NCERT Solutions Class 9 Maths Chapter 1 Number Systems Ex 1.3

NCERT Solutions Class 9 Maths
Chapter – 1 (Number Systems )

The NCERT Solutions in English Language for Class 9 Mathematics Chapter – 1 Number Systems Exercise 1.3 has been provided here to help the students in solving the questions from this exercise.

Chapter 1: Number Systems

Exercise – 1.3

1. Write the following in decimal form and say what kind of decimal expansion each has :

Answer –

(i) $\mathbf{\frac{36}{100}}$

= 0.36 (Terminating)

(ii) $\mathbf{\frac{1}{11}}$

= 0.0909… = $0.\overline{09}$ (Non Terminating and repeating)

(iii) $\mathbf{4\frac{1}{8}}$ = $\mathbf{\frac{33}{8}}$

NCERT Class 9 Solutions Maths
= 4.125 (Terminating)

(iv) $\mathbf{\frac{3}{13}}$

= 0.230769… = $0.\overline{230769}$ (Non Terminating and repeating)

(v) $\mathbf{\frac{2}{11}}$

= 0.1818… = $0.\overline{18}$ (Non Terminating and repeating)

(vi) $\mathbf{\frac{329}{400}}$

= 0.8225 (Terminating)

2. You know that $\mathbf{\frac{1}{7}}$ = $\mathbf{0.\overline{142857}}$ Can you predict what the decimal expansions of $\mathbf{\frac{2}{7}, \frac{3}{7},\frac{4}{7},\frac{5}{7},\frac{6}{7}}$ are, without actually doing the long division? If so, how?
[Hint: Study the remainders while finding the value of $\mathbf{\frac{1}{7}}$ carefully.]
Answer –
$\frac{1}{7} = 0.\overline{142857}$

$\frac{2}{7} = 2\times \frac{1}{7} =$ 2 × $0.\overline{142857}$ = $0.\overline{285714}$

$\frac{3}{7} = 3\times \frac{1}{7} =$ 3 × $0.\overline{142857}$ = $0.\overline{428571}$

$\frac{4}{7} = 4\times \frac{1}{7} =$ 4 × $0.\overline{142857}$ = $0.\overline{571428}$

$\frac{5}{7} = 5\times \frac{1}{7} =$ 5 × $0.\overline{142857}$ = $0.\overline{714285}$

$\frac{6}{7} = 6\times \frac{1}{7} =$ 6 × $0.\overline{142857}$ = $0.\overline{857142}$

3. Express the following in the form $\mathbf{\frac{p}{q}}$ , where p and q are integers and q ≠ 0.
(i) $\mathbf{0.\overline{6}}$
(ii) $\mathbf{0.4\overline{7}}$
(iii) $\mathbf{0.\overline{001}}$
Answer –
(i) $\mathbf{0.\overline{6}}$ = 0.666..
Assume that x = 0.666… ————- (i)
Multiplying both sides of the equation (i) by 10,
Then,
10x = 6.666… ————- (ii)
Subtracting the equation (i) from (ii),
10x = 6.666…
– x = 0.666…
9x = 6
Therefore, x = $\frac{2}{3}$
So, the decimal number becomes
$0.\overline{6}$ = $\frac{2}{3}$ and it is in the required $\frac{p}{q}$ form.

(ii) $\mathbf{0.4\overline{7}}$ = 0.47777…..
Assume that x = 0.4777… ————- (i)
Multiplying both sides of the equation (i) by 10,
Then,
10x = 4.777… ————- (ii)
Subtracting the equation (i) from (ii),
10x = 4.777…
– x = 0.4777…
9x = 4.3
Therefore, x = $\frac{4.3}{9}$ = $\frac{4.3}{9}$ × $\frac{10}{10}$ = $\frac{43}{10}$
So, the decimal number becomes
$0.4\overline{7}$ = $\frac{43}{10}$ and it is in the required $\frac{p}{q}$ form.

