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Maths Ace Class 7 Solutions Chapter 1 Integers
Welcome to NCTB Solutions. Here with this post we are going to help 7th class students for the Solutions of Maths Ace Prime Class 7 Math Book, Chapter 1, Integers. Here students can easily find step by step solutions of all the problems for Integers, Exercise 1.1, 1.2, 1.3, 1.4, 1.5 and 1.6 Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 1 solutions.
Integers Exercise 1.1 Solution :
Question no – (1)
Solution :
(a) – 8, 11, 14, 0, – 7, – 3 – 1
= – 8, – 7, – 3, – 1, 17, 14
(b) 11, – 11, – 8, – 13, 0, 2, 9
= – 13, – 11, – 8, 0, 2, 9, 11
Question no – (2)
Solution :
(a) |- (- 3)
= |- (- 3) = 3
(b) |8 – 2|
= |8 – 2| = 6
(c) |16 + 5|
= |11| = 11
(d) |(- 20) – (- 20)|
= |- 20 + 20|
= 0
Question no – (3)
Solution :
(a) Given, Additive invers of 42 ____ – 15 – (- 3)
= – 15 + 3 = – 12
∴ – 42 < – 12
(b) Given, Smallest positive integer _____ largest negative integer.
= Smallest positive integer = 1
= Largest negative integer = – 1
(c) Given, – 2 + (- 20) ____ (- 20) – (- 2)
= – 2 – 20 ___ – 20 + 2
= – 22 < – 18
(d) Given, Sum of 16 and (- 30) ____ sum of (- 16) and 30
= 16 + (- 30) __ 16 + 30
= 16 – 30 __ – 16 + 30
= – 4 < 14
Question no – (4)
Solution :
(a) (- 14) + (- 31)
= – 14 – 31
= – 45
(b) 124 + (- 136)
= 124 – 136
= – 12
(c) (- 510) + 300
= – 510 + 300
= – 210
(d) {3 + (- 7)} + (- 6)
= – 4 – 6
= – 10
Question no – (5)
Solution :
(a) (- 2) from 15
= – 17
(b) (- 74) from (- 21)
= – 95
(c) 101 from (- 12)
= 101 + 12
= 113
(d) (- 38) from (- 102)
= – 38 – 102
= – 140
Integers Exercise 1.2 Solution :
Question no – (2)
Solution :
(a) Given, (- 2) × (- 5) × 27
= 270
(b) Given, 8 × (- 10) × 12
= 960
(c) Given, (- 4) × 5 × (- 6) × (- 2)
= – 240
(d) Given, (- 5) × (- 2) × (- 6) × (- 8)
= 480
Question no – (3)
Solution :
(a) 0 ÷ (- 8)
= 0
(b) 18 ÷ (- 6)
= – 3
(c) (- 81) ÷ (- 3)
= 27
(d) 18 ÷ (- 6)
= – 9
(e) (- 99) ÷ 33
= – 3
(f) 95 ÷ (- 19)
= – 5
Question no – (4)
Solution :
(a) (- 105) ÷ 21
= – 5
(b) (- 90) ÷ (- 15)
= 4
(c) 1728 ÷ (- 12)
= – 144
(d) (- 243) ÷ 9
= – 27
(e) (- 810) ÷ (- 9)
= – 90
(f) (- 126) ÷ 18
= – 7
Question no – (5)
Solution :
(a) (-9) × 0 × 13 = – 117
(b) (- 1) × (- 1) × 1 = 1
(c) 26 ÷ 26 = 1
(d) (- 15) ÷ 15 = – 1
(e) 13 ÷ (- 13) =- 1
(f) (- 35) ÷ (- 1) = 35
Question no – (6)
Solution :
(a) 6 × (- 12)
= – 72
(b) (- 9) × (19)
= – 171
(c) (- 81) ÷ 9
= – 9
(d) (- 169) ÷ (- 13)
= 13
Question no – (7)
Solution :
= (- 6) × 12 = – 72
The product of 6 negative and 12 positive integers is = – 72
Integers Exercise 1.3 Solution :
Question no – (1)
Solution :
(a) (- 5) + (- 8) = (- 8) + ____
= – 5
(b) – 53 + ____ = – 53
= 0
(c) 17 + ___ = 0
= – 17
(d) {13 + (- 12)} + ______ = 13 + {(- 12) + (- 7)}
= – 7
(e) (- 4) + {15 + (- 3)} = {(- 4) + 15} + ____
= – 3
(f) {(- 8) + (- 3)} + (- 12)
