Rational Numbers Class 9 Worksheet

If you are a Class 9 student studying rational numbers and looking for a comprehensive worksheet on Rational Numbers , this worksheet will help you test your knowledge and understanding of the subject. Rational Numbers Class 9 Worksheet with answers will help you to improve your understanding of rational numbers.

This worksheet is filled with exercises that will require you to think critically and use your understanding of rational numbers. This worksheet is useful whether you are studying for a test or simply reviewing your understanding of rational numbers.

Whether you’re a student looking to revise your understanding of rational numbers or a teacher looking for an effective resource to help your students master this essential mathematical concept, this rational numbers class 9 worksheet with answers pdf worksheet is the perfect for you. So, get your pencils ready and start practicing rational numbers today!

Class 9 Rational Numbers Solved Examples

Example 1: Find 6 rational numbers between $\frac{3}{5}$ and $\frac{4}{5}$

Sol. To solve this problem we will use GAP Method to find rational numbers between two given numbers.

Let $a=\frac{3}{5}$ and $b=\frac{4}{5}$

, then $\frac{3}{5}< \frac{4}{5}$ . Gap between $\frac{3}{5}$

and $\frac{4}{5}$ = $\frac{4}{5}-\frac{3}{5}=\frac{1}{5}$

To find 6 rational numbers , divide by $6+1=7$ .

Dividing the gap by 7, we get $\frac{b-a}{n+1}=\frac{1}{35}$

Thus the 6 rational numbers between $\frac{3}{5}$ and $\frac{4}{5}$

are

$\frac{3}{5}+1\cdot \frac{1}{35}$ , $\frac{3}{5}+2\cdot \frac{1}{35}$

, $\frac{3}{5}+3\cdot \frac{1}{35}$ , $\frac{3}{5}+4\cdot \frac{1}{35}$

, $\frac{3}{5}+5\cdot \frac{1}{35}$ , and $\frac{3}{5}+6\cdot \frac{1}{35}$

i.e. $\frac{22}{35}, \frac{23}{35}, \frac{24}{35}, \frac{25}{35}, \frac{26}{35}\, \, and\, \, \, \frac{27}{35}$ .

Example 2 : Find a rational number between $-\frac{3}{4}$

and $\frac{2}{3}$ .

Sol. Suppose $x$

and $y$ are two rational numbers such that $x< y$

, then $\frac{x+y}{2}$ is a rational number lying between $x$

and $y$ .

Here $x=-\frac{3}{4}$

and $y= \frac{2}{3}$

$\therefore$

required rational number between $x$ and $y$

= $\frac{x+y}{2}$ = $\frac{1}{2}[\left ( -\frac{3}{4} \right )+\frac{2}{3}] = \frac{1}{2}(\frac{2}{3}-\frac{3}{4})$

= $\frac{1}{2}\times (-\frac{1}{12})= -\frac{1}{24}$

Hence $-\frac{1}{24}$

is a rational number between $-\frac{3}{4}$ and $\frac{2}{3}$

Example 3: Find six rational number between 4 and 5 .

Sol. To find six rational numbers between 4 and 5, we will convert given rational numbers into equivalent rational number by multiplying the numerator and denominator by a suitable number , usually (n+1). Here n= 6
We will convert 4 and 5 into equivalent rational numbers by multiplying 7 as multiplying factor.
Thus $4=\frac{4}{1}=\frac{4\times 7}{1\times 7}=\frac{28}{7}$ and $5=\frac{5}{1}=\frac{5\times 7}{1\times 7}=\frac{35}{7}$

Now $\frac{28}{7}< \frac{29}{7}<\frac{30}{7}<\frac{31}{7}<\frac{32}{7}<\frac{33}{7}<\frac{34}{7}<\frac{35}{7}$

or $4< \frac{29}{7}<\frac{30}{7}<\frac{31}{7}<\frac{32}{7}<\frac{33}{7}<\frac{34}{7}<5$

Hence six rational numbers between 4 and 5 are $\frac{29}{7}, \frac{30}{7}, \frac{31}{7}, \frac{32}{7}, \frac{33}{7} \, and\, \frac{34}{7}$ .

