (-82) × {-4 + 19}
By further calculation
= (-82) × {15}
= – 1230
Similarly RHS = (-82) × (- 4) + (-82) × 19
It can be written as
= (-82) × (-4 + 19)
By further calculation
= – 82 × 15
= – 1230
Therefore, LHS = RHS.
(iii) {7 – (-7)} × 7 = 7 × 7 – (-7) × 7
Consider LHS = {7 – (-7)} × 7
It can be written as
= {7 + 7} × 7
By further calculation
= {14} × 7
= 98
Similarly RHS = 7 × 7 – (-7) × 7
It can be written as
= 7 × 7 + 7 × 7
By further calculation
= 7 × (7 + 7)
= 7 × 14
= 98
Therefore, LHS = RHS.
(iv) {(-15) – 8} × -6 = (-15) × (-6) – 8 × (-6)
Consider LHS = {(-15) – 8} × -6
It can be written as
= {- 15 – 8} × -6
By further calculation
= {-23} × -6
= 138
Similarly RHS = (-15) × (-6) – 8 × (-6)
It can be written as
= – 6 × (- 15 – 8)
By further calculation
= – 6 × – 23
= 138
Therefore, LHS = RHS.
4. Evaluate:
(i) 15 × 8
(ii) 15 × (-8)
(iii) (-15) × 8
(iv) (-15) × -8
Solution:
(i) 15 × 8 = 120
(ii) 15 × (-8) = – 120
(iii) (-15) × 8 = – 120
(iv) (-15) × -8 = 120
5. Evaluate:
(i) 4 × 6 × 8
(ii) 4 × 6 × (-8)
(iii) 4 × (-6) × 8
(iv) (-4) × 6 × 8
(v) 4 × (-6) × (-8)
(vi) (-4) × (-6) × 8
(vii) (-4) × 6 × (-8)
(viii) (-4) × (-6) × (-8)
Solution:
(i) 4 × 6 × 8 = 192
(ii) 4 × 6 × (-8) = – 192
It has one negative factor
(iii) 4 × (-6) × 8 = – 192
It has one negative factor
(iv) (-4) × 6 × 8 = – 192
It has one negative factor
(v) 4 × (-6) × (-8) = 192
It has two negative factors
(vi) (-4) × (-6) × 8 = 192
It has two negative factors
(vii) (-4) × 6 × (-8) = 192
It has two negative factors
(viii) (-4) × (-6) × (-8) = – 192
It has three negative factors
6. Evaluate:
(i) 2 × 4 × 6 × 8
(ii) 2 × (-4) × 6 × 8
(iii) (-2) × 4 × (-6) × 8
(iv) (-2) × (-4) × 6 × (-8)
(v) (-2) × (-4) × (-6) × (-8)
Solution:
(i) 2 × 4 × 6 × 8 = 384
(ii) 2 × (-4) × 6 × 8 = -384
The number of negative integer in the product is odd
(iii) (-2) × 4 × (-6) × 8 = 384
The number of negative integer in the product is even
(iv) (-2) × (-4) × 6 × (-8) = -384
The number of negative integer in the product is odd
(v) (-2) × (-4) × (-6) × (-8) = 384
The number of negative integer in the product is even
7. Determine the integer whose product with ‘-1’ is:
(i) -47
(ii) 63
(iii) -1
(iv) 0
Solution:
(i) -47 = – 1 × 47
Therefore, the integer is 47.
(ii) 63 = – 1 × – 63
Therefore, the integer is – 63.
(iii) -1 = – 1 × 1
Therefore, the integer is 1.
(iv) 0 = -1 × 0
Therefore, the integer is 0.
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?
Solution:
(i) Out of 18 integers, 15 of them are negative which is odd number. Therefore, the sign of product is negative.
(ii) Out of 18 integers, 12 of them are negative which is even number. Therefore, the sign of product is positive.
(iii) Out of 18 integers, 9 of them are negative which is odd number. Therefore, the sign of product is negative.
(iv) All are negative which is even number. Therefore, sign of product is positive.
9. Find which is greater?
(i) (8 + 10) × 15 or 8 + 10 × 15
(ii) 12 × (6 – 8) or 12 × 6 – 8
(iii) {(-3) – 4} × (-5) or (-3) – 4 × (-5)
Solution:
(i) (8 + 10) × 15 or 8 + 10 × 15
We know that
(8 + 10) × 15 = 18 × 15 = 270
8 + 10 × 15 = 8 + 150 = 158
Therefore, (8 + 10) × 15 > 8 + 10 × 15.
(ii) 12 × (6 – 8) or 12 × 6 – 8
We know that
12 × (6 – 8) = 12 (-2) = – 24
12 × 6 – 8 = 72 – 8 = 64
Therefore, 12 × (6 – 8) < 12 × 6 – 8.
(iii) {(-3) – 4} × (-5) or (-3) – 4 × (-5)
We know that
{(-3) – 4} × (-5) = {- 3 – 4} × (-5)
By further calculation
= – 7 × -5 = 35
Similarly
(-3) – 4 × (-5) = – 3 + 20 = 17
Therefore, {(-3) – 4} × (-5) > (-3) – 4 × (-5)
10. State, true or false:
(i) product of two different integers can be zero.
(ii) product of 120 negative integers and 121 positive integers is negative.
(iii) a × (b + c) = a × b + c
(iv) (b – c) × a = b – c × a.
