Selina Solutions Concise Maths Class 7 Chapter 1 Integers Exercise 1A covers important concepts like introduction, multiplication of integers, properties of multiplication and examples. Students can obtain a hold on these concepts by practising the textbook problems on a daily basis. Practicing problems one day before the exam will reduce their confidence in answering concept based questions. The solutions are prepared for the exercise wise problems to help students boost their exam preparation. For further details about the exercise problems, students can access Selina Solutions Concise Maths Class 7 Chapter 1 Integers Exercise 1A PDF, from the links which are provided here.
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Access Selina Solutions Concise Maths Class 7 Chapter 1: Integers Exercise 1A
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
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