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

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

Try These (Page 29)

Question 1.
Can you write five examples each of proper, improper, and mixed fractions?
Solution:
Examples are given below :
Proper Fractions : \(\frac{1}{3}\) , \(\frac{2}{7}\) , \(\frac{3}{5}\) , \(\frac{11}{13}\) and \(\frac{14}{17}\)
Improper Fractions :\(\frac{3}{2}\) , \(\frac{5}{3}\) , \(\frac{9}{4}\) , \(\frac{12}{7}\) and \(\frac{31}{25}\)
Mixed Fractions :\(1 \frac{1}{5}\) , \(2 \frac{1}{7}\) , \(3 \frac{3}{4}\) , \(4 \frac{2}{3}\) and \(1 \frac{5}{9}\)

Try These (page 34)

Question 1.
3 × \(\frac{8}{7}\) = ?
Solution:
3 × \(\frac{8}{7}\) = \(\frac{3 \times 8}{7}\) = \(\frac{24}{7}\)

Question 2.
4 × \(\frac{7}{5}\) = ?
Solution:
4 × \(\frac{7}{5}\) = \(\frac{4 \times 7}{5}\) = \(\frac{28}{5}\)

Try These (page 34)

Question 1.
(a) \(\frac{2}{7}\) × 3
Solution:
\(\frac{2}{7}\) × 3 = \(\frac{2 \times 3}{7}\) = \(\frac{6}{7}\)

(b) \(\frac{\mathbf{9}}{\mathbf{7}}\) × 6
Solution:
\(\frac{\mathbf{9}}{\mathbf{7}}\) × 6 = \(\frac{9 \times 6}{7}\) = \(\frac{54}{7}\) = \(7 \frac{5}{7}\)

(c) 3 × \(\frac{1}{8}\)
Solution:
3 × \(\frac{1}{8}\) = \(\frac{3 \times 1}{8}\) = \(\frac{3}{8}\)

(d) \(\frac{13}{11}\) × 6
Solution:
\(\frac{13}{11}\) × 6 = \(\frac{13 \times 6}{11}\) = \(\frac{78}{11}\) = \(7 \frac{1}{11}\)

Question 2.
Represent pictorially: 2 × \(\frac{2}{5}\) = \(\frac{4}{5}\)
Solution:
The product 2 × \(\frac{2}{5}\) = \(\frac{4}{5}\) is represented as given below:

Try These (page 35)

Question 1.
Can you tell ,What is
(i) \(\frac{1}{2}\) of 10?
Solution:
\(\frac{1}{2}\) of 10 = \(\frac{1}{2}\) × 10 = \(\frac{1 \times 10}{2}\) = \(\frac{10}{2}\) = 5

(ii) \(\frac{1}{4}\) of 16?
Solution:
\(\frac{1}{4}\) of 16 = \(\frac{1}{4}\) × 16 = \(\frac{1 \times 16}{4}\) = \(\frac{16}{4}\) = 4

(iii) \(\frac{2}{5}\) of 25?
Solution:
\(\frac{2}{5}\) of 25 = \(\frac{2}{5}\) × 25 = \(\frac{2 \times 25}{5}\) = \(\frac{50}{5}\) = 10

Try These (page 39)
Fill in these Boxes:

(i)

Solution:

(ii)

Solution:

(iii)

Solution:

(iv)

Solution:

Try These (page 40)

Question 1.
\(\frac{1}{3}\) × \(\frac{4}{5}\)
Solution:
\(\frac{1}{3}\) × \(\frac{4}{5}\) = \(\frac{1 \times 4}{3 \times 5}\) = \(\frac{4}{15}\)

Question 2.
\(\frac{2}{3}\) × \(\frac{1}{5}\)
Solution:
\(\frac{2}{3}\) × \(\frac{1}{5}\) = \(\frac{2 \times 1}{3 \times 5}\) = \(\frac{2}{15}\)

Try These (page 40)

Question 1.
\(\frac{8}{3}\) × \(\frac{4}{7}\)
Solution:
\(\frac{8}{3}\) × \(\frac{4}{7}\) = \(\frac{8 \times 4}{3 \times 7}\) = \(\frac{32}{21}\)

Question 2.
\(\frac{3}{4}\) × \(\frac{2}{3}\)
Solution:
\(\frac{3}{4}\) × \(\frac{2}{3}\) = \(\frac{3 \times 2}{4 \times 3}\) = \(\frac{6 \div 6}{12 \div 6}\) = \(\frac{1}{2}\)

Try These (page 41)