Solution:
(i) True.
Example: 5 × 0 = 0, 0 × – 8 = 0
(ii) False.
The total number of negative integers is 120 which is an even number and we know that the product of even numbers of negative integers is always positive. Hence, the sign of the product will be positive.
(iii) False.
a × (b + c) ≠ a × b + c
ab + ac ≠ ab + c
(iv) False.
(b – c) × a ≠ b – c × a
ab – ac ≠ b – ca
Exercise 1B page: 11
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:
(i) 117 by 9
= 117/9
It can be written as
= (13 × 9)/ 9
= 13
(ii) (-117) by 9
= – 117/ 9
It can be written as
= (-13 × 9)/ 9
= -13
(iii) 117 by (-9)
= 117/ -9
It can be written as
= (13 × 9)/ -9
= – 13
(iv) (-117) by (-9)
= – 117/-9 = 117/9
It can be written as
= (13 × 9)/ 9
= 13
(v) 225 by (-15)
= 225/ (-15)
It can be written as
= (15 × 15)/ -15
= – 15
(vi) (-552) ÷ 24
= – 552/ 24
It can be written as
= (- 23 × 24)/ 24
= – 23
(vii) (-798) by (-21)
= – 798/ – 21 = 798/21
It can be written as
= (38 × 21)/ 21
= 38
(viii) (-910) ÷ 26
= – 910/ 26
It can be written as
= (- 35 × 26)/ 26
= – 35
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:
(i) (-234) ÷ 13
= – 234/ 13
It can be written as
= (-18 × 13)/ 13
= – 18
(ii) 234 ÷ (-13)
= 234/ -13
It can be written as
= (18 × 13)/ – 13
= – 18
(iii) (-234) ÷ (-13)
= – 234/ – 13 = 234/13
It can be written as
= (18 × 13)/ 13
= 18
(iv) 374 ÷ (-17)
= 374/ -17
It can be written as
= (22 × 17)/ (-17)
= – 22
(v) (-374) ÷ 17
= – 374/ 17
It can be written as
= (-22 × 17)/ (17)
= – 22
(vi) (-374) ÷ (-17)
= – 374/ -17 = 374/17
It can be written as
= (22 × 17)/ (17)
= 22
(vii) (-728) ÷ 14
= – 728/14
It can be written as
= (-52 × 14)/ 14
= -52
(viii) 272 ÷ (-17)
= 272/ -17
It can be written as
= (16 × 17)/ (-17)
= -16
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:
(i) 299 ÷ 23
= 299/23
It can be written as
= (23 × 13)/ 23
= 13
(ii) 299 ÷ (-23)
= 299/ -23
It can be written as
= (23 × 13)/ – 23
= – 13
(iii) (-384) ÷ 16
= – 384/16
It can be written as
= (- 24 × 16)/ 16
= – 24
(iv) (-572) ÷ (-22)
= – 572/ – 22 = 572/22
It can be written as
= (26 × 22)/ 22
= 26
(v) 408 ÷ (-17)
= 408/ -17
It can be written as
= (24 × 17)/ (-17)
= – 24
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:
(i) 204 by 17
= 204/17
It can be written as
= (12 × 17)/ 17
= 12
(ii) 152 by – 19
= 152/ -19
It can be written as
= (8 × 19)/ -19
= – 8
(iii) 0 by 35
= 0/35
= 0
(iv) 0 by (-82)
= 0/ -82
= 0
(v) 5490 by 10
= 5490/10
It can be written as
= (549 × 10)/ 10
= 549
(vi) 762800 by 100
= 762800/100
It can be written as
= (7628 x 100)/ 100
= 7628
5. State, true or false:
(i) 0 ÷ 32 = 0
(ii) 0 ÷ (-9) = 0
(iii) (-37) ÷ 0 = 0
(iv) 0 ÷ 0 = 0
Solution:
(i) True.
(ii) True.
(iii) False. It is not defined.
(iv) False. It is not defined.
