Number Systems (Chapter 1, Exercise 1.3) – Class 9 Maths Solutions) NCERT

This page contains step by step solutions of Exercise 1.3 of Chapter 1 of Class 9 Mathematics under NCERT Syllabus.

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

i. \frac{36}{100}

Answer: Terminating Decimal Expansion.

Explanation:

\frac{36}{100}=0.36

Since the value of \frac{36}{100} is 0.36, therefore it is a Terminating Decimal Expansion.

ii. \frac{1}{11}

Answer: Non-Terminating Recurring

Explanation:

\frac{1}{11}=0.090909090909….=0.\overline{09}

The number ‘09’ repeats itself after decimal if we keep on dividing 1 by 11 and it is never ending. Therefore, the given Decimal Expansion is Recurring and Non-Terminating.

iii. 4\frac{1}{8}

Answer: Terminating.

Explanation:

4\frac{1}{8}=\frac{32+1}{8}=\frac{33}{8}=4.125

The value of 4\frac{1}{8} is 4.125 which is a Terminating Number. Therefore, the above decimal expression is a Terminating Decimal Expansion.

iv. \frac{3}{13}

Answer: Non-Terminating Recurring.

Explanation:

\frac{3}{13}=0.230769230769….=0.\overline{230769}

If we convert the fraction \frac{3}{13} into decimal, then the number 230769 keeps on repeating after decimal and it never ends. Therefore, the above Decimal Expansion is called Recurring and Non-Terminating Decimal Expansion.

v. \frac{2}{11}

Answer: Non-Terminating Recurring.

Explanation:

\frac{2}{11}=0.181818181818….=0.\overline{18} If we find the decimal expansion for the fraction \frac{2}{11}, the number 18 will keep on repeating after decimal and never ends. Therefore, the Decimal Expansion is Recurring and Non-Terminating.

vi. \frac{329}{400}

Answer: Terminating.

Explanation:

\frac{329}{400}=0.8225

The decimal expression of the fraction \frac{329}{400} is 0.8225 which is a Terminating Decimal Expansion.

Question 2: You know that \frac{1}{7}=0.\overline{142857}. Can you what the decimal expansions of \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?

Solution:

Given, \frac{1}{7}=0.\overline{142857}

The fraction \frac{2}{7} can be written as, \frac{2}{7}=2\times \frac{1}{7}

Therefore, \frac{2}{7}=2\times 0.\overline{142857}=0.\overline{285714}

Similarly,

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

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

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

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

Question 3: Express the following in the form \frac{p}{q}, where p and q are integers and q\ne 0?

i. 0.\overline {6}

Solution:

We know that, 0.\overline {6}=0.666666…

Now, let us consider

x=0.666666…

Multiplying by 10 in both sides of the equation,

10\times x=10\times 0.666666…

\Rightarrow 10x=6.66666…

\Rightarrow 10x=6+0.66666…

\Rightarrow 10x=6+x    [since, x=0.66666…]

\Rightarrow 10x-x=6

\Rightarrow 9x=6

\Rightarrow x=\frac{6}{9}

\Rightarrow x=\frac{2}{3}

Therefore, the \frac{p}{q} form of 0.\overline{6} is \frac{2}{3}.

ii. 0.4\overline {7}

Solution:

We know that, 0.4\overline {7}=0.477777…

Let us consider,

x=0.477777…\,\,\,\,\,\,\,\,\,\,\, ………..(i) \therefore\,\,\,10x=4.7777…

Multiplying by 100 in both sides of the equation(i),

100x=47.7777…

\Rightarrow\,100x=43+4.7777…

\Rightarrow\,100x=43+10x

\Rightarrow\,100x-10x=43

\Rightarrow\,90x=43

\Rightarrow\,x=\frac {43}{90}

Therefore, the \frac{p}{q} form of 0.4\overline {7} is \frac{43}{90}.

iii. 0.\overline {001}

Solution:

We know that, 0.\overline {001} =0.001001001001…

Let us consider,

x=0.001001…

Multiplying by 1000 in both sides of the above equation,

1000x=1.001001…

\Rightarrow\,1000x=1+0.001001…

\Rightarrow\,1000x=1+x

\Rightarrow\,1000x-1x=1

\Rightarrow\,999x=1

\Rightarrow\,x=\frac{1}{999}

Therefore, the \frac{p}{q} form of 0.\overline{001} is \frac{1}{999}.

Question 4: Express 0.99999… in the form \frac{p}{q}.

Solution:

Let us consider, x=0.99999…

Multiplying both sides by 10,

10x=9.9999…

\Rightarrow\,10x=9+0.9999…

\Rightarrow\,10x=9+x

\Rightarrow\,10x-x=9

\Rightarrow\,9x=9

\Rightarrow\,x=\frac{9}{9}=1

Therefore, the \frac{p}{q} form of 0.99999 is 1.

Question 5: What can the maximum number of digits be in one repeating block of digits in the decimal expression of \frac {1}{17}?

Solution:

If we divide the number 1 by 17, then the Quotient will be,

\frac{1}{17}=0.05882352941176470588235294117647=0.\overline {0588235294117647}

From the above, it is clear that the sixteen numbers i.e., 0588235294117647 is repeating itself.

Therefore, the maximum number of digits in one repeating block is 16.

Question 6: Write three numbers whose decimal expansions are non-terminating non-recurring?

Solution:

Three numbers whose decimal expansions are non-terminating and non-recurring are:

  1. \pi=3.1415926536…
  2. \sqrt{2}=1.41421356…
  3. \sqrt{5}=2.23606798…

Question 7: Find three different irrational Numbers between the rational numbers \frac {5}{7} and \frac {9}{11}.

Solution:

The decimal expansion of \frac{5}{7} and \frac{9}{11} will be:

\frac{5}{7}=0.714285714285…=0.\overline{714285} \frac{9}{11}=0.818181…=0.\overline{81}

We know that the Decimal Expansion of an Irrational Number is Non-Terminating and Non-Recurring and three such numbers between \frac{5}{7} and \frac{9}{11} are,

  1. 0.730730073000730000…
  2. 0.750750075000750000…
  3. 0.791791179111791111…

Question 8: Classify the following Numbers as Rational or irrational

i. \sqrt{23}

Solution:

\sqrt{23}=4.795831523312… From above, we have seen that the value of \sqrt{23} is Non-Terminating and Non-Recurring. Therefore, \sqrt{23} is an Irrational Number.

ii. \sqrt{225}

Solution:

\sqrt{225}=15

The value of \sqrt{225} is Terminating. Hence, \sqrt {225} is a Rational Number.

iii. 0.3796

Solution: The given number 0.3796 is a Rational Number because it is Terminating.

iv. 7.478478…

Solution:

7.478478…=7.\overline{478}

Here, the number 7.478478… is Non-Terminating but Recurring. Therefore, the number is a Rational Number.

v. 1.101001000100001…

Solution: The given number is Non-Terminating and Non-Recurring. Therefore, it is an Irrational Number.

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