Practicing OP Malhotra Class 9 Solutions Chapter 1 Rational and Irrational Numbers Ex 1(A) is the ultimate need for students who intend to score good marks in examinations.

## S Chand Class 9 ICSE Maths Solutions Chapter 1 Rational and Irrational Numbers Ex 1(A)

Question 1.

(i) Find a rational number between \(\frac { 1 }{ 2 }\) and \(\frac { 3 }{ 4 }\).

(ii) Find two rational numbers between 0.1 and 0.2.

(iii) How many rational numbers can you find between two given rational numbers?

Solution:

(i) One rational number between \(\frac { 1 }{ 2 }\) and \(\frac { 3 }{ 4 }\)

= \(\frac{1}{2}\left[\frac{1}{2}+\frac{3}{4}\right]=\frac{1}{2}\left[\frac{2+3}{4}\right]\)

= \(\frac{1}{2} \times \frac{5}{4}=\frac{5}{8}\)

(ii) Two rational numbers between 0.1 and 0.2 1

First number = \(\frac { 1 }{ 2 }\) [0.1 + 0.2]

= \(\frac { 1 }{ 2 }\) x 0.3 = 0.15 or \(\frac { 15 }{ 100 }\) = \(\frac { 3 }{ 20 }\)

and second number = \(\frac{1}{2}\left[\frac{3}{20}+\frac{2}{10}\right]\)

= \(\frac{1}{2}\left[\frac{3+4}{20}\right]=\frac{1}{2} \times \frac{7}{20}=\frac{7}{40}\)

∴ Two numbers are \(\frac { 3 }{ 10 }\) and \(\frac { 7 }{ 40 }\)

(iii) We can find infinite numbers of rational numbers between two given rational numbers.

Question 2.

Find two rational numbers between

(i) \(\frac { 4 }{ 5 }\) and \(\frac { 7 }{ 13 }\)

(ii) \(\frac { 3 }{ 4 }\) and 1\(\frac { 1 }{ 5 }\)

Solution:

(i) One rational number between \(\frac { 4 }{ 5 }\) and \(\frac { 7 }{ 13 }\)

= \(\frac{1}{2}\left[\frac{4}{5}+\frac{7}{13}\right]\) \(\frac { 1 }{ 2 }\){a + b}

= \(\frac{1}{2}\left(\frac{52+35}{65}\right)=\frac{87}{130}\)

and second rational number

= \(\frac{1}{2}\left[\frac{87}{130}+\frac{7}{13}\right]=\frac{1}{2}\left[\frac{87+70}{130}\right]\)

= \(\frac{1}{2}\left[\frac{157}{130}\right]=\frac{157}{260}\)

(ii) One rational number between \(\frac { 3 }{ 4 }\) and 1\(\frac { 1 }{ 5 }\) or \(\frac { 3 }{ 4 }\) and \(\frac { 6 }{ 5 }\)

= \(\frac{1}{2}\left[\frac{3}{4}+\frac{6}{5}\right]=\frac{1}{2}\left[\frac{15+24}{20}\right]\)

= \(\frac{1}{2}\left[\frac{39}{20}\right]=\frac{39}{40}\)

and second rational number

= \(\frac{1}{2}\left[\frac{39}{40}+\frac{6}{5}\right]\)

= \(\frac{1}{2}\left[\frac{39+48}{40}\right]=\frac{1}{2} \times \frac{87}{40}=\frac{87}{80}\)

Question 3.

Find three rational numbers between 0 and 0.2.

Solution:

First rational number between 0 and 0.2 = \(\frac { 1 }{ 2 }\) [0 + 0.2] = 0.1

Second rational number = \(\frac { 1 }{ 2 }\) [0 + 0.1]

= \(\frac { 1 }{ 2 }\) [0.1] = 0.05

and third rational number

= \(\frac { 1 }{ 2 }\) [0.1 + 0.2] = \(\frac { 1 }{ 2 }\) [0.3]

= 0.15

Hence three rational numbers are 0.05, 0.1 and 0.15

Question 4.

Find three rational numbers between 3 and 4.

