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

NCERT Solutions Class 7 Mathematics 
Chapter – 2 (Fractions and Decimals)

The NCERT Solutions in English Language for Class 7 Mathematics Chapter – 2 Fractions and Decimals Exercise 2.6 has been provided here to help the students in solving the questions from this exercise. 

Chapter : 2 Fractions and Decimals

Exercise – 2.6

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
= \frac{2}{10} × 6 = \frac{12}{10}
On dividing a decimal by 10, the decimal point is shifted to the left by one place.
Then, = 1.2

(ii) 8 × 4.6
= 8 × \frac{46}{10}  = \frac{368}{10}
On dividing a decimal by 10, the decimal point is shifted to the left by one place.
Then, = 36.8

(iii) 2.71 × 5
= \frac{271}{100} × 5 = \frac{1355}{100}
On dividing a decimal by 100, the decimal point is shifted to the left by two places.
Then, = 13.55

(iv) 20.1 × 4
= \frac{201}{10} × 4 = \frac{804}{10}
On dividing a decimal by 10, the decimal point is shifted to the left by one place.
Then, = 80.4

(v) 0.05 × 7
= \frac{5}{100} × 7  = \frac{35}{100}
On dividing a decimal by 100, the decimal point is shifted to the left by two places.
Then, = 0.35

(vi) 211.02 × 4
= \frac{21102}{100} × 4 = \frac{84408}{100}
On dividing a decimal by 100, the decimal point is shifted to the left by two places.
Then, = 844.08

(vii) 2 × 0.86
= 2 × \frac{86}{100} = \frac{172}{100}
On dividing a decimal by 100, the decimal point is shifted to the left by two places.
Then, = 1.72

2. Find the area of rectangle whose length is 5.7cm and breadth is 3 cm.
Solution –
Length of the rectangle = 5.7 cm
Breadth of the rectangle = 3 cm
Area of the rectangle = Length × Breadth
= 5.7 cm × 3 cm
= 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
Solution –
(i) 1.3 × 10
On multiplying a decimal by 10, the decimal point is shifted to the right by one place.
= 1.3 × 10 = 13

(ii) 36.8 × 10
On multiplying a decimal by 10, the decimal point is shifted to the right by one place.
= 36.8 × 10 = 368

(iii) 153.7 × 10
On multiplying a decimal by 10, the decimal point is shifted to the right by one place.
= 153.7 × 10 = 1537

(iv) 168.07 × 10
On multiplying a decimal by 10, the decimal point is shifted to the right by one place.
= 168.07 × 10 = 1680.7

(v) 31.1 × 100
On multiplying a decimal by 100, the decimal point is shifted to the right by two places.
= 31.1 × 100 = 3110

(vi) 156.1 × 100
On multiplying a decimal by 100, the decimal point is shifted to the right by two places.
= 156.1 × 100 = 15610

(vii) 3.62 × 100
On multiplying a decimal by 100, the decimal point is shifted to the right by two places.
= 3.62 × 100 = 362

(viii) 43.07 × 100
On multiplying a decimal by 100, the decimal point is shifted to the right by two places.
= 43.07 × 100 = 4307

(ix) 0.5 × 10
On multiplying a decimal by 10, the decimal point is shifted to the right by one place.
= 0.5 × 10 = 5

(x) 0.08 × 10
On multiplying a decimal by 10, the decimal point is shifted to the right by one place.
= 0.08 × 10 = 0.8

(xi) 0.9 × 100
On multiplying a decimal by 100, the decimal point is shifted to the right by two places.
= 0.9 × 100 = 90

(xii) 0.03 × 1000
On multiplying a decimal by 1000, the decimal point is shifted to the right by three places.
= 0.03 × 1000 = 30

4. A two-wheeler covers a distance of 55.3 km in one liter of petrol. How much distance will it cover in 10 liters of petrol?
Solution –
Distance covered by two-wheeler in 1 Liter of petrol = 55.3 km
Distance covered by two wheeler in 10 Liter of petrol = (10 × 55.3) = 553 km
∴ Two-wheeler covers a distance in 10L of petrol is 553 km.

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 –
(i) 2.5 × 0.3
= \frac{25}{10} × \frac{3}{10} = \frac{75}{100}
On dividing a decimal by 100, the decimal point is shifted to the left by two places.
Then, = 0.75

(ii) 0.1 × 51.7
= \frac{1}{10} × \frac{517}{10} = \frac{517}{100}
On dividing a decimal by 100, the decimal point is shifted to the left by two places.
Then, = 5.17

(iii) 0.2 × 316.8
= \frac{2}{10} × \frac{3168}{10} = \frac{6336}{100}
On dividing a decimal by 100, the decimal point is shifted to the left by two places.
Then, = 63.36

(iv) 1.3 × 3.1
= \frac{13}{10} × \frac{31}{10} = \frac{403}{100}
On dividing a decimal by 100, the decimal point is shifted to the left by two places.
Then, = 4.03

(v) 0.5 × 0.05
= \frac{5}{10} × \frac{5}{100} = \frac{25}{1000}
On dividing a decimal by 1000, the decimal point is shifted to the left by three places.
Then, = 0.025

(vi) 11.2 × 0.15
= \frac{112}{10} × \frac{15}{100} = \frac{1680}{1000}
On dividing a decimal by 1000, the decimal point is shifted to the left by three places.
Then, = 1.680

(vii) 1.07 × 0.02
= \frac{107}{100} × \frac{2}{100} = \frac{214}{10000}
On dividing a decimal by 10000, the decimal point is shifted to the left by four places.
Then, = 0.0214

(viii) 10.05 × 1.05
= \frac{1005}{100} × \frac{105}{100} = \frac{105525}{10000}
On dividing a decimal by 10000, the decimal point is shifted to the left by four places.
Then, = 10.5525

(ix) 101.01 × 0.01
= \frac{10101}{100} × \frac{1}{100} = \frac{10101}{10000}
On dividing a decimal by 10000, the decimal point is shifted to the left by four places.
Then, = 1.0101

(x) 100.01 × 1.1
= \frac{10001}{100} × \frac{11}{10} = \frac{110011}{1000}
On dividing a decimal by 1000, the decimal point is shifted to the left by three places.
Then, = 110.011

 

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