AP State Syllabus AP Board 7th Class Maths Solutions Chapter 1 Integers Ex 5 Textbook Questions and Answers.

## AP State Syllabus 7th Class Maths Solutions 1st Lesson Integers Exercise 5

Question 1.

Verify the following.

(i) 18 × [7 + (-3)] = [18 × 7] + [18 x (-3)]

(ii) (-21) × [(-4) + (-6)] = [(-21) × (-4) ] + [(-21) × (-6)]

Solution:

(i) 18 × [7 + (-3)] = [18 x 7] + [18 x (-3)]

LHS: 18 × (7 + (-3)] = 18 × (4) = 72

RHS:[18 × 7] + [18 × (-3)] = 126 + (-54) = 72

∴ LHS = RHS

(ii) (-21) × [(-4) + (-6)] = [(-21) × (-4) ] + [(-21) x (-6)]

LHS:(-21) × [(-4) + (-6)] = (-21) × (-10) = 210

RHS: [(-21) × (-4)] + [(-21) × (-6)] = (84) + (126)= 210

∴ LHS=RHS

Question 2.

(1) For any integer a, what is(-1) × a equal to9

(ii) Determine the integer whose product with (-1) is 5

Solution:

i) For any integer a

(- 1) × a = -a

ii) …………. × (-1) = 5

∴ (-5) × (-1) = 5

Question 3.

Find the product, using suitable properties.

(i) 26 × (-48)+(-48) × (-36)

(ii) 8 × 53 × (-125)

(iii) 15 × (-25) × (4) × (-10)

(iv) (-41) × 102

(v) 625 × (-35) +(625) × 65

(vi) 7 × (50 – 2)

(vii) (-17) × (-29)

(viii) (-57) × (-19) + 57

Solution:

(i) 26 × (-48) + (-48) × (-36)

= -48 × [26 + (-36)]

= -48 × (-10)

= 480

(ii) 8 × 53 × (- 125)

424 × (- 125)

= -53000

(iii) 15 × (-25) × (-4) × (-10)

= (-375) × (40)

= -15000

(iv) (-41) × 102

= -4182

(v) 625 × (-35) + (-625) × 65

= 625 × [(-35) + (-65)]

=625 × (-100)

= – 62500

(vi) 7 × (50 – 2)

= 7 × 48

= 336

(vii) (-17) × (-29)

= 493

(viii) (-57) × (-19) + 57

=1083 + 57

= 1140