Solution:

First rational number between 3 and 4

= \(\frac { 1 }{ 2 }\) [3 + 4] = \(\frac { 1 }{ 2 }\) x 7 = \(\frac { 7 }{ 2 }\)

Second rational number between 3 and \(\frac { 7 }{ 2 }\)

= \(\frac{1}{2}\left[3+\frac{7}{2}\right]=\frac{1}{2} \times \frac{6+7}{2}=\frac{13}{4}\)

and third number between \(\frac { 7 }{ 2 }\) and 4

= \(\frac{1}{2}\left[\frac{7}{2}+4\right]=\frac{1}{2}\left[\frac{7+8}{2}\right]=\frac{1}{2} \times \frac{15}{2}=\frac{15}{4}\)

Hence three rational number are

\(\frac { 13 }{ 4 }\), \(\frac { 7 }{ 2 }\) and \(\frac { 15 }{ 4 }\)

Question 5.

Find the rational number that is one seventh of the way from 1\(\frac { 3 }{ 4 }\) to 4\(\frac { 3 }{ 8 }\).

Solution:

1\(\frac{3}{4} \text { to } 4 \frac{3}{8} \Rightarrow \frac{7}{4} \text { to } \frac{35}{8}\)

⇒ \(\frac { 14 }{ 8 }\) to \(\frac { 35 }{ 8 }\)

Between 14 and 35, there are 21 terms i.e.

\(\frac{15}{8}, \frac{16}{8}, \frac{17}{8}, \frac{18}{8} \ldots . ., \frac{34}{8}\)

∴ \(\frac { 1 }{ 7 }\)th of 21 terms = 21 x \(\frac { 1 }{ 7 }\) = 3rd

∴ 7th term = \(\frac { 17 }{ 8 }\) i.e. 2\(\frac { 1 }{ 8 }\).

Question 6.

Find four rational numbers between – 1 and \(\frac { 1 }{ -2 }\).

Solution:

First rational number between – 1 and \(\frac { 1 }{ -2 }\)

Question 7.

Express \(\frac { 12 }{ 125 }\) as decimal fraction.

Solution:

\(\frac { 12 }{ 125 }\) = 0.096

(Dividing 12 by 125 by long division)

Question 8.

Find a vulgar fraction equivalent to 0.0

Solution:

0.03 = 0.033333

Let x = 0.033333 ….

10x = 3.3333 …. (i)

100x = 3.3333 …. (ii)

Subtracting (i) from (ii)

99x = 3.00 ….

x = \(\frac { 3 }{ 90 }\) = \(\frac { 1 }{ 30 }\)

∴ Required vulgar fraction = \(\frac { 1 }{ 10 }\)

Question 9.

Express the following rational numbers in the form \(\frac { p }{ q }\), p, q are integers, q ≠ 0.

(i) 6.\(\overline{46}\)

(ii) 0.1\(\overline{36}\)

(iii) 3.\(\overline{146}\)

(iv) – 5.\(\overline{12}\)

Solution:

(i) 6.\(\overline{46}\) = 6.464646 ….

Let x = 6.46464646 …. (i)

100x = 646.46464646 ….(ii)

Subtracting (i) from (ii)

99x = 646 – 6 = 640

x = \(\frac { 640 }{ 99 }\)

∴ Fraction = \(\frac { 640 }{ 99 }\)

(ii) 0.1\(\overline{36}\) = 0.1363636…

Let x = 0.1363636….

10x = 1.363636… (i)

and 1000c = 136.363636 ….(ii)

Subtracting (i) from (ii)

990x = 135

x = \(\frac{135}{990}=\frac{27}{198}=\frac{3}{22}\)

∴ Fraction = \(\frac { 3 }{ 22 }\)

(iii) 3.\(\overline{146}\)

Let x = 3.\(\overline{146}\) = 3.146146146… (i)

1000x = 3146.146146146…. (ii)

Subtracting (i) from (ii)

999x = 3143

x = \(\frac { 3143 }{ 999 }\)

Hence fraction = \(\frac { 3143 }{ 999 }\)

(iv) – 5.\(\overline{12}\)

Let x = – 5.\(\overline{12}\) = – 5.121212 … (i)

100x = – 512.121212 ….(ii)

Subtracting (i) from (ii)

99x = 507

⇒ x = \(\frac{-507}{99}=\frac{-169}{33}\)

∴ Fraction = \(\frac { -169 }{ 33 }\)

Question 10.

Write the terminating decimal numeral for the given rational number :

(i) \(\frac { 7 }{ 4 }\)

(ii) \(\frac { 29 }{ 50 }\)

(iii) \(\frac { 17 }{ 32 }\)

Solution:

Question 11.

Write the repeating decimal for each of the following and use a bar to show the repetend.

(i) \(\frac { 1 }{ 9 }\)

(ii) \(\frac { -4 }{ 3 }\)

(iii) \(\frac { 1 }{ 6 }\)

Solution: