NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.3 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.3.

Board CBSE
Textbook NCERT
Class Class 9
Subject Maths
Chapter Chapter 1
Chapter Name Number Systems
Exercise Ex 1.3
Number of Questions Solved 9
Category NCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.3

Question 1.
Write the following in decimal form and say what kind of decimal expansion each has
NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.3 img 1
Solution:
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Question 2.
You know that \(\frac { 1 }{ 7 }\) = \(\bar { 0.142857 }\). Can you predict what the decimal expansions of \(\frac { 2 }{ 7 }\) , \(\frac { 13 }{ 7 }\) , \(\frac { 4 }{ 7 }\) , \(\frac { 5 }{ 7 }\) , \(\frac { 6 }{ 7 }\) are , without actually doing the long division? If so, how?
Solution:
NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.3 img 5

Question 3.
Express the following in the form \(\frac { p }{ q }\)where p and q are integers and q ≠ 0.
(i) 0.\(\bar { 6 }\)
(ii) 0.4\(\bar { 7 }\)
(iii) 0.\(\overline { 001 }\)
Solution:
(i)Let x= 0.\(\bar { 6 }\) = 0.666… ….(i)
Multiplying Eq. (i) by 10, we get
10x = 6.666.. ….(ii)
On subtracting Eq. (ii) from Eq. (i), we get
(10x- x)=(6.666…) – (0.666…)
9x = 6
x= 6/9
⇒ x=2/3

(ii)
Let x = 0.4\(\bar { 7 }\) = 0.4777… …(iii)
Multiplying Eq. (iii) by 10. we get
10x = 4.777… . …(iv)
Multiptying Eq. (iv) by 10, we get
100x = 47.777 ….. (v)
On subtracting Eq. (v) from Eq. (iv), we get
(100 x – 10x)=(47.777….)-(4.777…)
90x =43
⇒ x = \(\frac { 43 }{ 90 }\)

(iii)
Let x = 0.\(\overline { 001 }\)= 0.001001001… …(vI)
Multiplying Eq. (vi) by (1000), we get
1000x = 1.001001001… .. .(vii)
On subtracting Eq. (vii) by Eq. (vi), we get
(1000x—x)=(1.001001001….) – (0.001001001……)
999x = 1
⇒ x = \(\frac { 1 }{ 999 }\)

Question 4.
Express 0.99999… in the form \(\frac { p }{ q }\)Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.
Solution:
Let x = 0.99999… ………..(i)
Multiplying Eq. (i) by 10, we get
10x = 9.99999… …(ii)
On subtracting Eq. (ii) by Eq. (i), we get
(10 x – x) = (9.99999..) – (0.99999…)
9x = 9
⇒ x = \(\frac { 9 }{ 9 }\)
x = 1

Question 5.
What can the maximum number of digits be in the repeating block of digits in the decimal expansion of \(\frac { 1 }{ 17 }\)? Perform the division to check your answer.
Solution:
The maximum number of digits in the repeating block of digits in the decimal expansion of \(\frac { 1 }{ 17 }\) is 17-1 = 16 we have,
NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.3 img 6
Thus,\(\frac { 1 }{ 17 }\) = 0.\(\overline { 0588235294117647….., }\)a block of 16-digits is repeated.

Question 6.
Look at several examples of rational numbers in the form \(\frac { p }{ q }\) (q≠ 0). Where, p and q are integers with no common factors other that 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy?
Solution:
Consider many rational numbers in the form \(\frac { p }{ q }\) (q≠ 0). where p and q are integers with no common factors other that 1 and having terminating decimal representations.
Let the various such rational numbers be \(\frac { 1 }{ 2 }\), \(\frac { 1 }{ 4 }\), \(\frac { 5 }{ 8 }\), \(\frac { 36 }{ 25 }\), \(\frac { 7 }{ 125 }\), \(\frac { 19 }{ 20 }\), \(\frac { 29 }{ 16 }\) etc.
In all cases, we think of the natural number which when multiplied by their respective denominators gives 10 or a power of 10.
NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.3 img 7
From the above, we find that the decimal expansion of above numbers are terminating. Along with we see that the denominator of above numbers are in the form 2m x 5n, where m and n are natural numbers. So, the decimal representation of rational numbers can be represented as a terminating decimal.

Question 7.
Write three numbers whose decimal expansions are non-terminating non-recurring.
Solution:
0.74074007400074000074…
0.6650665006650006650000…
0.70700700070000…

Question 8.
Find three different irrational numbers between the rational numbers \(\frac { 5 }{ 7 }\) and \(\frac { 9 }{ 11 }\) .
Solution:
To find irrational numbers, firstly we shall divide 5 by 7 and 9 by 11,
so,
NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.3 img 8

Question 9.
Classify the following numbers as rational or irrational
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Solution:
(i) \( \sqrt{23} \) (irrational ∵ it is not a perfect square.)
(ii) \( \sqrt{225} \) = 15 (rational) (whole number.)
(iii) 0.3796 = rational (terminating.)
(iv) 7.478478… =7.\(\bar { 478 }\) = rational (non-terminating repeating.)
(v) 1.101001000100001… = irrational (non-terminating non-repeating.)

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