These NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals InText Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals InText Questions

NCERT In-text Question Page No. 34

Question 1.

Find:

(a) \(\frac { 1 }{ 2 }\) × 3

(b) \(\frac { 9 }{ 7 }\) × 6

(c) 3 × \(\frac { 1 }{ 8 }\)

(d) \(\frac { 13 }{ 11 }\) × 6

if the product is an improper fraction express it as a mixed fraction

Answer:

Question 2.

Represent pictorially: 2 × \(\frac{2}{5}=\frac{4}{5}\)

Answer:

NCERT In-text Question Page No. 34

Question 1.

(i) 5 × 2 \(\frac{3}{7}\)

(ii) 1 \(\frac{4}{9}\) × 6

Answer:

NCERT In-text Question Page No. 35

Question 1.

Can you tell, what is (i) \(\frac{1}{2}\) of 10?

(i) \(\frac{1}{4}\) of 16? (iii) (i) \(\frac{2}{5}\) of 25?

Answer:

NCERT In-text Question Page No. 39

Question 1.

Fill in these boxes:

NCERT In-text Question Page No. 40

Question 1.

Find:

\(\frac{1}{3} \times \frac{4}{5} ; \frac{2}{3} \times \frac{1}{5}\)

Answer:

Question 2.

Find:

\(\frac{8}{3} \times \frac{4}{7} \times \frac{3}{4} \times \frac{2}{3}\)

Answer:

NCERT In-text Question Page No. 44

Question 1.

(i) Will the reciprocal of a proper fraction be again a proper fraction?

(ii) Will the reciprocal of an improper fraction be again an improper fraction?

Answer:

(i) No, the reciprocal of an improper fraction is a improper fraction.

(ii) No, the reciprocal of an improper fraction is a proper fraction.

Now, we can say that

(a) 1 ÷ \(\frac { 1 }{ 2 }\) = 1 × \(\frac { 2 }{ 1 }\) = 1 × reciprocal of \(\frac { 1 }{ 2 }\)

(b) 3 ÷ \(\frac { 1 }{ 4 }\) = 3 × \(\frac { 4 }{ 1 }\) = 3 × reciprocal of \(\frac { 1 }{ 4 }\)

(c) 3 ÷ \(\frac { 1 }{ 2 }\) = ………….. = …………….. 3 ÷ \(\frac { 1 }{ 2 }\) = 3 × \(\frac { 2 }{ 1 }\) = 3 × reciprocal of \(\frac { 1 }{ 2 }\)

And, 2 ÷ \(\frac { 3 }{ 4 }\) = 2 × reciprocal of \(\frac { 3 }{ 4 }\) = 2 × \(\frac { 4 }{ 3 }\)

(d) 5 ÷ \(\frac { 2 }{ 9 }\) = 5 × …………… = 5 × ……………..

∴ 5 ÷ \(\frac { 2 }{ 9 }\) = 5 × \(\frac { 9 }{ 2 }\) = 5 × reciprocal of \(\frac { 2 }{ 9 }\)

NCERT In-text Question Page No. 45

Question 1.

Find:

(i) 7 ÷ \(\frac { 2 }{ 5 }\)

(ii) 7 ÷ \(\frac { 4 }{ 7 }\)

(iii) 7 ÷ \(\frac { 8 }{ 9 }\)

Answer:

NCERT In-text Question Page No. 45

Question 1.

Find:

(i) 6 ÷ 5\(\frac { 1 }{ 3 }\)

(ii) 7 ÷ 2\(\frac { 4 }{ 7 }\)

Answer:

NCERT In-text Question Page No. 45

Question 1.

Find:

(i) \(\frac{3}{5} \div \frac{1}{2}\)

(ii) \(\frac{1}{2} \div \frac{3}{5}\)

(iii) 2\(\frac{1}{3} \div \frac{3}{5}\)

(iv) 5\frac{1}{6} \div \frac{9}{2}

Answer:

NCERT In-text Question Page No. 50

Question 1.

