Class 7 Mathematics - Exercise 2.4 (Fraction and Decimal)

Understanding Fraction and Decimal - Exercise 2.4 Solutions

In this section, we will explore Exercise 2.4 from Chapter 2: Fraction and Decimal of Class 7 General Mathematics. This exercise focuses on multiplication of decimals and real-life applications of decimal operations. Let’s go through the questions along with their solutions step by step.

SCERT Solution for Class 7 Maths Chapter 2 Fractions and Decimals

Chapter Name	Solution Link
Exercise 2.1	Click Here
Exercise 2.2	Click Here
Exercise 2.3	Click Here
Exercise 2.4	Click Here
Exercise 2.5	Click Here

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Question 1: Find the Product

Multiply the given decimal numbers:

(i) 0.01 × 5 (ii) 6 × 2.7 (iii) 3.89 × 4 (iv) 7.21 × 9 (v) 8 × 11.7

Solution:

(i) 0. 01 × 5 = 0 .05

(ii) 6 × 2.7 = 16.2

(iii) 3. 89 × 4 = 15 .56

(iv) 7 .21 × 9 = 64 .89

(v) 8× 11.7 = 93. 6

 

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Question 2: Evaluate the Following

(i) 0.6 × 10  (ii) 2.8 × 10 (iii) 5.7 × 100 

(iv) 3.79 × 100 (v) 4.286 × 100 (vi) 12.54 × 100 

(vii) 2.234 × 1000 (viii) 3.9524 × 1000 

(ix) 0.08 × 10 (x) 1.05 × 10

Solution:

(i) 0 .6 × 10 = 6 .0 = 6

(ii) 2 .8 × 10 = 28

(iii) 5.7 × 100 = 570

(iv) 3 .79 × 100 = 379

(v) 4.286× 100 = 428.6

(vi) 12.54 × 100 = 1254

(vii) 2.234 × 1000 = 2234

(viii) 3.9524 × 1000 = 3952.4

(ix) 0 .08 × 10 = 0.8

(x) 1.05 × 10 = 10.5

 

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Question 3: Multiply the Following

(i) 0.51 × 0.5 (ii) 0.25 × 0.25 

(iii) 1.57 × 3.55 (iv) 5.7 × 3.25 

(v) 100.03 × 2.2 (vi) 101.01 × 1.01 

(vii) 0.5 × 0.05 (viii) 1.51 × 5.15

Solution:

(i) 0 .51 × 0. 5 = 0.255

(ii) 0.25 × 0.25 = 0.0625

(iii) 1.57 × 3.55 = 5.5735

(iv) 5.7 ×3.25 = 18.525

(v) 100.03 × 2.2 = 220.066

(vi) 101 .01 × 1.01 = 102 .0201

(vii) 0.5 × 0 .05 = 0 .025

(viii) 1.51 × 5.15 = 7.7765

Question 4. Bullbuli's father bought a plot of land of measure 18.25m ×15 .75m in city. If he wants to cover the area by a boundary wall. What will be the length of the boundary wall?

Solution:

A plot of land of measure = 18.25m ×15.75m

Its perimeter = 2(18.25+15.75)

     = (2×34) m

    = 68m

 :. The length of the boundary wall is 68metrs.

 

Question 5. What is the area of a square of side 2.4 cm?

Solution:

Area of a square = side×side

                                = 2.4cm 2 ×4cm

 = 5.76 sq.cm.

 

Question 6. A four-wheeler vehicle covers 15.5 km distance by 1 litre of petrol. How much distance would it cover in 100 liters of petrol?

Solution:

Distance covered in 100 liters = 15.5 × 100

= 1550 km

:. The vehicle will cover 1550 km.

 

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Conclusion

This exercise helps students strengthen their understanding of multiplication of decimals and real-world applications. Mastering these concepts is essential for solving practical mathematical problems efficiently.

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