Class-7 Subject-Maths Chapter-1 Exercise1.3 INTEGERS

 
EVENTS CONVENT HIGH SCHOOL
10/08/2021          CLASS-7         SESSION2021-22(SLOT-1)
Maths
Chapter-1
INTEGERS EXERCISE 1.3
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Question 1.Find each of the following products:

(a) 3 × (-1)

(b) (-1) × 225

(c) (-21) × (-30)

(d) (-316) × (-1)

(e) (-15) × 0 × (-18)

(f) (-12) × (-11) × (10)

(g) 9 × (-3) × (-6)

(h) (-18) × (-5) × (-4)

(i) (-1) ×(-2) × (-3) × 4

(j) (-3) × (-6) × (-2) × (-1)

Solution:

(a) 3 × (-1) = -3 × 1 = -3

(b) (-1) × 225 = -1 × 225 = -225

(c) (-21) × (-30) = (-) × (-) × 21 × 30 = 630

(d) (-316) × (-1) = (-) × (-) × 316 × 1 = 316

(e) (-15) × 0 × (-18) = 0 [∵ a × 0 = a]

(f) (-12) × (-11) × (10)

= (-) × (-) × 12 × 11 × 10 = 1320

(g) 9 × (-3) × (-6) = (-3) × (-6) × 9

= (—) × (-) × 3 × 6 × 9 = 162

(h) (-18) × (-5) × (-4)

= (-) × (-) × (-) × 18 × 5 × 4 = -360

(i) (-1) × (-2) × (-3) × 4

= (-) × (-) × (-) × 1 × 2 × 3 × 4 = -24

(j) (-3) × (-6) × (-2) × (-1)

= (-) × (-) × (-) × (-) × 3 × 6 × 2 × 1 = 36

 

Question 2.Verify the following:

(a) 18 × [7 + (-3)] = [18 × 7] + [18 × (-3)]

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

Solution:

(a) 18 × [7 + (-3)] = [18 × 7] + [18 × (-3)]

LHS = 18 × [7 + (-3)] = 18 × 4 = 72

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

= 126 – 54 = 72

LHS = RHS

Hence, verified.

 

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

LHS = (-21) × [(-4) + (-6)]

= (-21) × (-10)

= (-) × (-) × 21 × 10 = 210

RHS = [(-21) × (-4)] + [(-21) × (-6)]

= (84) + (126) = 84 + 126 = 210

LHS = RHS

Hence, verified.

 

Question 3.(i) For any integer a, what is (-1) × a equal to?

(ii) Determine the integer whose product with (-1) is 0.

(a) -22

(b) 37

(c) 0

Solution:

(i) (-1) × a = -a

(ii) (-1) × 0 = 0 [∵ a × 0 = 0]

Hence (c) 0 is the required integer.

 

Question 5.Find the product, using suitable properties:

(a) 26 × (-48) + (-48) × (-36)

(b) 8 × 53 × (-125)

(c) 15 × (-25) × (-4) × (-10)

(d) (-41) × 102

(e) 625 × (-35) + (-625) × 65

(f) 7 × (50 – 2)

(g) (-17) × (-29)

(h) (-57) × (-19) + 57

Solution:

(a) 26 × (-48) + (-48) × (-36)

= -48 × [26 + (-36)] = -48 × [26 – 36] = -48 × -10 = 480 [Distributive property of multiplication over

addition]

 

(b) 8 × 53 × (-125) = 53 × [8 × (-125)]

[Associative property of multiplication] = 53 × (-1000) = -53000

 

(c) 15 × (-25) × (-4) × (-10)

= [(-25) × (-4)] × [15 × (-10)]

[Regrouping the terms] = 100 × (-150) = -15000

 

(d) (-41) × 102 = (-41) × [100 + 2]

= (-41) × 100 + (-41) × 2

[Distributive property of multiplication over addition] = -4100 – 82 = -4182

 

(e) 625 × (-35) + (-625) × 65

= 625 × [(-35) + (-65)]

[Distributive property of multiplication over addition]

= 625 × (-100) = -62500

 

(f) 7 × (50 – 2) = 7 × 48 = 336 or

7 × (50 – 2) = 7 × 50 -7 × 2 = 350 – 14 = 336 [Distributive property of multiplication over addition]

 

(g) (-17) × (-29) = (-17) × [30 + (-1)]

= (-17) × 30 + (-17) × (-1)

= -510 + 17 = -493

[Distributive property of multiplication over addition]

 

(h) (-57) × (-19) + 57 = 57 × 19 + 57

= 57 × 19 + 57 × 1 [Y (-) × (-) = (+)] [Distributive property of multiplication over addition]

= 57 × (19 + 1) = 57 × 20 = 1140

 

Question 6.A certain freezing process requires that room temperature be lowered from 40°C at the rate of 5°C every hour. What will be the room temperature 10 hours after the process begins?

Solution:

Temperature of the room in the beginning = 40°C

Temperature after 1 hour

= 40°C – 1 × 5°C = 40°C – 5°C – 35°C

Similarly, temperature of the room after 10 hours

= 40°C – 10 × 5°C = 40°C – 50°C = -10°C