These Class 7 Maths Extra Questions Chapter 1 Integers will help students prepare well for the exams.
Class 7 Maths Chapter 1 Extra Questions Integers
Class 7 Maths Integers Extra Questions – 2 Marks
Question 1.
Given that a, b are integers such that a – (–b) = a + b, then verify the following.
i) a = –2, b = 3
ii) a = 7, b = –4
iii) a = 1, b = 6
iv) a = 10, b = 5
Solution:
Given a – (–b) = a + b
i) a = –2, b = 3
a – (–b) = –2 – (–3) = –2 + 3 = 1
a + b = –2 + 3 = 1
∴ True
ii) a = 7, b = –4
a – (–b) = 7 – (– (–4)) = 7 – 4 = 3
a + b = 7 – 4 = B
∴ True
iii) a = 1, b = 6
a – (–b) = 1 – (–6) = 1 + 6 = 7
a + b = 1 + 6 = 7
∴ True
iv) a = 10, b = 5
a – (–b) = 10 – (–5) = 10 + 5 = 15
a + b = 10 + 5 = 15
∴ True
Question 2.
Evaluate the following
i) – 1 + 1 – 1 + 1 ……………. 9 times
Solution:
– 1 + 1 – 1 + 1 + …………… 9 times 9 is odd
∴ – 1 + 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1
= 0 + 0 + 0 + 0 – 1 = – 1
ii) 1 – 1 + 1 – 1 + ………. 13 times
Solution:
1 – 1 + 1 – 1 + …………… 13 times 13 is odd
1 – 1 + 1 – 1 ……….. = 1
Question 3.
Evaluate the following
i)– 1 + 1 – 1 + 1 – 1 + 1 20 times
Solution:
– 1 + 1 – 1 + 1 – 1 + 1 ………. 20 times
20 is even
– 1 + 1 – 1 + 1 = 0
ii) 1 – 1 + 1 – 1 + 1 – 1 ………….. 16 times
Solution:
1 – 1 + 1 – 1 + 1 – 1 …….16 times 16 is even
1 – 1 + 1 – 1 ……….. 16 times = 0
Question 4.
Evaluate the following
– 115 + 11 – 32 – 74
Solution:
–115 + 11 – 32 – 74 = – 104 – 32 – 74
= – 104 – 106 = – 210
Question 5.
Determine the integer whose product with ‘–1 ‘ is
i) 48
ii) –210
Solution:
Question 6.
State commutative property in integers under multiplication. Write an example also.
Solution:
Commutative property : If a, b are any two integers then a × b = b × a.
Example :
a = 7, b = –4
a × b = 7 × –4 = – 28
b × a = –4 × 7 = – 28
∴ a × b = b × a
Question 7.
Identify true statement from the following :
i) –13 – 7 + 20 = 1
ii) 8 – (–8) – 8 = 0
Solution:
i) – 13 – 7 + 20 = – 20 + 20 = 0 (False)
ii) 8 – (–8) – 8 = 8 + 8 – 8 = 8 ≠ 0 (False)
Question 8.
Simplify 20 – 4 ÷ 2 × 1
Solution:
20 – 4 ÷ 2 × 1 = 20 – 2 × 1 = 20 – 2 = 18
Question 9.
Divide :
i) –49 by 7
ii) 100 by –100
Solution:
i) –49 ÷ 7 = \(\frac{–49}{7}\) = –7
ii) 100 ÷ –100 = \(\frac{100}{–100}\) = –1
Question 10.
Simplify 8 – (5 – 6 ÷ 2)
Solution:
8 –(5 – 6 ÷ 2) = 8 – (5 – 3) = 8 – 2 = 6
Question 11.
What must be subtracted from -100 to get 897?
Solution:
Let the number to be subtracted = x
–100 – x = 897
–100 – 897 = x
–997 = x
∴ –997 is to be subtracted.
Question 12.
Find the value of – 28 × 7 + 16 × 10.
Solution:
–28 × 7 + 16 × 10 = – 196 + 160 = – 36.
Integers Extra Questions Class 7 – 3 Marks
Question 1.
State associative property in integer under addition and also prove with an example.
Solution:
If a, b & c are any three integers, then
a + (b + c) = (a+b) + c
Let a = 3, b = –2, c = 4
LHS a + (b + c) = 3 + (– 2 + 4) = 3 + 2 = 5
RHS(a+b) + c = (3 – 2) + 4 = 1 + 4 = 5
LHS = RHS
Question 2.
Verify a × (b – c) = a × b – a × c, if a = 4, b = 3, c = –1.