6. Evaluate:
(i) 42 ÷ 7 + 4
(ii) 12 + 18 ÷ 3
(iii) 19 – 20 ÷ 4
(iv) 16 – 5 × 3 + 4
(v) 6 – 8 – (-6) ÷ 2
(vi) 13 – 12 ÷ 4 × 2
(vii) 16 + 8 ÷ 4 – 2 × 3
(viii) 16 ÷ 8 + 4 – 2 × 3
(ix) 16 – 8 + 4 ÷ 2 × 3
(x) (-4) + (-12) ÷ (-6)
(xi) (-18) + 6 ÷ 3 + 5
(xii) (-20) × (-1) + 14 ÷ 7
Solution:
(i) 42 ÷ 7 + 4
It can be written as
= 42/7 + 4
By further calculation
= 6 + 4
= 10
(ii) 12 + 18 ÷ 3
It can be written as
= 12 + 18/3
By further calculation
= 12 + 6
= 18
(iii) 19 – 20 ÷ 4
It can be written as
= 19 – 20/4
By further calculation
= 19 – 5
= 14
(iv) 16 – 5 × 3 + 4
It can be written as
= 16 – 15 + 4
By further calculation
= 20 – 15
= 5
(v) 6 – 8 – (-6) ÷ 2
It can be written as
= 6 – 8 – (-6/2)
By further calculation
= 6 – 8 – (-3)
So we get
= 6 – 8 + 3
= 9 – 8
= 1
(vi) 13 – 12 ÷ 4 × 2
It can be written as
= 13 – 12/4 × 2
By further calculation
= 13 – 3 × 2
= 13 – 6
= 7
(vii) 16 + 8 ÷ 4 – 2 × 3
It can be written as
= 16 + 8/4 – 2 × 3
By further calculation
= 16 + 2 – 2 × 3
So we get
= 16 + 2 – 6
= 18 – 6
= 12
(viii) 16 ÷ 8 + 4 – 2 × 3
It can be written as
= 16/8 + 4 – 2 × 3
By further calculation
= 2 + 4 – 6
So we get
= 6 – 6
= 0
(ix) 16 – 8 + 4 ÷ 2 × 3
It can be written as
= 16 – 8 + 4/2 × 3
By further calculation
= 16 – 8 + 2 × 3
So we get
= 16 – 8 + 6
Here
= 22 – 8
= 14
(x) (-4) + (-12) ÷ (-6)
It can be written as
= (-4) + (-12/-6)
By further calculation
= – 4 + 2
= -2
(xi) (-18) + 6 ÷ 3 + 5
It can be written as
= (-18) + 6/3 + 5
By further calculation
= (-18) + 2 + 5
So we get
= – 18 + 7
= – 11
(xii) (-20) × (-1) + 14 ÷ 7
It can be written as
= (-20) × (-1) + 14/7
By further calculation
= (-20) × (-1) + 2
So we get
= 20 + 2
= 22
Exercise 1C page: 15
Evaluate:
1. 18 – (20 – 15 ÷ 3)
Solution:
18 – (20 – 15 ÷ 3)
It can be written as
= 18 – (20 – 15/3)
By further calculation
= 18 – (20 – 5)
= 18 – 20 + 5
So we get
= 18 + 5 – 20
= 23 – 20
= 3
2. – 15 + 24 ÷ (15 – 13)
Solution:
– 15 + 24 ÷ (15 – 13)
It can be written as
= – 15 + 24 ÷ 2
So we get
= – 15 + 12
= – 3
3. 35 – {15 + 14 – (13 + )}
Solution:
35 – {15 + 14 – (13 +
)}
It can be written as
= 35 – [15 + 14 – (13 + 4)]
By further calculation
= 35 – [15 + 14 – 17]
Multiplying the negative sign
= 35 – 15 – 14 + 17
So we get
= 52 – 29
= 23
4. 27 – {13 + 4 – (8 + 4 – )}
Solution:
27 – {13 + 4 – (8 + 4 –
)}
It can be written as
= 27 – {13 + 4 – (8 + 4 – 4)}
By further calculation
= 27 – {13 + 4 – 8}
So we get
= 27 – {13 + (-4)}
= 27 – 9
= 18
5. 32 – [43 – {51 – (20 – )}]
Solution:
32 – [43 – {51 – (20 –
)}]
It can be written as
= 32 – [43 – {51 – (20 – 11)}]
By further calculation
= 32 – [43 – {51 – 9}]
So we get
= 32 – [43 – 42]
= 32 – 1
= 31
6. 46 – [26 – {14 – (15 – 4 ÷ 2 × 2)}]
Solution:
46 – [26 – {14 – (15 – 4 ÷ 2 × 2)}]
It can be written as
= 46 – [26 – {14 – (15 – 2 × 2)}]
By further calculation
= 46 – [26 – {14 – (15 – 4)}]
So we get
= 46 – [26 – {14 – 11}]
= 46 – [26 – 3]
Here
= 46 – 23
= 23
7. 45 – [38 – {60 ÷ 3 – (6 – 9 ÷ 3) ÷ 3}]
Solution:
45 – [38 – {60 ÷ 3 – (6 – 9 ÷ 3) ÷ 3}]
It can be written as
= 45 – [38 – {60 ÷ 3 – (6 – 3) ÷ 3}]
By further calculation
= 45 – [38 – {20 – 3 ÷ 3}]
So we get
= 45 – [38 – {20 – 1}]
By subtraction
= 45 – [38 – 19]
= 45 – 19
= 26
8. 17 – [17 – {17 – (17 – )}]
Solution:
17 – [17 – {17 – (17 –
)}]
It can be written as
= 17 – [17 – {17 – (17 – 0)}]
By further calculation
= 17 – [17 – {17 – 17}]
So we get
= 17 – [17 – 0]
= 17 – 17
= 0
9. 2550 – [510 – {270 – (90 – )}]
Solution:
2550 – [510 – {270 – (90 –
)}]
It can be written as
= 2550 – [510 – {270 – (90 – 87)}]
By further calculation
= 2550 – [510 – {270 – 3}]
So we get
= 2550 – [510 – 267]
= 2550 – 243
= 2307
10. 30 + [{-2 × (25 – )}]
Solution:
30 + [{-2 × (25 –
)}]
It can be written as
= 30 + [{-2 × (25 – 10)}]
By further calculation
= 30 + [{-2 × 15}]
So we get
= 30 + [-30]
= 30 – 30
= 0
11. 88 – {5 – (-48) ÷ (-16)}
Solution:
88 – {5 – (-48) ÷ (-16)}
It can be written as
= 88 – {5 – (- 48/ -16)}
By further calculation
= 88 – {5 – 3}
So we get
= 88 – 2
= 86
12. 9 × (8 – ) – 2 (2 + )
Solution:
9 × (8 –
) – 2 (2 +
)
It can be written as
= 9 × (8 – 5) – 2 (2 + 6)
By further calculation
= 9 × 3 – 2 × 8
So we get
= 27 – 16
= 11
13. 2 – [3 – {6 – (5 – )}]
Solution:
2 – [3 – {6 – (5 –
)}]
It can be written as
= 2 – [3 – {6 – (5 – 1)}]
By further calculation
= 2 – [3 – {6 – 4}]
So we get
= 2 – [3 – 2]
