**Fractions and Decimals Class 7 Ex. 2.4;**

**New Ncert Solution for Class 7 Maths Chapter 2 Fractions and Decimals Free Solution;**

**Exercise 2.4**

**Question 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**

**Solution 1:-**

(i) 0.2 × 6

Since, 0.2 × 6 = 1.2 [1 digit right to the decimal point in 0.2.]

Thus 0.2 × 6 = 1.2

(ii) 8 × 4.6

Since, 8 × 4.6 = 36.8 [one digit right to the decimal point in 4.6.]

Thus 8 × 4.6 = 36.8

(iii) 2.71 × 5

Since, 2.71 × 5 = 13.55 [two digits right to the decimal point in 2.71.]

Thus, 2.71 × 5 = 13.55

(iv) 20.1 × 4

Since, 20.1 × 4 = 80.4 [one digit right to the decimal point in 20.1.]

Thus, 20.1 × 4 = 80.4

(v) 0.05 × 7

Since, 0.05 × 7 = 0.35 [2 digits right to the decimal point in 0.05.]

Thus 0.05 × 7 = 0.35

(vi) 211.02 × 4

Since, 211.02 × 4 = 844.08 [2 digits right to the decimal point in 211.02.]

Thus 211.02 × 4 = 844.08

(vii) 2 × 0.86

Since, 2 × 0.86 = 1.72 [2 digits right to the decimal point in 0.86.]

Thus 2 × 0.86 = 1.72

**Question 2 :- Find the area of rectangle, whose length is 5.7cm and breadth is 3cm.**

**Solution 2:- **

Length = 5.7 cm

Breadth = 3 cm

Area of rectangle = length × breadth = 5.7 × 3 = 17.1

Hence, the required area =17.1 .

**Question 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**

**Solution 3:- **

**Question 4 :- A two-wheeler, covers a distance; 55.3 km in one-litre of petrol. How much distance will it cover is 10-litres of petrol?**

**Solution 4:- **

Distance covered by two-wheeler, in 1-litre petrol = 55.3 km;

Distance covered by two-wheeler, in 10-litres petrol = 55.3 × 10 km;

Hence, the required distance = 553 km.

**Question 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**

**Solution 5:- **

(i) 2.5 × 0.3

Since, 2.5 × 0.3 = 0.75 [2 digits (1 + 1) right to the decimal points in 2.5 and 0.3.]

Thus, 2.5 × 0.3 = 0.75

(ii) 0.1 × 51.7

Since, 0.1 × 51.7 = 5.17 [two digits (1 + 1) right to the decimal places in 0.1 and 51.7.

Thus 0.1 x 51.7 = 5.17.

(iii) 0.2 × 316.8

Since, 0.2 × 316.8 = 63.36 [2 digits (1 + 1) right to the decimal points in 0.2 and 316.8.

Thus 0.2 × 316.8 = 63.36.

(iv) 1.3 × 3.1

Since, 1.3 × 3.1 = 4.03 [2 digits (1 + 1) right to the decimal points in 1.3 and 3.1.

Thus 1.3 × 3.1 – 4.03

(v) 0.5 × 0.05

Since, 0.5 × 0.05 = 0.025 [ 3 digits (1 + 2) right to the decimal points in 0.5 and 0.05.

Thus 0.5 × 0.05 = 0.025

(vi) 11.2 × 0.15

Since, 11.2 × 0.15 = 1.680 [3 digits (1 + 2) right to the decimal points in 11.2 and 0.15.

Thus 11.2 × 0.15 = 1.680

(vii) 1.07 × 0.02

Since, 1.07 × 0.02 = 0.0214 [4-digits (2 + 2) right to the decimal places is 1.07 × 0.02.

Thus 1.07 × 0.02 = 0.0214.

(viii) 10.05 × 1.05

Since, 1005 × 1.05 = 10.5525 [4 digits (2 + 2) right to the decimal places in 10.05 × 1.05.

Thus 10.05 × 1.05 = 10.5525

(ix) 101.01 × 0.01

Since, 101.01 × 0.01 = 1.0101 [4 digits (2 + 2) right to the decimal places in 101.01 and 0.01.

Thus 101.01 × 0.01 = 1.0101

(x) 100.01 × 1.1

Since, 100.01 × 1.1 = 110.011 [3 digits (2 + 1) right to the decimal points in 100.01 and 1.1.

Thus 100.01 × 1.1 = 110.011.

**Chapter-2, Fractions and Decimals Class-7 Exercise-2.4 for Free**