Example 4 : Without actual division , find which of the following is a terminating decimal (i) $\frac{5}{32}$

(ii) $\frac{5}{8}$ (iii) $\frac{11}{24}$

Sol. A rational number $\frac{p}{q}$ is a terminating decimal only when prime factors of $q$

are 2 and 5 only.
(i) Denominator of $\frac{5}{32}$ is 32 and $32=2^{5}$

$\therefore$ 32 has no prime factors other than 2.
So, $\frac{5}{32}$

is a terminating decimal.

(ii) Denominator of $\frac{5}{8}$ is 8 and $8=2^{3}$

$\therefore$ 8 has no prime factors other than 2.
So, $\frac{5}{8}$

is a terminating decimal.

(iii) Denominator of $\frac{11}{24}$ is 24 and $24=2^{3}\times 3$

$\therefore$ 24 has a prime factor 3, which is other than 2 and 5.
So, $\frac{11}{24}$

is not a terminating decimal.

Example 5: Write the following in decimal form and say what kind of decimal expansion each has :

(a) $\frac{3}{14}$
Sol. By actual division , we have $\frac{3}{14}$

= 0.2142857142857………….. = $0.2\overline{142857}$
Clearly , $\frac{3}{14}$

has a non-terminating recurring decimal representation.

(b) $\frac{1}{11}$
Sol. By actual division , we have $\frac{1}{11}$

= 0.09090909………….. = $0.\overline{09}$
Clearly , $\frac{1}{11}$

has a non-terminating recurring decimal representation.

(c) $5\frac{1}{2}$
Sol. By actual division , we have $5\frac{1}{2}=\frac{11}{2} = 5.5$

Clearly , $5\frac{1}{2}$ has a terminating decimal representation.

Example 6: Is zero a rational number? Can it be written in the form p/q, where p and q are integers and q ≠ 0?

Solution: Yes, $0=\frac{0}{1}=\frac{0}{2}= ......$

Thus zero is a rational number. It can be written in the form of $\frac{p}{q}$ where p and q are integers and q ≠ 0.

Example 7: State whether the following numbers are rational or not:

(i) (𝟐 + √𝟐)^𝟐(ii) (𝟑 − √𝟑)^𝟐(iii) (𝟓 + √𝟓)(𝟓 − √𝟓)
(iv) (√𝟑 − √𝟐)^𝟐

Sol. (i) By using the formula $(a+b)^{2}=a^{2}+b^{2}+2ab$

(2 + √2)² = 2² + 2(2)(√2) + (√2)² = 4 + 4√2 + 2
= 6 + 4√2
Hence, it is irrational

(ii) (3 – √3)² = (3)² – 2(3)( √3) + (√3)²= 9 – 6√3 + 3
= 12 – 6√3
= 6(2 – √3)
Therefore, it is irrational.

(iii) Using the formula $(a-b)(a+b)=a^{2}-b^{2}$
(5 + √5)(5 – √5) = (5)² – (√5)²= 25 – 5
= 20
Hence, it is rational.

(iv) (√3 – √2)² = (√3)² – 2(√3)(√2) + (√2)²= 3 – 2√6 + 2
= 5 – 2√6
Therefore, it is irrational.

Example 8: Insert a rational number between and 2/9 and 3/8 arrange in descending order.

Solution: Given Rational numbers are $\frac{2}{9}$

and $\frac{3}{8}$ .

Let us rationalize the numbers, by taking LCM for denominators 9 and 8 which is 72.

$\frac{2}{9}=\frac{2 \times 8}{9\times 8}=\frac{16}{72}$

, $\frac{3}{8}=\frac{3 \times 9}{8\times 9}=\frac{27}{72}$

Since $\frac{16}{72}<\frac{27}{72}$

So, 2/9 < 3/8. The rational number between $\frac{2}{9}$ and $\frac{3}{8}$

is $\frac{\frac{2}{9}+\frac{3}{8}}{2}= \frac{\frac{16+27}{72}}{2} = \frac{16+27}{2\times 72}$ = $\frac{43}{144}$

Hence, $\frac{3}{8}>\frac{43}{144}>\frac{2}{9}$ .