= 2 – 1
= 1
Exercise 1D page: 15
1. The sum of two integers is – 15. If one of them is 9, find the other.
Solution:
It is given that
Sum of two integers = – 15
One integer = 9
Other integer = – 15 – 9
Taking negative sign as common
= – (15 + 9)
= – 24
2. The difference between integers x and – 6 is – 5. Find the values of x.
x – (-6) = – 5 or – 6 – x = – 5
Solution:
It is given that
The difference between integers x and -6 is -5
x – (-6) = – 5 or – 6 – x = – 5
So the value of x is
x + 6 = – 5 or – x = – 5 + 6
By further calculation
x = – 5 – 6 or – x = 1
x = – 11 or x = – 1
3. The sum of two integers is 28. If one integer is – 45, find the other.
Solution:
It is given that
Sum of two integers = 28
One integer = – 45
Other integer = 28 – (-45)
By further calculation
= 28 + 45
= 73
4. The sum of two integers is – 56. If one integer is – 42, find the other.
Solution:
It is given that
Sum of two integers = – 56
One integer = – 42
Other integer = – 56 – (- 42)
By further calculation
= – 56 + 42
= – 14
5. The difference between an integer x and (-9) is 6. Find all possible values of x.
Solution:
It is given that
The difference between an integer x and (-9) is 6
x – (-9) = 6 or – 9 – x = 6
So the value of x is
x – (-9) = 6 or – 9 – x = 6
By further calculation
x + 9 = 6 or – x = 6 + 9
So we get
x = 6 – 9 or – x = 15
Here
x = – 3 or x = – 15
Therefore, possible values of x are – 3 and – 15.
6. Evaluate:
(i) (-1) × (-1) × (-1) × ……… 60 times.
(ii) (-1) × (-1) × (-1) × (-1) × …….. 75 times.
Solution:
(i) (-1) × (-1) × (-1) × ……… 60 times = 1 because (-1) is multiplied even number of times.
(ii) (-1) × (-1) × (-1) × (-1) × ……… 75 times = -1 because (-1) is multiplied odd number of times.
7. Evaluate:
(i) (-2) × (-3) × (-4) × (-5) × (-6)
(ii) (-3) × (-6) × (-9) × (-12)
(iii) (-11) × (-15) × (-11) × (-25)
(iv) 10 × (-12) + 5 × (-12)
Solution:
(i) (-2) × (-3) × (-4) × (-5) × (-6)
= 6 × 20 × (-6)
By further calculation
= 120 × (-6)
= – 720
(ii) (-3) × (-6) × (-9) × (-12)
= 18 × 108
By further calculation
= 1944
(iii) (-11) × (-15) + (-11) × (-25)
= 165 + 275
By further calculation
= 440
(iv) 10 × (-12) + 5 × (-12)
= – 120 – 60
By further calculation
= – 180
8. (i) If x × (-1) = – 36, is x positive or negative?
(ii) If x × (-1) = 36, is x positive or negative?
Solution:
(i) x × (-1) = – 36
So we get
– x = – 36
By further simplification
x = 36
Hence, it is a positive integer.
(ii) x × (-1) = 36
So we get
– x = 36
By further simplification
x = – 36
Hence, it is a negative integer.
9. Write all the integers between – 15 and 15, which are divisible by 2 and 3.
Solution:
Here the integers between – 15 and 15 are
– 12, – 6, 0, 6 and 12 which are divisible by 2 and 3.
10. Write all the integers between – 5 and 5, which are divisible by 2 or 3.
Solution:
Here the integers between – 5 and 5 are
– 4, – 3, – 2, 0, 2, 3 and 4 which are divisible by 2 or 3.