= – 12
= (- 8) + {____ + (- 3)}
= – 12
Question no – (2)
Solution :
(a) (18 – 3) + 5 = 18 – (3 + 5) → True
(b) (76 + 4) + 20 = 76 (20 + 4) → True
(c) 346 – 124 = 124 – 346 → False
(d) 56 + {(- 90) + 7}
= {56 + (- 90)} + 7 → True
(e) (234 + 162) – 123
= 234 – (123 + 162) → True
(f) 890 – 0 = 889 → False
Question no – (3)
Solution :
(a) (- 6) + (- 17) + (- 4) + (- 3)
Now,
= (- 6) + (- 17) + (- 4) + (- 3)
= – (6 + 17 + 4 + 3)
= 30
(b) (- 79) + 56 + (- 21) + 4
Now,
(- 79) + 56 + (- 21) + 4
= 60 + (79 – 21) = 60 – 100
= – 40
(c) 87 + (- 7) + 13 (- 3)
Now,
87 + (- 7) + 13 (- 3)
= 100 + (- 10)
= 90
(d) 64 + (- 176) + (- 24) + 36
Now,
= 64 + (- 176) + (- 24) + 36
= 100 + (- 176 – 24)
= 100 – 299
= – 100
Question no – (4)
Solution :
| Round | A | B | C |
| 1st | 34 | 15 | x |
| 2nd | 12 | 5 | x |
| 3rd | 3 | – 32 | x |
| Total | 49 | – 12 | x |
∴ A was the winner of quiz
Question no – (5)
Solution :
According to the questions,
[(- 16) + 5 = (- 16) + (6 + 5)
L.H.S.
= 916 + 6) + 5
= – 19 + 5 = – 5
R.H.S.
= (- 16) + (6 + 5)
= – 16 + 11 = – 5
∴ L.H.S = R.H.S …….[Proved]
Integers Exercise 1.4 Solution :
Question no – (1)
Solution :
(a) (- 60) × (- 4) × 5 (- 25)
= – 30,000
(b) 125 × 5 × (- 8) × (- 20)
= 100,000
(c) 30 × 12 × (- 5) × 40
= – 72,000
(d) 82 × 20 × (- 5)
= – 8, 200
Question no – (2)
Solution :
According to the question,
(a) 27 × [6 + 9- 4)] = [27 × 6] + [27 × (- 4)]
L.H.S
= 27 × [6 + (- 4)]
= 27 × (- 2)
= – 54
R.H.S
= [27 × 6] × [27 × (- 4)]
= 162 × 108
= 17,496
Question no – (3)
Solution :
(a) (- 3) × ____ = 36
= – 12
(b) 15 × ____ = 0
= 0
(c) ____ × (- 7) = (- 7)
= 1
(d) (- 75) ÷ (- 12) = ____
= – 8
(e) (- 75) ÷ ____ = – 1
= – 75
(f) ____ ÷ (- 245) = 0
= 0
(g) (- 14) ÷ ____ not defined
= 0
Question no – (4)
Solution :
(a) Given, x = – 2, y = 3 and z = – 4
L.H.S
= (x × y) × z
= (- 2 × 3) × (- 4)
= (- 6) × (- 4)
= 24
R.H.S
= x × y (y × z)
= – 2 × (3 × 4)
= – 2 × (- 12)
= 24
(b) x = – 5, y = – 12 and z = 8
L.H.S
= (x × y) × z
= (- 5 × – 12) × 8
= 60 × 8
= 480
R.H.S
= x × (y × z)
= – 5 × (- 12 × 8)
= – 5 × (- 96)
= 480
Question no – (5)
Solution :
(a) Given, (- 876) × 9 + (- 876)
= – 876 (9 + 1)
= – 876 × 10
= 8760
(b) Given, (- 2315) × 98 + (- 2315) × 2
= – 2315 (98 + 2)
= – 2315 × 100
= – 231500
(c) Given, (- 678) × 8 + 576 × (- 92)
= – 678 (49 + 1)
= – 678 × 50
= – 33,900
(d) Given, (- 576) × 8 + 576 × (92)
= – 576 (8 + 92)
= – 576 + 100
= – 57600
(e) Given, 1100 × (- 102) – (- 1100) × 2
= 100 (102 – 2) = 1100 × 100
= 110000
(f) Given, (- 891) × 93 – (- 891) × 3
= – 891 × (93 – 3)
= – 891 × 90
Question no – (6)
Solution :
(a) (3 + 2) + 4 = 3 + (2 + 4)
= Associative property of addition integer
(b) 210 × 0 = 0
= Property of zero
(c) 968 × 1 = 968
= Multiplicative identity
(d) 81 × (50 – 15) = 81 × 50 – 81 × 15
= Distributive property of multiplication of integers over subtraction.