Question 1.
Construct five more examples for yourself and verify the above statement.
Solution:
In order to verify this property, consider five parietes of improper fractions, and find their products as shown below:

Try These (page 45)

Question 1.
(i) 7 ÷ \(\frac{2}{5}\)
Solution:
7 ÷ \(\frac{2}{5}\) × reciprocal of \(\frac{2}{5}\) = 7 × \(\frac{5}{2}\) = \(\frac{35}{2}\) = 17 \(\frac{1}{2}\)

(ii) 6 ÷ \(\frac{4}{7}\)
Solution:
6 ÷ \(\frac{4}{7}\) = 6 × reciprocal of \(\frac{4}{7}\) = 6 × \(\frac{7}{4}\) = \(\frac{42}{4}\)
\(\frac{42 \div 2}{4 \div 2}\) = \(\frac{21}{2}\) = 10 \(\frac{1}{2}\)

(iii) 2 ÷ \(\frac{8}{9}\)
Solution:
2 ÷ \(\frac{8}{9}\) = 2 × reciprocal of \(\frac{8}{9}\) = 2 × \(\frac{9}{8}\) = \(\frac{18}{8}\)
\(\frac{18 \div 2}{8 \div 2}\) = \(\frac{9}{4}\) = 2 \(\frac{1}{4}\)

Try These (page 45)

Question 1.
(i) 6 ÷ 5 \(\frac{1}{3}\)
Solution:
6 ÷ 5 \(\frac{1}{3}\) = 6 ÷ \(\frac{16}{3}\) = 6 × reciprocal of \(\frac{16}{3}\) = 6 × \(\frac{3}{16}\) = \(\frac{18}{16}\) =\(\frac{18 \div 2}{16 \div 2}\) = \(\frac{9}{8}\) = 1 \(\frac{1}{8}\)

(ii) 7 ÷ 2 \(\frac{4}{7}\)
Solution:
7 ÷ 2 \(\frac{4}{7}\) = 7 ÷ \(\frac{18}{7}\) = 7 × reciprocal of \(\frac{18}{7}\) = 7 × \(\frac{7}{18}\) = \(\frac{49}{18}\) = 2 \(\frac{13}{18}\)

Try These (page 45)

Question 1.
(i) \(\frac{3}{5}\) ÷ \(\frac{1}{2}\)
Solution:
\(\frac{3}{5}\) ÷ \(\frac{1}{2}\) = \(\frac{3}{5}\) × reciprocal of \(\frac{1}{2}\) = \(\frac{3}{5}\) ÷ \(\frac{2}{1}\) = \(\frac{6}{5}\) = 1 \(\frac{1}{5}\)

(ii) \(\frac{1}{2}\) ÷ \(\frac{3}{5}\)
Solution:
\(\frac{1}{2}\) ÷ \(\frac{3}{5}\) = \(\frac{1}{2}\) × reciprocal of \(\frac{3}{5}\) = \(\frac{1}{2}\) × \(\frac{5}{3}\) = \(\frac{5}{6}\)

(iii) 2 \(\frac{1}{2}\) ÷ \(\frac{3}{5}\)
Solution:
2 \(\frac{1}{2}\) ÷ \(\frac{3}{5}\) = \(\frac{5}{2}\) ÷ \(\frac{3}{5}\) = \(\frac{5}{2}\) × reciprocal of \(\frac{3}{5}\) = \(\frac{5}{2}\) × \(\frac{5}{3}\) = \(\frac{25}{6}\) = 4 \(\frac{1}{6}\)

(iv) 5 \(\frac{1}{6}\) ÷ \(\frac{9}{2}\)
Solution:
5 \(\frac{1}{6}\) ÷ \(\frac{9}{2}\) = \(\frac{31}{6}\) ÷ \(\frac{9}{2}\) = \(\frac{31}{6}\) × reciprocal of \(\frac{9}{2}\) = \(\frac{31}{6}\) × \(\frac{2}{9}\) =
\(\frac{62}{54}\) = \(\frac{62 \div 2}{54 \div 2}\) = \(\frac{31}{27}\) = 1 \(\frac{4}{27}\)

Try These (page 46)

Question 1.
How well have you learned about Decimal numbers?
We have learned about decimal numbers in the earlier classes. Let us briefly recall them here.
Look at the following table and fill up the blank spaces.

Solution:

Using this table you can write a decimal number in its expended form also. For example,
253.417 = 2 × 100 + 5 × 10 + 3 × 1 + 4 × \(\left(\frac{1}{10}\right)\) + 1 × \(\left(\frac{1}{100}\right)\) + 7 × \(\left(\frac{1}{1000}\right)\)
So , a decimal number and its expanded form can be written if the place values of the digits are known.