Solution:
a × (b – c) = a × b – a × c
a = 4, b = 3, c = –1
LHS a × (b – c) = 4 × (3 –(– 1))
= 4 × (3+1) = 4 × 4 = 16
RHS a × b – a × c = 4 × 3 – 4 × (-1)
= 12 + 4 = 16
LHS = RHS
Question 3.
Find the following products
(–1) × (–3) × (–10) × (–7)
Solution:
(–1) × (–3) × (–10) × (–7)
=3 × 70 = 210
Question 4.
A certain freezing process requires that room temperature be lowered from 30°C at the rate of 3°C every hour. What will be the room temperature 4 hours after the process begins?
Solution:
Starting Temperature = 30°C
Decrease of temperature per hour = 3°C
Temperature after 1 hour
= 30°C – 3°C = 27°C
Temperature after 2 hours
= 27°C – 3°C = 24°C
Temperature after 3 hours
= 24°C – 3°C = 21°C
Question 5.
Verify 9 × [6 + (–1)] = 9 × 6 + 9 × (–1)
Solution:
LHS 9 × [6+(–1)] = 9 × 5 = 45
RHS 9 × 6 + 9(–1) = 54 – 9 = 45
LHS = RHS
Question 6.
Complete the following multiplication table.
| × | 2 | 3 | 4 |
| 4 | |||
| 3 | |||
| –2 |
Solution:
| × | 2 | 3 | 4 |
| 4 | 8 | 12 | 16 |
| 3 | 6 | 9 | 12 |
| –2 | –4 | –6 | –8 |
Question 7.
Complete the following addition table.
| + | –3 | 1 | 7 |
| 5 | |||
| –2 | |||
| 0 |
Solution:
| + | –3 | 1 | 7 |
| 5 | 2 | 6 | 12 |
| –2 | –5 | –1 | 5 |
| 0 | –3 | 1 | 7 |
Question 8.
Which is greater
(13+12) × 14 or 13 + 12 × 14?
Solution:
(13+12) × 14 = 25 × 14 = 350
13 + 12 × 14 = 13 + 168 = 181
(13+12) × 14 is greater.
Question 9.
Find the product by convenient grouping (–16) × 20 × (–4) × (–2).
Solution:
(–16) × 20 × (–4) × (–2)
= (–16) × (20 × –2) × (–4)
= (– 16 × – 40) × –4
= 16 × (40 × –4) = –16 × 160 = –2560
Question 10.
Prepare 3 pairs of integers whose sum is ‘ –9 ‘.
Solution:
Sum is –9, Given a + b = –9
i) –10 + 1 = –9
(a, b) = (–10, 1)
ii) – 1 – 8 = –9
(a, b) = (–1, –8)
iii) – 18 + 9 = – 9
(a,b) = (–18, 9)
Question 11.
Identify the letters given in the number line.
Solution:
I = 6 ; N = –6 ; D = –10
Question 12.
What is the
i) additive identity in integers?
ii) multiplicative identity in integers?
iii) additive inverse of 2?
Solution:
i) 0
ii) 1
iii) 2 – 2 = 0
–2 is the additive inverse of 2.
Question 13.
Evaluate the following
i) 0 ÷ 2011
ii) –191 ÷ 191
ii) 216 ÷ –6
Solution:
i) 0 ÷ 2011 = \(\frac{0}{2011}\) = 0
ii) -191 ÷ –191 = \(\frac{–191}{–191}\) = 1
iii) \(\frac{216}{–6}\) = \(\frac{6 \times 36}{–6}\) = –36
Extra Questions of Integers Class 7 – 5 Marks
Question 1.
Complete the following addition table.
| + | 0 | –1 | 2 | –3 | –4 |
| 2 | |||||
| 7 | |||||
| 9 | |||||
| 6 | |||||
| –3 |
Solution:
| + | 0 | –1 | 2 | –3 | –4 |
| 2 | 2 | 1 | 4 | –1 | –2 |
| 7 | 7 | 6 | 9 | 4 | 3 |
| 9 | 9 | 8 | 11 | 6 | 5 |
| 6 | 6 | 5 | 8 | 3 | 2 |
| –3 | –3 | –4 | –1 | –6 | –7 |
Question 2.
Complete the following multiplication table
| × | 0 | –1 | –2 | –3 | –4 |
| 7 | |||||
| 2 | |||||
| 3 | |||||
| 1 | |||||
| 0 |
Solution:
| × | 0 | –1 | –2– | 3 | –4 |
| 7 | 0 | –7 | –14 | –21 | –28 |
| 2 | 0 | –2 | –4 | –6 | –8 |
| 3 | 0 | –3 | –6 | –9 | –12 |
| 1 | 0 | –1 | –2 | –3 | –4 |
| 0 | 0 | 0 | 0 | 0 | 0 |
Question 3.
