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Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 2 Real Numbers Ex 2.2
Question 1.
Express the following rational numbers into decimal and state the kind of decimal expression.
(i) \(\frac{2}{7}\)
(ii) -5\(\frac{3}{11}\)
(iii) \(\frac{22}{3}\)
(iv) \(\frac{327}{200}\)
Solution:
(i) \(\frac{2}{7}\) = 0.2857142….
= 0.\(\overline {285714}\)
Non-terminating and recurring decimal expansion.
(ii) -5\(\frac{3}{11}\) = -5 + 0.272 = -5.272……..
= -5.\(\overline {27}\)
Non-terminating and recurring decimal expansion.
(iii) \(\frac{22}{3}\) = 7.333……..
= 7.\(\overline {3}\)
Non-terminating and recurring decimal expansion.
(iv) \(\frac{327}{200}\) = \(\frac{327}{2×100}\)
= \(\frac{3.27}{2}\)
= 1.635
Terminating decimal expansion.
Question 2.
Express \(\frac{1}{13}\) in decimal form. Find the length of the period of decimals.
Solution:
\(\frac{1}{13}\) = 0.07692307
= 0.\(\overline {076923}\)
Length of the period of decimal is 6.
Question 3.
Express the rational number \(\frac{1}{33}\) in recurring decimal form by using the recurring decimal expansion of \(\frac{1}{11}\). Hence write \(\frac{71}{33}\) in recurring decimal form.
Solution:
\(\frac{1}{11}\) = 0.0909……… = 0.\(\overline {09}\)
∴ \(\frac{1}{33}\) = \(\frac{1}{3}\) × \(\frac{1}{11}\)
= \(\frac{1}{3}\) × 0.0909 ……..
= 0.0303 …… = 0.\(\overline {03}\)
\(\frac{71}{33}\) = 2\(\frac{5}{33}\) = 2 + \(\frac{5}{33}\) = 2 + 5 × \(\frac{1}{33}\)
= 2 + 5 × 0.\(\overline {03}\)
2 + (5 × 0.030303 ……..)
2 + 0.151515 ………
2+ 0.\(\overline {15}\)
2.\(\overline {15}\)
Question 4.
Express the following decimal expression into rational numbers.
(i) 0.24
Solution:
Let x = 0.242424 ………. →(1)
100 x = 24.2424 ……… →(2)
(2) – (1) ⇒ 100 x – x = 24.2424 ……….. (-)
0.2424 ……..
99 x = 24.0000
x = \(\frac{24}{99}\)
(or)
\(\frac{8}{33}\)
(ii) 2.327
Solution:
Let x = 2.327327327 ………. →(1)
1000 x = 2327.327327 ……… →(2)
(2) – (1) ⇒ 1000 x – x = 2327.327327 ……….. (-)
2.327327 ……..
999 x = 2325.000
x = \(\frac{2325}{999}\)
(or)
\(\frac{775}{333}\)
(iii) – 5.132
Solution:
– 5.132 = -5 + \(\frac{1}{10}\) + \(\frac{3}{100}\) + \(\frac{2}{1000}\)
= \(\frac{-5000 + 100 +30 + 2}{1000}\) = \(\frac{-4868}{1000}\)
(or)
\(\frac{-1217}{250}\)
(iv) 3.17
Solution:
Let x = 3.1777 ………. →(1)
10 x = 31.777 ……… →(2)
100 x = 317.77 …….. →(3)
(3) – (2) ⇒ 100 x – 10 x = 317.77 ……….. (-)
31.777 ……..
90 x = 286.000
x = \(\frac{286}{90}\)
(or)
\(\frac{143}{45}\)
(v) 17.215
Solution:
Let x = 17.2151515 ………. →(1)
10 x = 172.151515 ……… →(2)
100 x = 17215.1515 …….. →(3)
(3) – (2) ⇒ 1000 x – 10 x = 17215.1515 ……….. (-)
17215.1515 ……..
990 x = 17043
x = \(\frac{17043}{990}\)
(or)
\(\frac{5681}{330}\)
(vi) -21.2137
Solution:
Let x = -21.213777 ………. →(1)
1000 x = -21213.777 ……… →(2)
100 x = -212137.77 …….. →(3)
(3) – (2) ⇒ 10000 x – 1000 x = -21213.777 ……….. (-)
-21213.777 ……..
9000 x = -190924
x = \(\frac{-190924}{9000}\)
(or)
\(\frac{-47731}{2250}\)
Question 5.
Without actual division, find which of the following rational numbers have terminating decimal expression.
(i) \(\frac{7}{128}\)
Solution:
\(\frac{7}{128}\) = \(\frac{7}{2^{7}}\)
∴ \(\frac{7}{128}\) has terminating decimal expression.
(ii) \(\frac{21}{15}\)
Solution:
\(\frac{21}{15}\) = \(\frac{7}{5}\) = \(\frac{7}{5^1}\)
\(\frac{21}{15}\) has terminating decimal expression.
(iii) 4\(\frac{9}{35}\)
Solution:
4\(\frac{9}{35}\) = \(\frac{149}{35}\)
4\(\frac{149}{5×7}\) (It is not in the form of \(\frac{P}{2^{m} × 5^{n}}\)
∴ 4\(\frac{9}{35}\) has non-terminating recurring decimal expression.
(iv) \(\frac{219}{2200}\)
Solution:
\(\frac{219}{2200}\) = \(\frac{219}{2^{3} × 5^{2} × 11}\) (It is not in the form of \(\frac{P}{2^{m} × 5^{n}}\)
∴ \(\frac{219}{2200}\) has non-terminating recurring decimal expression.