11. Evaluate:
(i) (-20) + (-8) ÷ (-2) × 3
(ii) (-5) – (-48) ÷ (-16) + (-2) × 6
(iii) 16 + 8 ÷ 4 – 2 × 3
(iv) 16 ÷ 8 × 4 – 2 × 3
(v) 27 – [5 + {28 – (29 – 7)}]
(vi) 48 – [18 – {16 – (5 – )}]
(vii) – 8 – {- 6 (9 – 11) + 18 ÷ -3}
(viii) (24 ÷ – 12) – (3 × 8 ÷ 4 + 1)
Solution:
(i) (-20) + (-8) ÷ (-2) × 3
It can be written as
= – 20 + 4 × 3
By further calculation
= – 20 + 12
= – 8
(ii) (-5) – (-48) ÷ (-16) + (-2) × 6
It can be written as
= (-5) – 3 + (-2) × 6
By further calculation
= – 5 – 3 – 12
So we get
= – 8 – 12
= – 20
(iii) 16 + 8 ÷ 4 – 2 × 3
It can be written as
= 16 + 2 – 2 × 3
By further calculation
= 16 + 2 – 6
So we get
= 18 – 6
= 12
(iv) 16 ÷ 8 × 4 – 2 × 3
It can be written as
= 2 × 4 – 2 × 3
By further calculation
= 8 – 6
= 2
(v) 27 – [5 + {28 – (29 – 7)}]
It can be written as
= 27 – [5 + {28 – 22}]
By further calculation
= 27 – [5 + 6]
So we get
= 27 – 11
= 16
(vi) 48 – [18 – {16 – (5 – 
)}]
It can be written as
= 48 – [18 – {16 – (5 – 5)}]
By further calculation
= 48 – [18 – {16 – 0}]
So we get
= 48 – [18 – 16]
= 48 – 2
= 46
(vii) – 8 – {- 6 (9 – 11) + 18 ÷ -3}
It can be written as
= – 8 – {- 6 (-2) – 6}
By further calculation
= – 8 – {12 – 6}
So we get
= – 8 – 6
= – 14
(viii) (24 ÷ 
– 12) – (3 × 8 ÷ 4 + 1)
It can be written as
= (24 ÷ 3 – 12) – (3 × 2 + 1)
By further calculation
= (8 – 12) – (6 + 1)
So we get
= – 4 – 7
= – 11
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:
Here the required number = sum of all integers from 20 to 30 – sum of all integers between 20 and 30
Substituting the values
= (20 + 21 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29 + 30) – (21 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29)
On further calculation
= 20 + 30 = 50
Hence, the required number is 50.
13. Add the product of (-13) and (-17) to the quotient of (-187) and 11.
Solution:
It is given that
(-13) × (-17) + (-187 ÷ 11)
By further calculation
= (-13) × (-17) + (-17)
So we get
= 221 – 17
= 204
14. The product of two integers is – 180. If one of them is 12, find the other.
Solution:
Product of two integers = – 180
One integer = 12
Other integer = – 180/ 12
By division we get
= – 15
15. (i) A number changes from – 20 to 30. What is the increase or decrease in the number?
(ii) A number changes from 40 to – 30. What is the increase or decrease in the number?
Solution:
(i) A number changes from – 20 to 30
It can be written as
– 20 – 30 = – 50
Hence, – 50 will be the increase in the number.
(ii) A number changes from 40 to – 30
It can be written as
40 – (-30) = 40 + 30 = 70
Hence, 70 will be the decrease in the number.
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:
(i) 117 by 9
= 117/9
It can be written as
= (13 × 9)/ 9
= 13
(ii) (-117) by 9
= – 117/ 9
It can be written as
= (-13 × 9)/ 9
= -13
(iii) 117 by (-9)
= 117/ -9
It can be written as
= (13 × 9)/ -9
= – 13
(iv) (-117) by (-9)
= – 117/-9 = 117/9
It can be written as
= (13 × 9)/ 9
= 13
(v) 225 by (-15)
= 225/ (-15)
It can be written as
= (15 × 15)/ -15
= – 15
(vi) (-552) ÷ 24
= – 552/ 24
It can be written as
= (- 23 × 24)/ 24
= – 23
(vii) (-798) by (-21)
= – 798/ – 21 = 798/21
It can be written as
= (38 × 21)/ 21
= 38
(viii) (-910) ÷ 26
= – 910/ 26
It can be written as
= (- 35 × 26)/ 26
= – 35
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:
(i) (-234) ÷ 13
= – 234/ 13
It can be written as
= (-18 × 13)/ 13
= – 18
(ii) 234 ÷ (-13)
= 234/ -13
It can be written as
= (18 × 13)/ – 13
= – 18
(iii) (-234) ÷ (-13)
= – 234/ – 13 = 234/13
It can be written as
= (18 × 13)/ 13
= 18
(iv) 374 ÷ (-17)
= 374/ -17
It can be written as
= (22 × 17)/ (-17)
= – 22
(v) (-374) ÷ 17
= – 374/ 17
It can be written as
= (-22 × 17)/ (17)
= – 22
(vi) (-374) ÷ (-17)
= – 374/ -17 = 374/17
It can be written as
= (22 × 17)/ (17)
= 22
(vii) (-728) ÷ 14
= – 728/14
It can be written as
= (-52 × 14)/ 14
= -52
(viii) 272 ÷ (-17)
= 272/ -17
It can be written as
= (16 × 17)/ (-17)
= -16
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:
(i) 299 ÷ 23
= 299/23
It can be written as
= (23 × 13)/ 23
= 13
(ii) 299 ÷ (-23)
= 299/ -23
It can be written as
= (23 × 13)/ – 23
= – 13
(iii) (-384) ÷ 16
= – 384/16
It can be written as
= (- 24 × 16)/ 16
= – 24
(iv) (-572) ÷ (-22)
= – 572/ – 22 = 572/22
It can be written as
= (26 × 22)/ 22
= 26
(v) 408 ÷ (-17)
= 408/ -17
It can be written as
= (24 × 17)/ (-17)
= – 24
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:
(i) 204 by 17
= 204/17
It can be written as
= (12 × 17)/ 17
= 12
(ii) 152 by – 19
= 152/ -19
It can be written as
= (8 × 19)/ -19
= – 8
(iii) 0 by 35
= 0/35
= 0
(iv) 0 by (-82)
= 0/ -82
= 0
(v) 5490 by 10
= 5490/10
It can be written as
= (549 × 10)/ 10
= 549
(vi) 762800 by 100
= 762800/100
It can be written as
= (7628 x 100)/ 100
= 7628
5. State, true or false:
(i) 0 ÷ 32 = 0
(ii) 0 ÷ (-9) = 0
(iii) (-37) ÷ 0 = 0
(iv) 0 ÷ 0 = 0
Solution:
(i) True.
