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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.