Selina Solutions Concise Maths Class 7 Chapter 4: Decimal Fractions (Decimals) Exercise 4F

Selina Solutions Concise Maths Class 7 Chapter 4 Decimal Fractions (Decimals) Exercise 4F are prepared by teachers at BYJU’S having experience in the education industry. The main aim of providing solutions is to help students, irrespective of their intelligence quotient. The solutions have explanations in simple language to help students perform well in the annual exam. Students are advised to solve the exercise wise problems on a daily basis to understand the concepts in a better way. Students can download Selina Solutions Concise Maths Class 7 Chapter 4 Decimal Fractions (Decimals) Exercise 4F PDF from the links, which are given here.

Selina Solutions Concise Maths Class 7 Chapter 4: Decimal Fractions (Decimals) Exercise 4F Download PDF

 

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Access other exercises of Selina Solutions Concise Maths Class 7 Chapter 4: Decimal Fractions (Decimals)

Exercise 4A Solutions

Exercise 4B Solutions

Exercise 4C Solutions

Exercise 4D Solutions

Exercise 4E Solutions

Access Selina Solutions Concise Maths Class 7 Chapter 4: Decimal Fractions (Decimals) Exercise 4F

page: 67

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

Solution:

It is given that

Weight of an object = 3.06 kg

So the weight of 48 objects = 3.06 × 48 = 146.88 kg

Selina Solutions Concise Maths Class 7 Chapter 4 Image 128

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

Solution:

It is given that

Cost of cloth per metre = ₹ 112.50

So the cost of 17.5 m cloth = ₹ 112.50 × 17.5

On further calculation

= ₹ 1968.750

= ₹ 1968.75

Selina Solutions Concise Maths Class 7 Chapter 4 Image 129

3. One kilogramme of oil costs ₹ 73.40. Find the cost of 9.75 kilogramme of the oil.

Solution:

It is given that

Cost of 1 kg oil = ₹ 73.40

So the cost of 9.75 kg oil = ₹ 73.40 × 9.75

On further calculation

= ₹ 715.6500

= ₹ 715.65

Selina Solutions Concise Maths Class 7 Chapter 4 Image 130

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

Solution:

It is given that

Weight of 8 identical objects = 51.2 kg

So the weight of 1 object = 51.2 ÷ 8 = 6.4 kg

5. 18.5 m of cloth costs ₹ 666. Find the cost of 3.8 m cloth.

Solution:

It is given that

Cost of 18.5 m cloth = ₹ 666

So the cost of 1m cloth = ₹ 666 ÷ 18.5 and cost of 3.8 m cloth

We can write it as

= (666 ÷ 18.5) × 3.8

Multiplying by 10

= (6660 ÷ 185) × 3.8

= 36 × 3.8

So we get

= ₹ 136.80

Selina Solutions Concise Maths Class 7 Chapter 4 Image 131

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

Solution:

(i) 0.5 of ₹ 7.60 + 1.62 of ₹ 30

It can be written as

= ₹ 3.80 + ₹ 48.60

So we get

= ₹ 52.40

Selina Solutions Concise Maths Class 7 Chapter 4 Image 132

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

It can be written as

= 16.79 kg + 0.432 kg

So we get

= 17.222 kg

Selina Solutions Concise Maths Class 7 Chapter 4 Image 133

(iii) 6.25 of 8.4 – 4.7 of 3.24

It can be written as

= 52.500 – 15.228

So we get

= 37.272

Selina Solutions Concise Maths Class 7 Chapter 4 Image 134

(iv) 0.98 of 235 – 0.09 of 3.2

It can be written as

= 230.30 – 0.288

So we get

= 230.012

Selina Solutions Concise Maths Class 7 Chapter 4 Image 135

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

Solution:

(i) 5.6 – 1.5 of 3.4

It can be written as

= 5.6 – 5.1

So we get

= 0.5

Selina Solutions Concise Maths Class 7 Chapter 4 Image 136

(ii) 4.8 ÷ 0.04 of 5

It can be written as

= 4.8 ÷ 0.20

Multiplying by 10

= 48 ÷ 2

= 24

(iii) 0.72 of 80 ÷ 0.2

It can be written as

= 57.60 ÷ 0.2

Multiplying by 10

= 576 ÷ 2

= 288

(iv) 0.72 ÷ 80 of 0.2

It can be written as

= 0.72 ÷ 16.0

Multiplying by 100

= 72 ÷ 1600

= 0.045

Selina Solutions Concise Maths Class 7 Chapter 4 Image 137

(v) 6.45 ÷ (3.9 – 1.75)

It can be written as

= 6.45 ÷ 2.15

Multiplying by 100

= 645 ÷ 215

= 3

Selina Solutions Concise Maths Class 7 Chapter 4 Image 138

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

It can be written as

= 0.12 of 0.084 + 0.36 × 0.5

So we get

= 0.01008 + 0.180

= 0.19008

Selina Solutions Concise Maths Class 7 Chapter 4 Image 139

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