Samacheer Kalvi 9th Maths Solutions Chapter 2 Real Numbers Ex 2.2

You can Download Samacheer Kalvi 9th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 2 Real Numbers Ex 2.2

9th Maths Exercise 2.2 Samacheer Kalvi 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) \(\frac { 2 }{ 7 }\)
9th Maths Exercise 2.2 Samacheer Kalvi
\(\frac{2}{7}=0 . \overline{285714}\)
Nen-terminating and recurring

(ii) \(-5 \frac{3}{11}\)
9th Maths Exercise 2.2 In Tamil Samacheer Kalvi
\(-5 \frac{3}{11}=-5 . \overline{27}\)
Nen-terminating and recurring

(iii) \(\frac { 22 }{ 3 }\)
9th Standard Maths Exercise 2.2 Samacheer Kalvi
\(\frac{22}{3}=7 . \overline{3}\)
Nen-terminating and recurring

(iv) \(\frac { 327 }{ 200 }\)
Maths 9th Class Chapter 2 Real Numbers Samacheer Kalvi
\(\frac { 327 }{ 200 }\) = 1.635, Terminating.

9th Maths Exercise 2.2 In Tamil Question 2.
Express \(\frac { 1 }{ 13 }\) in decimal form. Find the length of the period of decimals.
Solution:
Class 9 Maths Chapter 2 Real Numbers Samacheer Kalvi
\(\frac{1}{13}=0 . \overline{076923}\) has the length of the period of decimals = 6.

9th Standard Maths Exercise 2.2 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…. = \(0.\overline { 09 }\)
Exercise 2.2 Class 9 Maths Solutions Samacheer Kalvi

Maths 9th Class Chapter 2 Real Numbers 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 = –\(\frac { 190924 }{ 9000 }\)

Class 9 Maths Chapter 2 Real Numbers 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) \(\frac { 7 }{ 128 }\)
9th Class Maths Exercise 2.2 Samacheer Kalvi
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) \(\frac { 21 }{ 15 }\)
9th Maths Samacheer Kalvi Chapter 2 Real Numbers Ex 2.2
So \(\frac { 21 }{ 15 }\) has a terminating decimal expansion.

(iii) 4\(\frac { 9 }{ 35 }\) = \(\frac { 149 }{ 35 }\)
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Ex 2.2
\(\frac{49}{35}=\frac{149}{5^{1} 7^{1}}\)
∴ 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) \(\frac { 219 }{ 2200 }\)
Samacheer Kalvi 9th Maths Book Solutions Chapter 2 Real Numbers Ex 2.2
\(\frac{219}{2200}=\frac{219}{2^{3} 5^{2} 11^{1}}\)
∴ This is not of the form \(\frac{p}{2^{m} 5^{n}}\)
So \(\frac { 219 }{ 2200 }\) has a non-terminating recurring decimal expansion.