(ii) True.
(iii) False. It is not defined.
(iv) False. It is not defined.
6. Evaluate:
(i) 42 ÷ 7 + 4
(ii) 12 + 18 ÷ 3
(iii) 19 – 20 ÷ 4
(iv) 16 – 5 × 3 + 4
(v) 6 – 8 – (-6) ÷ 2
(vi) 13 – 12 ÷ 4 × 2
(vii) 16 + 8 ÷ 4 – 2 × 3
(viii) 16 ÷ 8 + 4 – 2 × 3
(ix) 16 – 8 + 4 ÷ 2 × 3
(x) (-4) + (-12) ÷ (-6)
(xi) (-18) + 6 ÷ 3 + 5
(xii) (-20) × (-1) + 14 ÷ 7
Solution:
(i) 42 ÷ 7 + 4
It can be written as
= 42/7 + 4
By further calculation
= 6 + 4
= 10
(ii) 12 + 18 ÷ 3
It can be written as
= 12 + 18/3
By further calculation
= 12 + 6
= 18
(iii) 19 – 20 ÷ 4
It can be written as
= 19 – 20/4
By further calculation
= 19 – 5
= 14
(iv) 16 – 5 × 3 + 4
It can be written as
= 16 – 15 + 4
By further calculation
= 20 – 15
= 5
(v) 6 – 8 – (-6) ÷ 2
It can be written as
= 6 – 8 – (-6/2)
By further calculation
= 6 – 8 – (-3)
So we get
= 6 – 8 + 3
= 9 – 8
= 1
(vi) 13 – 12 ÷ 4 × 2
It can be written as
= 13 – 12/4 × 2
By further calculation
= 13 – 3 × 2
= 13 – 6
= 7
(vii) 16 + 8 ÷ 4 – 2 × 3
It can be written as
= 16 + 8/4 – 2 × 3
By further calculation
= 16 + 2 – 2 × 3
So we get
= 16 + 2 – 6
= 18 – 6
= 12
(viii) 16 ÷ 8 + 4 – 2 × 3
It can be written as
= 16/8 + 4 – 2 × 3
By further calculation
= 2 + 4 – 6
So we get
= 6 – 6
= 0
(ix) 16 – 8 + 4 ÷ 2 × 3
It can be written as
= 16 – 8 + 4/2 × 3
By further calculation
= 16 – 8 + 2 × 3
So we get
= 16 – 8 + 6
Here
= 22 – 8
= 14
(x) (-4) + (-12) ÷ (-6)
It can be written as
= (-4) + (-12/-6)
By further calculation
= – 4 + 2
= -2
(xi) (-18) + 6 ÷ 3 + 5
It can be written as
= (-18) + 6/3 + 5
By further calculation
= (-18) + 2 + 5
So we get
= – 18 + 7
= – 11
(xii) (-20) × (-1) + 14 ÷ 7
It can be written as
= (-20) × (-1) + 14/7
By further calculation
= (-20) × (-1) + 2
So we get
= 20 + 2
= 22
Exercise 1C
Evaluate:
1. 18 – (20 – 15 ÷ 3)
Solution:
18 – (20 – 15 ÷ 3)
It can be written as
= 18 – (20 – 15/3)
By further calculation
= 18 – (20 – 5)
= 18 – 20 + 5
So we get
= 18 + 5 – 20
= 23 – 20
= 3
2. – 15 + 24 ÷ (15 – 13)
Solution:
– 15 + 24 ÷ (15 – 13)
It can be written as
= – 15 + 24 ÷ 2
So we get
= – 15 + 12
= – 3
3. 35 – {15 + 14 – (13 + )}
Solution:
35 – {15 + 14 – (13 +)}
It can be written as
= 35 – [15 + 14 – (13 + 4)]
By further calculation
= 35 – [15 + 14 – 17]
Multiplying the negative sign
= 35 – 15 – 14 + 17
So we get
= 52 – 29
= 23
4. 27 – {13 + 4 – (8 + 4 – )}
Solution:
27 – {13 + 4 – (8 + 4 –)}
It can be written as
= 27 – {13 + 4 – (8 + 4 – 4)}
By further calculation
= 27 – {13 + 4 – 8}
So we get
= 27 – {13 + (-4)}
= 27 – 9
= 18
5. 32 – [43 – {51 – (20 – )}]
Solution:
32 – [43 – {51 – (20 –)}]
It can be written as
= 32 – [43 – {51 – (20 – 11)}]
By further calculation
= 32 – [43 – {51 – 9}]
So we get
= 32 – [43 – 42]
= 32 – 1
= 31
6. 46 – [26 – {14 – (15 – 4 ÷ 2 × 2)}]
Solution:
46 – [26 – {14 – (15 – 4 ÷ 2 × 2)}]
It can be written as
= 46 – [26 – {14 – (15 – 2 × 2)}]
By further calculation
= 46 – [26 – {14 – (15 – 4)}]
So we get
= 46 – [26 – {14 – 11}]
= 46 – [26 – 3]
Here
= 46 – 23
= 23
7. 45 – [38 – {60 ÷ 3 – (6 – 9 ÷ 3) ÷ 3}]
Solution:
45 – [38 – {60 ÷ 3 – (6 – 9 ÷ 3) ÷ 3}]
It can be written as
= 45 – [38 – {60 ÷ 3 – (6 – 3) ÷ 3}]
By further calculation
= 45 – [38 – {20 – 3 ÷ 3}]
So we get
= 45 – [38 – {20 – 1}]
By subtraction
= 45 – [38 – 19]
= 45 – 19
= 26
8. 17 – [17 – {17 – (17 – )}]
Solution:
17 – [17 – {17 – (17 –)}]
It can be written as
= 17 – [17 – {17 – (17 – 0)}]
By further calculation