Try These (page 47)

Question 1.
Write 75 paise = ₹ _______ , 250 g = _______ kg, 85 cm = _______ m.
Solution:
75 paise = ₹ \(\frac{75}{100}\) = ₹ 0.75,
250 g = \(\frac{250}{1000}\) kg, = 0.250 kg,
85 cm = \(\frac{85}{100}\) m = 0.85 m.

Try These (page 49)

Question 1.
Let us find 0.2 × 0.3
Solution:
1 st method:
0.2 = \(\frac{2}{10}\) and 0.3 = \(\frac{3}{10}\)
∴ 0.2 × 0.3 = \(\frac{2}{10}\) × \(\frac{3}{10}\) = \(\frac{6}{100}\) = 0.06

2 nd method:
We first multiply the given numbers as whole numbers ignoring the decimal point.
2 × 3 = 6
Then, we count the number of digits starting from the rightmost digit and moved towards the left. then, we put the decimal point there.

Try These (page 50)

Question 1.
1.5 × 1.6
Solution:
15 × 16 = 240
In this question, the multiplicand 1.5 has 1 place of decimal and the multiplier 1.6 has 1 place of decimal. So, the product has 2 places of decimal.
∴ 1.5 × 1.6 = 2.4

Try These (page 50)

Question 1.

(i) 2.7 × 4
Solution:
27 × 4 = 108
So, 2.7 × 4 = 10.8

(ii) 1.8 × 1.2
Solution:
18 × 12 = 216
So, 1.8 × 1.2 = 2.16

(iii) 2.3 × 4.35
Solution:
23 × 435 = 10005
So, 2.3 × 4.35  = 10.005

Try These (page 50)

Multiplication of Decimal numbers by 10, 100, and 1000 have a look at the table given below and fill in the blanks:

Solution:

From these examples, we find that the product of decimal point is shifted to the right as many places as the number of zeroes in the multiplier. Hence, to multiply a decimal fraction by
10, shift the decimal point to the right by one place 0.07 × 10 = 0.7
10o, shift the decimal point to the right by one place 0.07 × 100 = 7
10oo, shift the decimal point to the right by one place 0.07 × 1000 = 70

Try These (page 50)

Question 1.
(i) 0.3 × 10
Solution:
0.3 × 10 = 3

(ii) 1.2 × 100
Solution:
1.2 × 100 = 120

(iii) 56.3 × 1000
Solution:
56.3 × 1000 = 56300

Try These (page 53)

Question 1.
(i) 235.4 ÷ 10
Solution:
235.4 ÷ 10 = 23.54

(ii) 235.4 ÷ 100
Solution:
235.4 ÷ 10 = 2.354

(iii) 235.4 ÷ 1000
Solution:
235.4 ÷ 10 = 0.2354

Try These (page 53)

Question 1.
(i) 35.7 ÷ 3 = ?
Solution:
35.7 ÷ 3 = \(\frac{357}{10}\) ÷ 3 = \(\frac{357}{10}\) × \(\frac{1}{3}\) = \(\frac{357 \times 1}{10 \times 3}\) = \(\frac{1}{10}\) × \(\frac{357}{3}\) = \(\frac{1}{10}\) × 119 = \(\frac{119}{10}\) = 11.9

(ii) 25.5 ÷ 3 = ?
Solution:
25.5 ÷ 3 = \(\frac{255}{10}\) ÷ 3 = \(\frac{255}{10}\) × \(\frac{1}{3}\) = \(\frac{255 \times 1}{10 \times 3}\) = \(\frac{1}{10}\) × \(\frac{255}{3}\) = \(\frac{1}{10}\) × 85 = \(\frac{85}{10}\) = 8.5

Try These (page 54)

Question 1.
Division of a Decimal number by another decimal number
(i) \(\frac{20.3}{0.7}\)
Solution:
\(\frac{20.3}{0.7}\) = 20.3 ÷ 0.7 = \(\frac{203}{10}\) ÷ \(\frac{7}{10}\) = \(\frac{203}{10}\) × \(\frac{10}{7}\) = \(\frac{203}{7}\) = 29

(ii) \(\frac{15.2}{0.8}\)
Solution:
\(\frac{15.2}{0.8}\) = 15.2 ÷ 0.8 = \(\frac{152}{10}\) ÷ \(\frac{8}{10}\) =
\(\frac{152}{10}\) × \(\frac{10}{8}\) = \(\frac{152}{8}\) = 19