These NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.3 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Exercise 1.3

Question 1.

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

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

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

(iii) \(4 \frac{1}{8}\)

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

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

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

Solution:

(i) By actual division we have

The decimal form of \(\frac{36}{100}\) is 0.36

Here, the remainder becomes 0, therefore it has a terminating decimal expansion.

(ii) By actual division we have

The decimal form of \(\frac{1}{11}\) is 0.090909….. of \(0 . \overline{09}\).

Hence remainder never becomes zero but repeats. Therefore, it has non terminating but repeating decimal expansion.

(iii) \(4 \frac{1}{8}\)

We can also write \(\frac{33}{8}\)

By actual division we have,

The decimal form of \(4 \frac{1}{8}\) is 4.125.

Here remainder becomes zero after certain steps. Therefore it has a terminating decimal expansion.

(iv) By actual division we have

The decimal form of \(\frac{3}{13}\) is 0.2307692307….. or \(0 . \overline{230769}\).

Here, the remainder never becomes zero but repeats after some steps.

Therefore, it has a non terminating but repeating decimal expansion.

(v) By actual division we have

The decimal form of \(\frac{2}{11}\) is 0.181818…. or \(0 . \overline{18}\). Here remainder never becomes zero but repeat after some steps.

Therefore, it has non terminating but repeating decimal expansion.

(vi) By actual division we have,

The decimal form of \(\frac{329}{400}\) is 0.8225. Here remainder becomes zero after some steps. Therefore, it has terminating decimal expansion.

5/9 as a decimal is 0.56

Question 2.

You know that \(\frac{1}{7}=0 . \overline{142857}\). Can you predict 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 devision? If so how?

Solution:

We have given that

\(\frac{1}{7}=0 . \overline{142857}\)

Question 3.

Express the following in the form \(\frac {p}{q}\), where p and q are integers and q ≠ 0.

(i) \(0 . \overline{6}\)

(ii) \(0 . \overline{47}\)

(iii) \(0 . \overline{001}\)

Solution:

Let x = \(0 . \overline{6}\)

or, x = 0.66666…… (i)

Now, add 6 both side in equation (i)

x + 6 = 0.66666…… + 6

or x + 6 = 6.6666……. (ii)

Again, multiply 10 both side in equation (i)

10x = 0.66666…. × 10

or, 10x = 6.6666…. (iii)

Now, from equation (ii) and (iii)

x + 6 = 10x

9x = 6

∴ x = \(\frac{6}{9}\)

Therefore, \(0 . \overline{6}\) = x = \(\frac{2}{3}\)

(ii) Let x = \(0.4 \overline{7}\)

or x = 0.47777……. (i)

Now, multiply 10 both side in equation (i)

10x = 0.47777……. × 10

or, 10x = 4.77777…… (ii)

Again, multiply 100 both side in equation (i)

100x = 0.47777 × 100

or, 100x = 4.77777…….. (iii)

Subtract equation (ii) from equation (iii)

90x = 43.0000

or 90x = 43

∴ x = \(\frac{43}{90}\)

Therefore, \(0.4 \overline{7}\) = x = \(\frac{43}{90}\)

which is in the form of \(\frac {p}{q}\)

(iii) Let x = \(0 . \overline{001}\)

or, x = 0.001001001……. (i)

Now, add 1 both side in equation (i)

x + 1 = 0.001001001…… + 1

or x + 1 = 1.001001001……. (ii)

Again, multiply 1000 both side in equation (i)

1000 × x = 1000 × 0.001001001……..

or, 1000x = 1.001001001….. (iii)

From equation (ii) and (iii) we get

x +1 = 1000x

or 999x = 1

or x = \(\frac{1}{999}\)

Therefore, \(0 . \overline{001}=x \times \frac{1}{999}\)

which is in the form of \(\frac {p}{q}\)

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 = 0.9999…… (i)

Add 9 both side in equation (i)

x + 9 = 9 + 0.99999………

or, x + 9 = 9.99999…… (ii)

Again, multiply 10 both side in equation (i)

10x = 9.9999……. (iii)

From equation (ii) and (iii) we get

10x = x + 9

or, 9x = 9

or, x = 1

Therefore, 0.9999…… = 1

It is because there is infinite 9 comes after the point; which is very-very close to 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:

By actual division we have

It is clear that, \(\frac{1}{17}=0 . \overline{0588235294117647}\)

It is clear that, \(\frac{1}{17}=0 . \overline{0588235294117647}\)

So, the maximum number of digits be in the repeating block of digits in die decimal expansion of \(\frac{1}{17}\) is 16.

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 than 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy?

Solution:

Some examples of a rationed number having terminating decimal representations are:

(i) \(\frac{1}{2}\) = 0.5

(ii) \(\frac{7}{4}\) = 1.75

(iii) \(\frac{7}{8}\) = 0.875

(iv) \(\frac{2}{5}\) = 0.4

It is clear that the prime factorization of q has only power of 2 or power of 5 or both.

Question 7.

Write three numbers whose decimal expansions are non-terminating non-recurring.

Solution:

We knew that the decimal expansions of an irrational number are non-terminating non-recurring. Three examples of such numbers are:

√2 = 1.4142135…….

√3 = 1.732050807………

π = 3.1415926535………

Question 8.

Find three different irrational numbers between the rational numbers \(\frac{5}{7}\) and \(\frac{9}{11}\).

Solution:

To find an irrational numbers between \(\frac{5}{7}\) and \(\frac{9}{11}\) is non terminating non recurring lying between them.

Here, \(\frac{5}{7}=0 . \overline{714285}\) and \(\frac{9}{11}=0 . \overline{81}\)

Therefore, the required three different irrational number which is lying between \(\frac{5}{7}\) and \(\frac{9}{11}\) are 0.720720072000…….., 0.730730073000……., and 0.740740074000………

Question 9.

Classify the following number as rational or irrational:

(i) √23

(ii) √225

(iii) 0.3796

(iv) 7.478478…….

(v) 1.101001000100001…….

Solution:

(i) We have

√23 = 4.795831523……..

It is non-terminating non-recurring.

So, √23 is an irrational number

(ii) We have

√225 = 15

or √225 = \(\frac{15}{1}\)

which is a rational number

(iii) We have,

0.3796 = \(\frac{3796}{10000}\)

which is in the form of \(\frac{p}{q}\)

So, 0.3796 is a rational number.

(iv) We have

7.478478……. = \(7 . \overline{478}\) which is non terminating recurring.

Therefore, 7.478478…….. is an irrational number.

(v) We have, 1.101001000100001……..

which is non-terminating non-recurring.

Therefore, 1.101001000100001……… is an irrational number.