# NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.6

**NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Exercise 2.6**

Ex 2.6 Class 7 Maths 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:

(i) 0.2 × 6

∵ 2 × 6 = 12 and we have 1 digit right to the decimal point in 0.2.

Thus 0.2 × 6 = 1.2

(ii) 8 × 4.6

∵ 8 × 46 = 368 and there is one digit right to the decimal point in 4.6.

Thus 8 × 4.6 = 36.8

(iii) 2.71 × 5

∵ 271 × 5 = 1355 and there are two digits right to the decimal point in 2.71.

Thus 2.71 × 5 = 13.55

(iv) 20.1 × 4

∵ 201 × 4 = 804 and there is one digit right to the decimal point in 20.1.

∵ 20.1 × 4 = 80.4

(v) 0.05 × 7

∵ 5 × 7 = 35 and there are 2 digits right to the decimal point in 0.05.

Thus 0.05 × 7 = 0.35

(vi) 211.02 × 4

∵ 21102 × 4 = 84408 and there are 2 digits right to the decimal point in 211.02.

Thus 211.02 × 4 = 844.08

(vii) 2 × 0.86

∵ 2 × 86 = 172 and there are 2 digits right to the decimal point in 0.86.

Thus 2 × 0.86 = 1.72

Ex 2.6 Class 7 Maths Question 2.

Find the area of rectangle whose length is 5.7 cm and breadth is 3 cm.

Solution:

Length = 5.7 cm

Breadth = 3 cm

Area of rectangle = length × breadth

= 5.7 × 3 = 17.1 cm^{2}

Hence, the required area =17.1 cm^{2}

Ex 2.6 Class 7 Maths 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:

Ex 2.6 Class 7 Maths Question 4.

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

Solution:

Distance covered in 1 litre = 55.3 km

Distance covered in 10 litres = 55.3 × 10 km

Hence, the required distance = 553 km

Ex 2.6 Class 7 Maths 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) 100.01 × 1.1

(x) 100.01 × 1.1

Solution:

(i) 0 2.5 × 0.3

∵ 25 × 3 = 75 and there are 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

∵ 1 × 517 = 517 and there are two digits (1 + 1) right to the decimal places in 0.1 and 51.7.

(iii) 0.2 × 316.8

∵ 2 × 3168 = 6336 and there are 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

∵ 13 × 31 = 403 and there are 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

∵ 5 × 5 = 25 and there are 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

∵ 112 × 15 = 1680 and there are 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

∵ 107 × 2 = 214 and there are 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

∵ 1005 × 105 = 105525 and there are 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

∵ 10101 × 1 = 10101 and there are 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

∵ 10001 × 11 = 110011 and there are 3 digits (2 + 1) right to the decimal points in 100.01 and 1.1.

Thus 100.01 × 1.1 = 110.011.