= 17 – [17 – {17 – 17}]
So we get
= 17 – [17 – 0]
= 17 – 17
= 0
9. 2550 – [510 – {270 – (90 – )}]
Solution:
2550 – [510 – {270 – (90 –)}]
It can be written as
= 2550 – [510 – {270 – (90 – 87)}]
By further calculation
= 2550 – [510 – {270 – 3}]
So we get
= 2550 – [510 – 267]
= 2550 – 243
= 2307
10. 30 + [{-2 × (25 – )}]
Solution:
30 + [{-2 × (25 –)}]
It can be written as
= 30 + [{-2 × (25 – 10)}]
By further calculation
= 30 + [{-2 × 15}]
So we get
= 30 + [-30]
= 30 – 30
= 0
11. 88 – {5 – (-48) ÷ (-16)}
Solution:
88 – {5 – (-48) ÷ (-16)}
It can be written as
= 88 – {5 – (- 48/ -16)}
By further calculation
= 88 – {5 – 3}
So we get
= 88 – 2
= 86
12. 9 × (8 – ) – 2 (2 + )
Solution:
9 × (8 –) – 2 (2 +)
It can be written as
= 9 × (8 – 5) – 2 (2 + 6)
By further calculation
= 9 × 3 – 2 × 8
So we get
= 27 – 16
= 11
13. 2 – [3 – {6 – (5 – )}]
Solution:
2 – [3 – {6 – (5 –)}]
It can be written as
= 2 – [3 – {6 – (5 – 1)}]
By further calculation
= 2 – [3 – {6 – 4}]
So we get
= 2 – [3 – 2]
= 2 – 1
= 1
Exercise 1D
1. The sum of two integers is – 15. If one of them is 9, find the other.
Solution:
It is given that
Sum of two integers = – 15
One integer = 9
Other integer = – 15 – 9
Taking negative sign as common
= – (15 + 9)
= – 24
2. The difference between integers x and – 6 is – 5. Find the values of x.
x – (-6) = – 5 or – 6 – x = – 5
Solution:
It is given that
The difference between integers x and -6 is -5
⇒ x – (-6) = – 5 or – 6 – x = – 5
So the value of x is
x + 6 = – 5 or – x = – 5 + 6
By further calculation
x = – 5 – 6 or – x = 1
x = – 11 or x = – 1
3. The sum of two integers is 28. If one integer is – 45, find the other.
Solution:
It is given that
Sum of two integers = 28
One integer = – 45
Other integer = 28 – (-45)
By further calculation
= 28 + 45
= 73
4. The sum of two integers is – 56. If one integer is – 42, find the other.
Solution:
It is given that
Sum of two integers = – 56
One integer = – 42
Other integer = – 56 – (- 42)
By further calculation
= – 56 + 42
= – 14
5. The difference between an integer x and (-9) is 6. Find all possible values of x.
Solution:
It is given that
The difference between an integer x and (-9) is 6
x – (-9) = 6 or – 9 – x = 6
So the value of x is
x – (-9) = 6 or – 9 – x = 6
By further calculation
x + 9 = 6 or – x = 6 + 9
So we get
x = 6 – 9 or – x = 15
Here
x = – 3 or x = – 15
Therefore, possible values of x are – 3 and – 15.
6. Evaluate:
(i) (-1) × (-1) × (-1) × ……… 60 times.
(ii) (-1) × (-1) × (-1) × (-1) × …….. 75 times.
Solution:
(i) (-1) × (-1) × (-1) × ……. 60 times = 1 because (-1) is multiplied even number of times.
(ii) (-1) × (-1) × (-1) × (-1) × …….. 75 times = -1 because (-1) is multiplied odd number of times.
7. Evaluate:
(i) (-2) × (-3) × (-4) × (-5) × (-6)
(ii) (-3) × (-6) × (-9) × (-12)
(iii) (-11) × (-15) × (-11) × (-25)
(iv) 10 × (-12) + 5 × (-12)
Solution:
(i) (-2) × (-3) × (-4) × (-5) × (-6)
= 6 × 20 × (-6)
By further calculation
= 120 × (-6)
= – 720
(ii) (-3) × (-6) × (-9) × (-12)
= 18 × 108
By further calculation
= 1944
(iii) (-11) × (-15) + (-11) × (-25)
= 165 + 275
By further calculation
= 440
(iv) 10 × (-12) + 5 × (-12)
= – 120 – 60
By further calculation
= – 180
8. (i) If x × (-1) = – 36, is x positive or negative?
(ii) If x × (-1) = 36, is x positive or negative?
Solution:
(i) x × (-1) = – 36
So we get
– x = – 36
By further simplification
x = 36
Hence, it is a positive integer.