Find:

(i) 2.7 × 4

(ii) 1.8 × 1.2

(iii) 2.3 × 4.35

Answer:

(i) 27 × 4 = 108 and there is one digit to the right of the decimal point in 27 2.7 × 4 = 10.8

(ii) 18 × 12 = 216 and number of digits to the right of decimal point is (1 + 1) 2 = 2

1.8 × 1.2 = 2.16

(iii) 23 × 435 = 10005 and there are 1 + 2 = 3 digits to the right of decimal point

2.3 × 4.35 = 10.005

Question 2.

Arrange the products obtained in Question in descending order.

Answer:

The products are 10.8, 2.16, 10.005. Comparing 10.8 and 10.005, we have :

10 = 10, 8 > 0, i.e. 10.005 < 10.8

Here, the smallest number = 2.16

and, the largest number = 10.8

Thus the required descending order is : 10.8, 10.005,2.16.

NCERT In-text Question Page No. 51

Question 1.

Find:

(i) 0.3 × 10

(ii) 1.2 × 100

(iii) 56.3 × 1000

Answer:

(i) 0.3 × 10

There is one zero in 10

∵ The decimal point is shifted to the right by one place

Thus, 0.3 × 10 = 3

(ii) There are 2 zeroes in 100

∵ The decimal point is shifted to the right by 2 places

= 1.20 × 100= 120

Thus, 1.2 × 100 = 120

(iii) There are 3 zeros in 1000

∵ The decimal point is shifted to the right by 3 places

Thus, 56.3 × 1000 = 56.300 × 1000 = 56300

NCERT In-text Question Page No. 53

Question 1.

Find:

(i) 235.4 ÷ 10

(ii) 235.4 ÷ 100

(iii) 235.4 ÷ 1000

Answer:

(i) 235.4 ÷ 10

Since, there is one zero in 10.

∴ The decimal point in the quotient is shifted to the left by one place.

∴ 235.4 ÷ 10 = 23.54

(ii) 235.4 ÷ 100

Since, there are two zeros in 100

∴ The decimal point in the quotient is

shifted to the left by three places.

∴ 235.4 H- 1000 = 0.2354

NCERT In-text Question Page No. 53

Question 1.

Find:

(i) 35.7 ÷ 3 = ?

(ii) 25.5 ÷ 3 = ?

Answer:

(i) 35.7 ÷ 3

Since, \(\frac{357}{3}\) = 119 and there is one digit in the decimal part of the given decimal number.

∴ The decimal point is placed in the quotient after one digit from the right most digit.

∴ 35.7 ÷ 3 = 11.9

(ii) 25.5 ÷ 3

Since, 255 ÷ 3 = 85 and there is one digit in the decimal part of the given decimal number.

∴ The decimal is placed in the quotient after one digit from the right most digit.

∴ 25.5 ÷ 3 = 8.5

NCERT In-text Question Page No. 53

Question 1.

(i) 43.15 ÷ 5 = ?

(ii) 82.44 ÷ 6 = ?

Answer:

NCERT In-text Question Page No. 53

Question 1.

Find:

(i) 15.5 ÷ 5

(ii) 126.35 ÷ 7

Answer:

(i) 15.5 ÷ 5

Since 155 ÷ 5 = 31 and there is one digit in the decimal part of the given decimal number.

∴ Place the decimal point in 31 such that there is one digit to its right.

∴ 15.5 + 5 = 3.1

(ii) 126.37 + 7

Since 12635 + 7 = 1805 and there are two digits in the decimal part of the given decimal number.

∴ Place the decimal point in 1805 such that there are two digits to its right.

∴ 126.35 ÷ 7 = 18.05

NCERT In-text Question Page No. 54

Question 1.

Find:

(i) \(\frac{7.75}{0.25}\)

(ii) \(\frac{42.8}{0.02}\)

(iii) \(\frac{5.6}{1.4}\)

Answer: