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:
