**NCERT
Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5 ****(Rationalised Contents)**

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and
Decimals Ex 2.5 are the part of NCERT Solutions for Class 7 Maths (Rationalised Contents). Here you can
find the NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex
2.5.

**NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1****NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.2****NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.3****NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.4****NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5**

**Ex 2.5 Class 7 Maths Question 1.**

### Find:

(i) 0.4 ÷ 2

(ii) 0.35 ÷ 5

(iii) 2.48 ÷ 4

(iv) 65.4 ÷ 6

(v) 651.2 ÷ 4

(vi) 14.49 ÷ 7

(vii) 3.96 ÷ 4

(viii) 0.80 ÷ 5

**Solution:**

**Ex 2.5 Class 7 Maths Question 2.**

### Find:

(i) 4.8 ÷ 10

(ii) 52.5 ÷ 10

(iii) 0.7 ÷ 10

(iv) 33.1 ÷ 10

(v) 272.23 ÷ 10

(vi) 0.56 ÷ 10

(vii) 3.97 ÷ 10

**Solution:**We know that when we divide a decimal number by 10, the decimal point is shifted 1 place to the left.

(i) 4.8 ÷ 10 = 0.48 (By shifting the decimal point to the left by 1 place)

(ii) 52.5 ÷ 10 = 5.25 (By shifting the decimal point to the left by 1 place)

(iii) 0.7 ÷ 10 = 0.07 (By shifting the decimal point to the left by 1 place)

(iv) 33.1 ÷ 10 = 3.31 (By shifting the decimal point to the left by 1 place)

(v) 272.23 ÷ 10 = 27.223 (By shifting the decimal point to the left by 1 place)

(vi) 0.56 ÷ 10 = 0.056 (By shifting the decimal point to the left by 1 place)

(vii) 3.97 ÷ 10 = 0.397 (By shifting the decimal point to the left by 1 place)

**Ex 2.5 Class 7 Maths Question 3.**

### Find:

(i) 2.7 ÷ 100

(ii) 0.3 ÷ 100

(iii) 0.78 ÷ 100

(iv) 432.6 ÷ 100

(v) 23.6 ÷ 100

(vi) 98.53 ÷ 100

**Solution:**We know that when we divide a decimal number by 100, the decimal point is shifted 2 places to the left.

(i) 2.7 ÷ 100 = 0.027 (By shifting the decimal point to the left by 2 places)

(ii) 0.3 ÷ 100 = 0.003 (By shifting the decimal point to the left by 2 places)

(iii) 0.78 ÷ 100 = 0.0078 (By shifting the decimal point to the left by 2 places)

(iv) 432.6 ÷ 100 = 4.326 (By shifting the decimal point to the left by 2 places)

(v) 23.6 ÷ 100 = 0.236 (By shifting the decimal point to the left by 2 places)

(vi) 98.53 ÷ 100 = 0.9853 (By shifting the decimal point to the left by 2 places)

**Ex 2.5 Class 7 Maths Question 4.**

### Find:

(i) 7.9 ÷ 1000

(ii) 26.3 ÷ 1000

(iii) 38.53 ÷ 1000

(iv) 128.9 ÷ 1000

(v) 0.5 ÷ 1000

**Solution:**We know that when we divide a decimal number by 1000, the decimal point is shifted 3 places to the left.

(i) 7.9 ÷ 1000 = 0.0079 (By shifting the decimal point to the left by 3 places)

(ii) 26.3 ÷ 1000 = 0.0263 (By shifting the decimal point to the left by 3 places)

(iii) 38.53 ÷ 1000 = 0.03853 (By shifting the decimal point to the left by 3 places)

(iv) 128.9 ÷ 1000 = 0.1289 (By shifting the decimal point to the left by 3 places)

(v) 0.5 ÷ 1000 = 0.0005 (By shifting the decimal point to the left by 3 places)

**Ex 2.5 Class 7 Maths Question 5.**

### Find:

(i) 7 ÷ 3.5

(ii) 36 ÷ 0.2

(iii) 3.25 ÷ 0.5

(iv) 30.94 ÷ 0.7

(v) 0.5 ÷ 0.25

(vi) 7.75 ÷ 0.25

(vii) 76.5 ÷ 0.15

(viii) 37.8 ÷ 1.4

(ix) 2.73 ÷ 1.3

**Solution:**

**Ex 2.5 Class 7 Maths Question 6.**

### A vehicle covers a distance of 43.2 km in 2.4 litres of petrol. How much distance will it cover in one litre of petrol?

**Solution:**Since 2.4 litres of petrol is required to cover 43.2 km of distance.

∴ 1 litre of petrol will be required to cover

Hence, the required distance to be covered in 1 litre of petrol is 18 km.

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