(ii) x × (-1) = 36
So we get
– x = 36
By further simplification
x = – 36
Hence, it is a negative integer.
9. Write all the integers between – 15 and 15, which are divisible by 2 and 3.
Solution:
Here the integers between – 15 and 15 are
– 12, – 6, 0, 6 and 12 which are divisible by 2 and 3.
10. Write all the integers between – 5 and 5, which are divisible by 2 or 3.
Solution:
Here the integers between – 5 and 5 are
– 4, – 3, – 2, 0, 2, 3 and 4 which are divisible by 2 or 3.
11. Evaluate:
(i) (-20) + (-8) ÷ (-2) × 3
(ii) (-5) – (-48) ÷ (-16) + (-2) × 6
(iii) 16 + 8 ÷ 4 – 2 × 3
(iv) 16 ÷ 8 × 4 – 2 × 3
(v) 27 – [5 + {28 – (29 – 7)}]
(vi) 48 – [18 – {16 – (5 – )}]
(vii) – 8 – {- 6 (9 – 11) + 18 ÷ -3}
(viii) (24 ÷ – 12) – (3 × 8 ÷ 4 + 1)
Solution:
(i) (-20) + (-8) ÷ (-2) × 3
It can be written as
= – 20 + 4 × 3
By further calculation
= – 20 + 12
= – 8
(ii) (-5) – (-48) ÷ (-16) + (-2) × 6
It can be written as
= (-5) – 3 + (-2) × 6
By further calculation
= – 5 – 3 – 12
So we get
= – 8 – 12
= – 20
(iii) 16 + 8 ÷ 4 – 2 × 3
It can be written as
= 16 + 2 – 2 × 3
By further calculation
= 16 + 2 – 6
So we get
= 18 – 6
= 12
(iv) 16 ÷ 8 × 4 – 2 × 3
It can be written as
= 2 × 4 – 2 × 3
By further calculation
= 8 – 6
= 2
(v) 27 – [5 + {28 – (29 – 7)}]
It can be written as
= 27 – [5 + {28 – 22}]
By further calculation
= 27 – [5 + 6]
So we get
= 27 – 11
= 16
(vi) 48 – [18 – {16 – (5 – )}]
It can be written as
= 48 – [18 – {16 – (5 – 5)}]
By further calculation
= 48 – [18 – {16 – 0}]
So we get
= 48 – [18 – 16]
= 48 – 2
= 46
(vii) – 8 – {- 6 (9 – 11) + 18 ÷ -3}
It can be written as
= – 8 – {- 6 (-2) – 6}
By further calculation
= – 8 – {12 – 6}
So we get
= – 8 – 6
= – 14
(viii) (24 ÷ – 12) – (3 × 8 ÷ 4 + 1)
It can be written as
= (24 ÷ 3 – 12) – (3 × 2 + 1)
By further calculation
= (8 – 12) – (6 + 1)
So we get
= – 4 – 7
= – 11
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:
Here the required number = sum of all integers from 20 to 30 – sum of all integers between 20 and 30
Substituting the values
= (20 + 21 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29 + 30) – (21 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29)
On further calculation
= 20 + 30 = 50
Hence, the required number is 50.
13. Add the product of (-13) and (-17) to the quotient of (-187) and 11.
Solution:
It is given that
(-13) × (-17) + (-187 ÷ 11)
By further calculation
= (-13) × (-17) + (-17)
So we get
= 221 – 17
= 204
14. The product of two integers is – 180. If one of them is 12, find the other.
Solution:
Product of two integers = – 180
One integer = 12
Other integer = – 180/ 12
By division we get
= – 15
15. (i) A number changes from – 20 to 30. What is the increase or decrease in the number?
(ii) A number changes from 40 to – 30. What is the increase or decrease in the number?
Solution:
(i) A number changes from – 20 to 30
It can be written as
– 20 – 30 = – 50
Hence, – 50 will be the increase in the number.
(ii) A number changes from 40 to – 30
It can be written as
40 – (-30) = 40 + 30 = 70
Hence, 70 will be the decrease in the number.
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Selina Class 7 ICSE Solutions Mathematics : Chapter 1- Integers
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Chapterwise Selina Publishers ICSE Solutions for Class 7 Mathematics :
- Chapter 1- Integers
- Chapter 2- Rational Numbers
- Chapter 3- Fraction (Including Problems)
- Chapter 4- Decimal Fractions (Decimals)
- Chapter 5- Exponents (Including Laws of Exponents)
- Chapter 6- Ratio and Proportion (Including Sharing in a Ratio)
- Chapter 7- Unitary Method (Including Time and Work)
- Chapter 8- Percent and Percentage
- Chapter 9- Profit, Loss and Discount
- Chapter 10- Simple Interest
- Chapter 11- Fundamental Concepts (Including Fundamental Operations)
- Chapter 12- Simple Linear Equations (Including Word Problems)
- Chapter 13- Set Concepts
- Chapter 14- Lines and Angles (Including Construction of Angles)
- Chapter 15- Triangles
- Chapter 16- Pythagoras Theorem
- Chapter 17- Symmetry (Including Reflection and Rotation)
- Chapter 18- Recognition of Solids (Representing 3-D in 2-D)
- Chapter 19- Congruency: Congruent Triangles
- Chapter 20- Mensuration (Perimeter and Area of Plane Figures)
- Chapter 21- Data Handling
- Chapter 22- Probability
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