# Exercise 2.1 (Solutions)

### Question 1

Identify which of the following are rational and irrational numbers:

(i) $\sqrt{3}$ (ii) $\frac{1}{6}$ (iii) $\pi$ (iv) $\frac{15}{2}$ (v) $7.25$ (vi)$\sqrt{29}$

**Solution**

- Rational: $\frac{1}{6}$, $\frac{15}{2}$, $7.25$
- Irrational: $\sqrt{3}$, $\pi$, $\sqrt{29}$

### Question 2

Convert the following fraction into decimal fraction.

(i) $\frac{17}{25}$ (ii) $\frac{19}{4}$ (iii)$\frac{57}{8}$

(iv) $\frac{205}{18}$ (v) $\frac{5}{8}$ (vi) $\frac{25}{38}$

**Soluation**

- (i) o.68
- (ii) 4.75
- (iii) 7.125
- (iv) 11.3889
- (v) 0.625
- (vi) .65789

### Question 3

Which of the statements are true and which of the false?

- (i) $\frac{2}{3}$ is an irrational number.
- (ii) $\pi$ is an irrational number.
- (iii) $\frac{1}{9}$ is a terminating fraction.
- (iv) $\frac{3}{4}$ is a terminating fraction.
- (v) $\frac{4}{5}$ is a recurring fraction.

**Soluaton**

- (i) $\frac{2}{3}$ is an irrational number.
**False** - (ii) $\pi$ is an irrational number.
**True** - (iii) $\frac{1}{9}$ is a terminating fraction.
**False** - (iv) $\frac{3}{4}$ is a terminating fraction.
**True** - (v) $\frac{4}{5}$ is a recurring fraction.
**False**

### Question 4

Represent the following numbers on the number line.

(i) $\frac{2}{3}$ (ii) $-\frac{4}{5}$ (iii) $\frac{3}{4}$

(iv) $-2\frac{5}{8}$ (v) $2\frac{3}{4}$ (vi) $\sqrt{5}$

**Soluation**

(i)

- To represent the rational number $\frac{2}{3}$ , divide unit length between 0 and 1 into 3 equal parts.
- (ii) Take 2 parts on right of 0
- (iii) Point M represents $\frac{2}{3}$ on the number line in the following figure.

(ii)

- To represent the rational number $-\frac{4}{5}$ , divide unit length between 0 and -1 into 5 equal parts.
- Take 4 parts on right of 0
- Point M represents $-\frac{4}{5}$ on the number line in the following figure.

(iii)

- Rational number $\frac{3}{4}$ lies between 1 and 2 on the number line.
- The distance between 1 and 2 is divided into 4 equal parts , from L we take 3 parts.
- Point M represent $\frac{3}{4}$ on the number line.

(iv)

- Rational number $-2\frac{5}{8}$ lies between 2 and -3 on the number line.
- The distance between 2 and -3 is divided into 8 equal parts , from 2 we take 5 parts.

(v)

- Rational number $2\frac{3}{4}$ lies between 2 and 3 on the number line.
- The distance between 2 and 3 is divided into 4 equal parts , from 2 we take 3 parts.

(vi)

- (i) Construct a $\Delta OAB$ which $m \overline{OA} = 2$ unit and $\perp BA = 1$
- (ii) Take O as centre and draw an arc of radius $\overline{OB}$. It cuts the number line at M. $m \overline{OM} = \sqrt{5}$

$$(m OB)^2 = (m OA)^2 + (m AB)^2 \quad (Pythagorean\,\, Theorem)$$ $$(m \overline{OB})^2 = 4 + 1 = 5 $$ therefore $$m \overline{OB} = \sqrt{5}$$ thus $$m \overline{OB} = m \overline{OM} = \sqrt{5}$$

### Question 5

Give a rational number between $\frac{3}{4}$ and $\frac{5}{9}$.

**Soluation**

The mean of the numbers is between given numbers. Therefore $\frac{3/4+5/9}{2}= \frac{47}{72}$ is a number between required numbers.

### Question 6

Express the following recurring decimals as the rational number $\frac{p}{q}$, where p , q are integers and $q \neq 0$.

(i) $0.\overline{5}$ (ii) $0.\overline{13}$ (iii) $0.\overline{67}$

**Soluation**

(i) Let $x = 0.\overline{5}$. That is $$x = 0.5555....\qquad (1)$$

Only one digit 5 is being repeated, multiply 10 on both sides

$$10x = (0.5555.....) \times 10$$ $$10x = 5.5555..... \qquad (2)$$ Subtract (1) from (2), we get

$$9x = 5$$ $$\therefore \,\, x = \frac{5}{9}$$ $$0.\overline{5} = \frac{5}{9}$$

(ii) Let $x = 0.\overline{13}$. That is $$x = 0.13131313.... \qquad (1)$$ Only one digit 13 is being repeated, multiply 100 on both sides $$100x = (0.13131313.....) \times 100$$ $$100x = 13.13131313..... \qquad (2)$$ Subtract (1) from (2) $$99x = 13$$ $$\therefore \,\, x = \frac{13}{99}$$ $$0.\overline{13} = \frac{13}{99}$$

(iii) Let $x = 0.\overline{67}$. That is $$x = 0.67676767.... \qquad (1)$$ Only one digit 67 is being repeated, multiply 100 on both sides $$100x = (0.67676767.....) \times 100$$ $$100x = 67.67676767..... \qquad (2)$$ Subtract (1) from (2) $$99x = 67$$ $$\therefore\,\, x = \frac{67}{99}$$ $$0.\overline{67} = \frac{67}{99}$$