Tamilnadu State Board Class 9 Maths Solutions Chapter 2 Real Numbers Ex 2.2
Question 1.
Express the following rational numbers into decimal and state the kind of decimal expansion.
(i) \(\frac { 2 }{ 7 }\)
(ii) \(-5 \frac{3}{11}\)
(iii) \(\frac { 22 }{ 3 }\)
(iv) \(\frac { 327 }{ 200 }\)
Solution:
(i) \(\frac { 2 }{ 7 }\)
\(\frac{2}{7}=0 . \overline{285714}\)
Nen-terminating and recurring
(ii) \(-5 \frac{3}{11}\)
\(-5 \frac{3}{11}=-5 . \overline{27}\)
Nen-terminating and recurring
(iii) \(\frac { 22 }{ 3 }\)
\(\frac{22}{3}=7 . \overline{3}\)
Nen-terminating and recurring
(iv) \(\frac { 327 }{ 200 }\)
\(\frac { 327 }{ 200 }\) = 1.635, Terminating.
Question 2.
Express \(\frac { 1 }{ 13 }\) in decimal form. Find the length of the period of decimals.
Solution:
\(\frac{1}{13}=0 . \overline{076923}\) has the length of the period of decimals = 6.
Question 3.
Express the rational number \(\frac { 1 }{ 13 }\) in recurring decimal form by using the recurring decimal expansion of \(\frac { 1 }{ 11 }\) . Hence write \(\frac { 71 }{ 33 }\) in recurring decimal form.
Solution:
The recurring decimal expansion of \(\frac { 1 }{ 11 }\) = 0.09090909…. = \(0.\overline { 09 }\)
Question 4.
Express the following decimal expression into rational numbers.
(i) \(0.\overline { 24 }\)
(ii) \(2.\overline { 327 }\)
(iii) -5.132
(iv) \(3.1\overline { 7 }\)
(v) \(17.\overline { 215 }\)
(vi) \(-21.213\overline { 7 }\)
Solution:
(i) \(0.\overline { 24 }\)
Let x = \(0.\overline { 24 }\) = 0.24242424……… ….(1)
(Here period of decimal is 2, multiply equation (1) by 100)
100x = 24.242424 ………. ….(2)
(2) – (1)
100x – x = 24.242424…. – 0.242424….
99x = 24
x = \(\frac { 24 }{ 99 }\)
ii. \(2.\overline { 327 }\)
Let x = 2.327327327…… ….(1)
(Here period of decimal is 3, multiply equation (1) by 1000)
1000x = 2327.327…. ….(2)
(2) – (1)
1000x – x = 2327.327327… – 2.327327….
999x = 2325
x = \(\frac { 2325 }{ 999 }\)
(iii) -5.132
\(x=-5.132=\frac{-5132}{1000}=\frac{-1283}{250}\)
(iv) \(3.1\overline { 7 }\)
Let x = 3.1777 ……. ..(1)
(Here the repeating decimal digit is 7, which is the second digit after the decimal point, multiply equation (1) by 10)
10x = 31.7777 …….. ..(2)
(Now period of decimal is 1, multiply equation (2) by 10)
100x = 317.7777…….. …(3)
(3) – (2)
100x – 10x = 317.777…. – 31.777….
90x = 286
\(x=\frac{286}{90}=\frac{143}{45}\)
(v) \(17.\overline { 215 }\)
Let x = 17.215215 ……. ….(1)
1000x = 17215.215215…… ….(2)
(2) – (1)
1000x – x = 17215.215215… – 17.215…
999x = 17198
x = \(\frac { 17198 }{ 999 }\)
(vi) \(-21.213\overline { 7 }\)
Let x = -21.2137777… …(1)
10x = -212.137777…… …(2)
100x = -2121.37777…… …(3)
1000x = -21213.77777…. …(4)
10000x = 212137.77777….. ..(5)
(Now period of decimal is 1, multiply equation (4) it by 10)
(5) – (4)
10000x – 1000x = (-212137.7777…) – (-21213.7777…)
9000x = -190924
x = –\(\frac { 190924 }{ 9000 }\)
Question 5.
Without actual division, find which of the following rational numbers have terminating decimal expansion.
(i) \(\frac { 7 }{ 128 }\)
(ii) \(\frac { 21 }{ 15 }\)
(iii) 4\(\frac { 9 }{ 35 }\)
(iv) \(\frac { 219 }{ 2200 }\)
Solution:
(i) \(\frac { 7 }{ 128 }\)
So \(\frac{7}{128}=\frac{7}{2^{7} 5^{0}}\)
This of the form 4m, n ∈ W
So \(\frac { 7 }{ 128 }\) has a terminating decimal expansion.
(ii) \(\frac { 21 }{ 15 }\)
So \(\frac { 21 }{ 15 }\) has a terminating decimal expansion.
(iii) 4\(\frac { 9 }{ 35 }\) = \(\frac { 149 }{ 35 }\)
\(\frac{49}{35}=\frac{149}{5^{1} 7^{1}}\)
∴ This is not of the form \(\frac{p}{5^{1} 7^{1}}\)
So 4\(\frac { 9 }{ 35 }\) has a non-terminating recurring decimal expansion.
(iv) \(\frac { 219 }{ 2200 }\)
\(\frac{219}{2200}=\frac{219}{2^{3} 5^{2} 11^{1}}\)
∴ This is not of the form \(\frac{p}{2^{m} 5^{n}}\)
So \(\frac { 219 }{ 2200 }\) has a non-terminating recurring decimal expansion.