AP Board 7th Class Maths Solutions Chapter 1 Integers Ex 1.3

SCERT AP 7th Class Maths Solutions Pdf Chapter 1 Integers Ex 1.3 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 1st Lesson Integers Exercise 1.3

Question 1.
Identify the laws in the following statements :

(i) – 3 + 5 = 5 + (- 3)
Answer:
a + b = b + a (Additive commutative property)
– 3 + 5 = 5 – 3
2 = 2

AP Board 7th Class Maths Solutions Chapter 1 Integers Ex 1.3

(ii) – 2 × 1 = 1 × (-2) = – 2
Answer:
a × 1 = 1 × a = a
Multiplicative identity property
– 2 × 1 = 1 × (-2) = – 2
– 2 = – 2

(iii) [(-5) × 2)] × 3 = (-5) × [(2 × 3)]
Answer:
(a × b) × c = a × (b × c)
Multiplicative associative property.
(- 10) × 3 = (- 5) × (6)
– (10 × 3) = – (5 × 6)
– 30 = – 30

(iv) 18 × [7 +(- 3)] = [18 × 7] + [18 × (-3)]
Answer:
a × (b + c) = (a × b) + (a × c)
Distributive over addition property
18 × (7 – 3) = (126) + (- 54)
18 × 4= 126 – 54
72 = 72

(v) – 5 × 6 = – 30 .
Answer:
(- a) × b = – (ab)
Multiplicative closure property
– 30 = – 30

(vi) – 3 + 0 = 0 + (- 3) = – 3
Answer:
a + 0 = 0 + a = a
Additive identity property
-3 = 0 – 3 = -3
-3 = – 3 = – 3.

AP Board 7th Class Maths Solutions Chapter 1 Integers Ex 1.3

Question 2.
What will be the sign of the product of the following
(i) 24 times of negative integer?
(ii) 35 times of negative integer?
Answer:
(- 1) × (- 1) = + 1
(- 1) × (- 1) × (- 1) = – 1
(- 1) × (- 1) × (- 1) × (- 1) = + 1
(- 1) × (- 1) × (- 1) × (- 1) × (- 1) = – 1
If negative integer multiplied even number of times, the product is positive integer. If negative integer multiplied odd number of times, the product is negative integer.
(i) (-a) × (-a) × ………….. 24 times
= + a (Positive integer)

(ii) (-a) × (-a) × ……………. 35 times
= – a (Negative integer)

Question 3.
Write the suitable numbers in the blanks by using appropriate law.
(i) – 3 + ________= – 3
Answer:
a + 0 = a
– 3 + 0 = – 3

(ii) 2 × (-3) = (-3) × ________
Answer:
a × b = b × a
2 × (-3) = (-3) × 2

(iii) – 6 + [3 + (-2)] = [(- 6) + ________] + ________
Answer:
a + (b + c) = (a + b) + c
= [(-6) + 3] + (- 2)

(iv) – 6 × ________ = – 6
Answer:
a × 1 = a = 1 × a
-6 × ________ = – 6

(v) 5 × [(- 6) + 9] = ________ × (-6) + 5 × ________
Answer:
a × (b + c) = (a × b) + (a × c)
5×[(-6)+9] = 5 × (-6) + [(5 × 9 )]

AP Board 7th Class Maths Solutions Chapter 1 Integers Ex 1.3

Question 4.
State true or false. Write the reasons,
(i) 2 is the multiplicative identity of – 2.
Answer:
False.
2 is not the multiplicative identity of – 2. Multiplicative identity of – 2 is – \(\frac{1}{2}\).
2 is the additive inverse of – 2.

(ii) Integers are commutative under subtraction.
Answer:
False.
5 – 3 ≠ 3 – 5
2 ≠ – 2, i.e., a – b ≠ b – a.
Integers are not commutative under subtraction.

(iii) For any two integers a and b,
a × b = b × a
Answer:
True.
4 × (- 5) = (- 5) × 4
– (4 × 5) = -(5 × 4)
– 20 = – 20
So, for any two integers a and b,
a × b = b × a.

(iv) The division of integers by zero is not defined.
Answer:
True.
5 ÷ 0 is not defined.
So, the division of integers by zero is not defined.

(v) 6 + (-6) = (-6) + 6 = 0 indicates additive identity property. .
Answer:
False.
(-6) is the additive inverse of 6.
So, it indicates additive inverse property. But, not additive identity property.

AP Board 7th Class Maths Solutions Chapter 1 Integers Ex 1.3

Question 5.
Simplify the following using suitable laws.
(i) – 11 × (- 25) × (- 4)
Answer:
– 11 × (- 25) × (- 4)
a × (b × c)
= (- 11) × [(- 25) × (- 4)]
= (- 11) × (25 × 4)
= – 11 × 100
= – (11 × 100)
= – 1100

(ii)
3 × (- 18) + 3 × (- 32)
Answer:
3 × (- 18) + 3 × (- 32)
(a × b) + (a × c) = a × (b + c)
= 3 × [(- 18) + (- 32)]
= 3 × (- 18 – 32) .
= 3 × (- 50)
= – (3 × 50)
= – 150

Question 6.
Are the integers are associative under subtraction ? Explain by an example.
Answer:
No, integers are not associative under subtraction.
(a – b) – c ≠ a – (b – c)
(6 – 4) – 5 ≠ 6 – (4 – 5)
2 – 5 ≠ 6-( – 1)
– 3 ≠ 6 + 1
– 3 ≠ + 7

AP Board 7th Class Maths Solutions Chapter 1 Integers Ex 1.3

Question 7.
Verify [(-5) × 2)] × 3 = (- 5) × [(2 × 3)].
Answer:
[(- 5) × 2)] × 3 = (- 5) × [(2 × 3)]
(a × b) × c = a × (b × c)
[- (5 × 2)] × 3 = (- 5) × (6)
(- 10) × 3 = – (5 × 6)
– (10 × 3) = – 30
– 30 = – 30
L.H.S. = R.H.S.