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Chapter 4

Decimal Fractions — Exercise 4(D)

Class - 7 Concise Mathematics Selina



Exercise 4(D)

Question 1

The weight of an object is 3.06 kg. Find the total weight of 48 similar objects.

Answer

Total weight = Weight of one object × Number of objects

= 3.06 × 48 = 146.88 kg

Hence, the total weight of 48 objects is 146.88 kg.

Question 2

Find the cost of 17.5 m cloth at the rate of ₹ 112.50 per metre.

Answer

Cost of cloth = Rate per metre × Length

= ₹ 112.50 × 17.5 = ₹ 1968.75

Hence, the cost of 17.5 m cloth is ₹ 1968.75.

Question 3

One kilogram of oil costs ₹ 73.40. Find the cost of 9.75 kilograms of the oil.

Answer

Cost of oil = Cost per kg × Weight

= ₹ 73.40 × 9.75 = ₹ 715.65

Hence, the cost of 9.75 kg of oil is ₹ 715.65.

Question 4

Total weight of 8 identical objects is 51.2 kg. Find the weight of each object.

Answer

Weight of each object = Total weight ÷ Number of objects

= 51.2 ÷ 8

= 6.4 kg

Hence, the weight of each object is 6.4 kg.

Question 5

18.5 m of cloth costs ₹ 666. Find the cost of 3.8 m cloth.

Answer

Given,

Cost of 18.5 m of cloth = ₹ 666

Cost of 1 m cloth = ₹ 666 ÷ 18.5 = ₹ 36

Cost of 3.8 m cloth = ₹ 36 × 3.8 = ₹ 136.80

Hence, the cost of 3.8 m cloth is ₹ 136.80.

Question 6

Find the value of:

(i) 0.5 of ₹ 7.60 + 1.62 of ₹ 30

(ii) 2.3 of 7.3 kg + 0.9 of 0.48 kg

(iii) 6.25 of 8.4 − 4.7 of 3.24

(iv) 0.98 of 235 − 0.09 of 3.2

Answer

(i) Solving,

0.5 of ₹ 7.60 + 1.62 of ₹ 30 = (0.5 × 7.60) + (1.62 × 30)

= 3.80 + 48.60 = ₹ 52.40

Hence, the value is ₹ 52.40.

(ii) Solving,

2.3 of 7.3 kg + 0.9 of 0.48 kg = (2.3 × 7.3) + (0.9 × 0.48)

= 16.79 + 0.432 = 17.222 kg

Hence, the value is 17.222 kg.

(iii) Solving,

6.25 of 8.4 - 4.7 of 3.24 = (6.25 × 8.4) - (4.7 × 3.24)

= 52.5 - 15.228 = 37.272

Hence, the value is 37.272.

(iv) Solving,

0.98 of 235 - 0.09 of 3.2 = (0.98 × 235) - (0.09 × 3.2)

= 230.3 - 0.288 = 230.012

Hence, the value is 230.012.

Question 7

Evaluate:

(i) 5.6 − 1.5 of 3.4

(ii) 4.8 ÷ 0.04 of 5

(iii) 0.72 of 80 ÷ 0.2

(iv) 0.72 ÷ 80 of 0.2

(v) 6.45 ÷ (3.9 − 1.75)

(vi) 0.12 of (0.104 − 0.02) + 0.36 × 0.5

Answer

Following the order of operations (here 'of' is performed before ÷ and ×):

(i) Solving,

5.6 - 1.5 of 3.4 = 5.6 - (1.5 × 3.4)

= 5.6 - 5.1 = 0.5

Hence, 5.6 - 1.5 of 3.4 = 0.5

(ii) Solving,

4.8 ÷ 0.04 of 5 = 4.8 ÷ (0.04 × 5)

=4.8÷0.2=482=24= 4.8 \div 0.2 = \dfrac{48}{2} = 24

Hence, 4.8 ÷ 0.04 of 5 = 24

(iii) Solving,

0.72 of 80 ÷ 0.2 = (0.72 × 80) ÷ 0.2

=57.6÷0.2=5762=288= 57.6 \div 0.2 = \dfrac{576}{2} = 288

Hence, 0.72 of 80 ÷ 0.2 = 288

(iv) Solving,

0.72 ÷ 80 of 0.2 = 0.72 ÷ (80 × 0.2)

= 0.72 ÷ 16 = 0.045

Hence, 0.72 ÷ 80 of 0.2 = 0.045

(v) Solving,

6.45 ÷ (3.9 - 1.75) = 6.45 ÷ 2.15

=645215=3= \dfrac{645}{215} = 3

Hence, 6.45 ÷ (3.9 - 1.75) = 3

(vi) Solving,

0.12 of (0.104 - 0.02) + 0.36 × 0.5 = 0.12 × 0.084 + 0.18

= 0.01008 + 0.18 = 0.19008

Hence, 0.12 of (0.104 - 0.02) + 0.36 × 0.5 = 0.19008

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