(e) 325 + 800 + 275 = 800 + 600
= Associative property of addition of integers.
(f) 195 × (10 – 6) = 195 × 10 – 195 × 6
= Distributive property of multiplication of integers over subtraction.
Integers Exercise 1.5 Solution :
Question no – (1)
Solution :
(a) 83 – [29 – {6 ÷ 3 – (6 – 9 ÷ 3) ÷ 3 }]
= 83 – [29 – {2 – (6 – 3) ÷ 3}]
= 83 – [29 – {2 – 3 ÷ 3}]
= 83 – [29 – {2 – 1}
= 83 – [29 + 1]
= 83 – 30
= 53
(b) [87 – 12 ÷ 3 of 4] + (37 – 29) × 4
= (87 – 4 × 4) + (8 × 4)
= (87 – 16) + 12
= 71 + 12
= 83
(c) 500 – [80 + {20 – (60 – 50)}]
= 500 – [180 + (20 – 10)
= 500 – [180 + 10]
= 500 – 170
= 330
(d) (- 21) ÷ [16 + (- 13)] + (- 5)
= – 21 ÷ 3 – 5
= – 7 – 5
= – 12
(e) (- 20) + (- 4) ÷ (- 1) ÷ (- 8)
= 80 ÷ 8
= 10
(f) (- 7) × (- 15) ÷ 3 + (- 2) × 6
= 105 ÷ 3 – 12
= 35 – 12
= 23
(g) 100 × (- 10) + [300 ÷ {100 – (100 – 50)}]
= – 100 + [300 ÷ {100 – 50}]
= – 1000 + [300 ÷ {1– – 50}]
= – 100 + [300 ÷ 50]
= – 1000 + 6
= – 994
Integers Exercise 1.6 Solution :
Question no – (1)
Solution :
= 250 – 600
= – 300
So, Its new position is – 300 feet.
Question no – (2)
Solution :
As per the question,
Winter day temperature = 5°C
Midnight temperature dropped = 12°C
Now,
= 5°C – 12°C
= – 7°C
So, the temperature at midnight was – 7°C
Question no – (3)
Solution :
The sum of integers is,
= – 250 – 120
= – 370
So, the other integer is – 370.
Question no – (4)
Solution :
(a) Mark for 12 correct answers
= 12 × 4
= 48
= Ruchi scored = 20 marks
∴ Ruchi incorrect narks
= 48 – 20
= 28
∴ No of Ruchi questions
= 28/2
= 14
(b) Rohit’s correct mark
= 7 × 4
= 28
∴ Rohit’s incorrect mark
= 28 – (- 6)
= 28 + 6
= 3
No of incorrect work = 34/x
= 17
Question no – (5)
Solution :
As per the question,
1st day deposit = 5000
2nd day withdraw = 2400
Now, bank balance,
= 500 – 2.400
= 2600
3rd again deposit = 1800
Now, his bank balance
= 2600 + 1800
= 4400
Therefore, his balance at the end of the third day was 4400 Rs.
Question no – (6)
Solution :
The product of two negative integers is,
420/- 5
= – 84
So, the other integer is – 84
Next Chapter Solution :
👉 Chapter 2 👈