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

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. 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.

error: Content is protected !!