(iii) $\mathbf{0.\overline{001}}$ = 0.001001…
Assume that x = 0.001001… ————- (i)
Multiplying both sides of the equation (i) by 1000,
Then,
Then, 1000x = 1.001001… ————- (ii)
Subtracting the equation (i) from (ii)
1000x = 1.001001..
– x = 0.001001
999x = 1
Therefore, x = $\frac{1}{999}$
So, the decimal number becomes
$0.\overline{001}$ = $\frac{1}{999}$ and it is in the required $\frac{p}{q}$ form.

4. Express 0.99999…. in the form $\mathbf{\frac{p}{q}}$ . Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.
Answer – Assume that x = 0.9999 ….. ————– (i)
Multiplying both sides by 10,
10x = 9.9999…. ————– (ii)
Subtracting the equation (i) from (ii)
10x = 9.9999..
– x = 0.9999..
9x = 9
Therefore, x = 1
The difference between 1 and 0.999999 is 0.000001 which is negligible.
Hence, we can conclude that, 0.999 is too much near 1, therefore, 1 as the answer can be justified.

5. What can the maximum number of digits be in the repeating block of digits in the decimal expansion of $\mathbf{\frac{1}{17}}$ ? Perform the division to check your answer.
Answer – $\frac{1}{17}$
Dividing 1 by 17:
NCERT Class 9 Solutions Maths
$\frac{1}{17}$ = $0.\overline{0588235294117647}$
There are 16 digits in the repeating block of the decimal expansion of $\frac{1}{17}$ .

6. Look at several examples of rational numbers in the form $\mathbf{\frac{p}{q}}$ (q ≠ 0), where p and q are integers with no common factors other than 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy?
Answer – We observe that when q is 2, 4, 5, 8, 10… Then the decimal expansion is terminating.
For example:
$\frac{1}{2}$ = 0. 5, denominator q = 2¹

$\frac{7}{8}$ = 0. 875, denominator q =2³

$\frac{4}{5}$ = 0. 8, denominator q = 5¹We can observe that the terminating decimal may be obtained in the situation where prime factorization of the denominator of the given fractions has the power of only 2 or only 5 or both.

7. Write three numbers whose decimal expansions are non-terminating non-recurring.
Answer – We know that all irrational numbers are non-terminating non-recurring. three numbers with decimal expansions that are non-terminating non-recurring are:
√2 = 1.41421….
√3 = 1.73205….
√7 = 2.645751….

8. Find three different irrational numbers between the rational numbers $\mathbf{\frac{5}{7}}$ and $\mathbf{\frac{9}{11}}$ .
Answer – Converting $\frac{5}{7}$ and $\frac{9}{11}$ into the decimal form gives
$\frac{5}{7}$ = 0.714285…..

$\frac{9}{11}$ = 0.818181…..

Therefore, 3 irrational numbers that are contained between 0.714285….. and 0.818181….. are:
0.73073007300073000073…, 0.75075007300075000075… and 0.76076007600076000076…

9. Classify the following numbers as rational or irrational according to their type:
(i) $\mathbf{\sqrt{23}}$
(ii) $\mathbf{\sqrt{225}}$
(iii) 0.3796
(iv) 7.478478
(v) 1.101001000100001…
Answer –
(i) $\mathbf{\sqrt{23}}$ = 4.79583152331…
Since the number is non-terminating and non-recurring therefore, it is an irrational number.

(ii) $\mathbf{\sqrt{225}}$ = 15 = $\frac{15}{1}$
Since the number can be represented in $\frac{p}{q}$ form, it is a rational number.

(iii) 0.3796
Since the number,0.3796, is terminating, it is a rational number.

(iv) 7.478478
The number,7.478478, is non-terminating but recurring, it is a rational number.

(v) 1.101001000100001…
Since the number,1.101001000100001…, is non-terminating non-repeating (non-recurring), it is an irrational number.

NCERT Solutions Class 9 Maths Chapter 1 Number Systems Ex 1.3