The descending order of the numbers is $\frac{3}{8}>\frac{43}{144}>\frac{2}{9}$

Example 9: Write three numbers whose decimal expansion are non-terminating and non-repeating.
Sol. We know that the decimal expansion of an irrational number is non-terminating and non-repeating.
So, the required numbers are $\sqrt{2}, \sqrt{3} \, \mathrm{and }\, 2+ \sqrt{3}$ .

Example 10: Find two rational and irrational numbers between $\sqrt{2}$

and $\sqrt{3}$ .
Sol. We have $\sqrt{2}=1.41421356............$

and $\sqrt{3}= 1.732050807...........$
If we consider the numbers 1.5 and 1.6 between $\sqrt{2}$

and $\sqrt{3}$ , then both of them are rational numbers.
Hence two rational numbers between $\sqrt{2}$

and $\sqrt{3}$ are 1.5 and 1.6 .
Now consider two numbers a= 1.500067839402…………………….. and b= 1.647834934839048394……………..
Clearly , a and b are two irrational numbers between $\sqrt{2}$

and $\sqrt{3}$ .

Example 11. Insert two irrational numbers between 5 and 6.
Sol: Let’s write 5 and 6 as square root , then, 5 = √25 and 6 = √36
Now, take the numbers √25 < √26 < √27 < √28 < √29 < √30 < √31 < √32 < √33 < √34 < √35 < √36.
Hence, any two irrational numbers between 5 and 6 is √29 and √30

Example 12 : Are the square roots of all positive integers irrational? If not, give an example of the square root of a number that is a rational number.
Sol. No, the square roots of all positive integers are not irrational. For example,
√4 = 2 is rational.
√25 = 3 is rational.
Hence, the square roots of positive integers 4 and 25 are not irrational.

Example 13 : Find three different irrational numbers between the rational numbers 5/7 and 9/11.
Sol. We have $\frac{5}{7}= 0.\overline{714285}$

and $\frac{9}{11}=0.\overline{81}$
Three different irrational numbers are:
(i) 0.74074007400074000074…
(ii) 0.77077007700077000077…
(iii) 0.80080008000080000080…

Example 14: Express the following in the form $\frac{p}{q}$

, where $p$ and $q$

are integers and $q\neq 0$ .
(i)

Sol. Ncert solution class 9 chapter 1-11

Assume that $x = 0.666.......$

Then, $10x = 6.6666........$
$10x = 6 + x$

$9x = 6$
$\Rightarrow x = \frac{2}{3}$

(ii) $0.4\overline{7}$

\begin{array}{l} 0.4 \overset{―}{7} = 0.4777 . . \end{array}

$= \frac{4}{10}+\frac{0.777...}{10}$

Assume that $x = 0.777........$
Then, $10x = 7.777........$

$10x = 7 + x$
$x = \frac{7}{9}$

$= \frac{4}{10}+\frac{0.777...}{10}$ $= \frac{4}{10}+ \frac{7}{90}$

$= \frac{36}{90}+\frac{7}{90} = \frac{43}{90}$

Therefore $0.4\overline{7}$

= $\frac{43}{90}$ .

Rational numbers class 9 (Unsolved examples with answers)

(1) Express each of the following recurring decimals as a rational numbers.

(a) $0.\bar{5}$

(b) $0.\overline{13}$
(c) $0.\overline{341}$

(d) $1.\overline{27}$
(e) $0.2\overline{35}$

2) State whether the following statements are true or false.
(a) Every rational number is a whole number.
(b) There are infinitely many rational numbers between any two given rational numbers.
(c) $\sqrt{17}$ is a rational number.
(d) Every irrational number is a real number.
(e) Every point on the number line is of the form $\sqrt{m}$

, where m is a natural number.
(f) Every real number is an irrational number.
(g) A number whose decimal expansion is terminating or non-terminating recurring is rational.
(h)The decimal expansion of an irrational number is non-terminating non-recurring.
(i) There is no least or greatest rational number.
(j) Every integer is a rational number.
(k) Every rational number is an integer.
(l) A decimal that ends after a finite number of digits is called terminating decimal .
(m) Every rational is expressible either as a terminating decimal or non-terminating recurring decimal.
(n) A number whose decimal expansion is terminating or non-terminating is not a rational.
(o) Non- terminating and non-repeating decimals are irrational numbers.

3) Express $2.\overline{36}+ 0.\overline{23}$ as a fraction in simplest form .