Let’s try to solve all the questions of EXERCISE 2.4 from chapter 2, “Fractions and Decimals” in the NCERT book. In this chapter, we will apply basic arithmetic operations to solve problems with decimals. In one question, 2.5 × 0.3, we converted decimals as fractions — 25/10 × 3/10 to simplify the multiplication. After multiplying, we got 75/100, or 0.75 when converted back to a decimal.
This method of converting decimals to fractions and back again is the method we will use in the 5 questions we will solve in this exercise. The approach will give you a deeper understanding of the relationship between decimals and fractions in multiplication.
Class 8 Maths for Chapter 2 Exercise 2.4 Fractions and Decimals – NCERT Book Solutions
1. Find:
(i) 0.2 × 6 (ii) 8 × 4.6 (iii) 2.71 × 5 (iv) 20.1 × 4
(v) 0.05 × 7 (vi) 211.02 × 4 (vii) 2 × 0.86
(i) 0.2 × 6
Solution:
= 2/10 × 6 (Converting the decimal to a fraction by moving the decimal point one position to the left.)
= 12/10
= 1.2
(ii) 8 × 4.6
Solution:
= 8 × 46/10 (Converting 4.6 to a fraction for easier multiplication.)
= 368/10
= 36.8
(iii) 2.71 × 5
Solution:
= 271/100 × 5 (Representing the decimal 2.71 as a fraction by moving the decimal two places to the left.)
= 1355/100
= 13.55
(iv) 20.1 × 4
Solution:
= 201/10 × 4 (Transforming the decimal into a fraction to simplify the calculation.)
= 804/10
= 80.4
(v) 0.05 × 7
Solution:
= 5/100 × 7 (Expressing the decimal as a fraction to facilitate multiplication.)
= 35/100
= 0.35
(vi) 211.02 × 4
Solution:
= 21102/100 × 4 (Changing the decimal to a fraction for multiplication.)
= 84408/100
= 844.08
(vii) 2 × 0.86
Solution:
= 2 × 86/100 (The decimal is converted to a fraction by moving the decimal point two places to the left.)
= 172/100
= 1.72
2. Find the area of a rectangle whose length is 5.7 cm and breadth is 3 cm.
Solution:
To calculate the area of the rectangle, we multiply the length by the breadth.
Given:
Length (l) = 5.7 cm
Breadth (b) = 3 cm
Area (A) of the rectangle = l × b
= 5.7 cm × 3 cm
= 17.1 cm²
Therefore, the area of the rectangle is 17.1 cm².
3. Find:
(i) 1.3 × 10 (ii) 36.8 × 10 (iii) 153.7 × 10 (iv) 168.07 × 10
(v) 31.1 × 100 (vi) 156.1 × 100 (vii) 3.62 × 100 (viii) 43.07 × 100
(ix) 0.5 × 10 (x) 0.08 × 10 (xi) 0.9 × 100 (xii) 0.03 × 1000
(i) 1.3 × 10
= 13 (Moving the decimal point one place to the right.)
(ii) 36.8 × 10
= 368 (Shifting the decimal point one place to the right.)
(iii) 153.7 × 10
= 1537 (The decimal point moves one position to the right when multiplying by 10.)
(iv) 168.07 × 10
= 1680.7 (Multiplying by 10 shifts the decimal one place to the right.)
(v) 31.1 × 100
= 3110 (When multiplied by 100, the decimal point moves two places to the right.)
(vi) 156.1 × 100
= 15610 (The decimal moves two positions to the right with multiplication by 100.)
(vii) 3.62 × 100
= 362 (The decimal point is moved two places to the right as we multiply by 100.)
(viii) 43.07 × 100
= 4307 (Multiplying by 100 causes the decimal to shift two places to the right.)
(ix) 0.5 × 10
= 5 (The decimal point is shifted one place to the right when multiplying by 10.)
(x) 0.08 × 10
= 0.8 (The decimal point moves one place to the right with multiplication by 10.)
(xi) 0.9 × 100
= 90 (Multiplying by 100 shifts the decimal two places to the right.)
(xii) 0.03 × 1000
= 30 (When multiplied by 1000, the decimal point moves three places to the right.)
4. A two-wheeler covers a distance of 55.3 km in one litre of petrol. How much distance will it cover in 10 litres of petrol?
Solution:
To find the total distance covered by the two-wheeler using 10 litres of petrol, we will multiply the distance covered per litre by 10.
Distance covered per litre: = 55.3 km
Total distance covered in 10 litres: = Distance per litre × Number of litres
= 55.3 km × 10
= 553 km
Therefore, the two-wheeler will cover 553 km in 10 litres of petrol.
5. Find:
(i) 2.5 × 0.3 (ii) 0.1 × 51.7 (iii) 0.2 × 316.8 (iv) 1.3 × 3.1
(v) 0.5 × 0.05 (vi) 11.2 × 0.15 (vii) 1.07 × 0.02
(viii) 10.05 × 1.05 (ix) 101.01 × 0.01 (x) 100.01 × 1.1
(i) 2.5 × 0.3
Solution:
= (25/10) × (3/10) (Expressing each decimal as a fraction to facilitate multiplication.)
= 75/100
= 0.75 (Converting the fraction back to a decimal by placing the decimal after two digits from the right.)
(ii) 0.1 × 51.7
Solution:
= (1/10) × (517/10) (Transforming the decimal to a fraction, moving the decimal point one position to the right.)
= 517/100
= 5.17 (Reverting to decimal form moves the decimal two places to the left.)
(iii) 0.2 × 316.8
Solution:
= (2/10) × (3168/10) (Adjusting the decimal two places to the right for multiplication.)
= 6336/100
= 63.36 (Placing the decimal two places from the right gives us the decimal form.)
(iv) 1.3 × 3.1
Solution:
= (13/10) × (31/10) (Representing decimals as fractions to simplify the calculation.)
= 403/100
= 4.03 (The decimal point is placed two digits from the right to convert to decimal.)
(v) 0.5 × 0.05
Solution:
= (5/10) × (5/100) (Decimals converted to fractions for easier multiplication.)
= 25/1000
= 0.025 (The decimal point is placed three places from the right.)
(vi) 11.2 × 0.15
Solution:
= (112/10) × (15/100) (Decimals rewritten as fractions.)
= 1680/1000
= 1.68 (The decimal point is placed three digits from the right to return to decimal form.)
(vii) 1.07 × 0.02
Solution:
= (107/100) × (2/100) (Decimals rewritten as hundredths for multiplication.)
= 214/10000
= 0.0214 (The decimal point is placed four places from the right for the decimal result.)
(viii) 10.05 × 1.05
Solution:
= (1005/100) × (105/100) (Converting to a fraction to maintain place value during multiplication.)
= 105525/10000
= 10.5525 (The decimal is placed four digits from the right.)
(ix) 101.01 × 0.01
Solution:
= (10101/100) × (1/100) (Multiplying by a hundredth to shift the decimal place.)
= 10101/10000
= 1.0101 (The decimal point is placed four digits from the right.)
(x) 100.01 × 1.1
Solution:
= (10001/100) × (11/10) (Shifting the decimal place for easier computation.)
= 110011/1000
= 110.011 (Adjusting the decimal place for the final answer.)