Write 5 pairs negative integers whose difference gives 9.
Solution:
Pair 1 – 10 – (–19) = –10 + 19 = 9
(– 10, – 19)
Pair 2 11 – 2 = 9
(11,2)
Pair 3 – 1 – (–10) = – 1 + 10 = 9
(– 1, – 10)
Pair 4 89 – 80 = 9
(89, 80)
Pair 5 29 – 20 = 9
(29, 20)
Question 4.
Fill in the blanks to make the following statements true.
i) (–9) + (–11) = (–11) + ____
ii) –35 + ___ = – 35
iii) 27 + ____ = 0
iv) 13 + ( ) = 11
v) 115 – ( ) = ___
Solution:
i) (–9) + (–11) = (–11) + –9
ii) –35 + 0 = – 35
iii) 27 + (–27) = 0
iv) 13 + (–2) = 11
v) 115 – (115 ) = 0
Question 5.
Identify the property involved in the following
i) 3 × – 4 = – 4 × 3 = 12
ii) 19 + 0 = 0 + 19 = 19
ii) 131 × 1 = 131 × 1 = 131
iv) 1 + 2 = 3
v) (13 × 4) × 2 = 13 × (4 × 2)
Solution:
i) Commutative property under multiplication
ii) Additive identity
iii) Multiplicative identity
iv) Closure property
v) Associative property under multiplication
Question 6.
Name the property is the following statements.
i) 13 + (–13) = 0
ii) 3 + (2 + 1) = (3 + 2) + 1
iii) a × 1 = 1 × a = a
iv) a + 0 = 0 + a = a
v) 17 + (–17) = 0
Solution:
i) Inverse property under addition
ii) Associative property under addition
iii) Multiplicative identity property
iv) Identity property under addition
v) Inverse property under addition.
Question 7.
A shopkeeper earns a profit of ₹ 2 by selling one pen and incurs a loss of ₹ 1 per pencil while selling pencils of his old stock.
i) In a particular month he incurs a loss of ₹ 6. In this period, he sold 42 pens. How many pencils did he sell in this period?
Solution:
Profit earned on one pen = ₹1
Profit earned on selling 42 pens = ₹42
Total loss = ₹6, denoted as -₹6
Profit + Loss = Total loss
Loss incurred = Total loss – profit =₹(– 6 – 42) = –₹48
Number of pencils sold = \(\frac{-48}{-1}\) = –48 ÷ (–1) = 48
ii) In the next month he earns neither profit nor loss. If he sold 60 pens. How many pencils did he sell?
Solution:
Let the next month there is no profit nor loss
Profit earned + Loss incurred = 0
Profit earned = – Loss incurred
Profit earned by selling 60 pens = ₹60
∴ Loss incurred = – ₹60
Total number of pencils sold = – ₹60 ÷ (–1) = 60
Question 8.
In a Mathematics special periodic test (+ 4) marks are given for every correct answer (–1) mark are given for every incorrect answer.
i) Neeraj answered all the questions and scored 28 marks though he got 9 correct answers.
Solution:
Marks given for each correct answer = 4
Marks given for 9 correct answers = 9 × 4 = 36
Neeraj score = 28
Marks for incorrect answer = 28 – 36 = –8
Marks for each incorrect answer = – 1
Number of incorrect answers = (–8) ÷ (–1) = 8
ii) Zubeda also answered all the questions and scored (–10) marks though she got 6 correct answers. How many incorrect answers had she attempted?
Solution:
Marks for 6 correct answer = 6 × 4 = 24
Zubeda score = -10
Marks for incorrect answer = –10 – 24 = –34
Marks for each incorrect answer = –1
Number of incorrect answers = (–34) ÷ (–1) = 34
Question 9.
The temperature at 1 : 00 pm was 11°C above zero. If it decreases at the rate of 2°C per hour till 7 pm, find the temperature for each hour.
Solution:
Temperature at 1 : 00 pm = 11°C
Decrease of temperature for every hour =2°C
Temperature at 2 pm =11°C – 2°C = 9°C
Temperature at 3 pm =9°C – 2°C = 7°C
Temperature at 4 pm =7°C – 2°C = 5°C
Temperature at 5 pm =5°C – 2°C = 3°C
Temperature at 6 pm =3°C – 2°C = 1°C
Temperature at 7 pm =1°C – 2°C = – 1°C