Advertisement:


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)

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

(ii)

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

(iii)

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

(iv)

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

(v)

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

(vi)

  1. (i) Construct a $\Delta OAB$ which $m \overline{OA} = 2$ unit and $\perp BA = 1$
  2. (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 $$x = 0.5555\ldots.\qquad (1)$$

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

$$10x = (0.5555\ldots..) \times 10$$ $$10x = 5.5555\ldots.. \qquad (2)$$ 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 $$x = 0.13131313\ldots. \qquad (1)$$ Only one digit 13 is being repeated, multiply 100 on both sides $$100x = (0.13131313\ldots..) \times 100$$ $$100x = 13.13131313\ldots.. \qquad (2)$$ 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 $$x = 0.67676767\ldots. \qquad (1)$$ Only one digit 67 is being repeated, multiply 100 on both sides $$100x = (0.67676767\ldots..) \times 100$$ $$100x = 67.67676767\ldots.. \qquad (2)$$ Subtract (1) from (2) $$99x = 67$$ $$\therefore\,\, x = \frac{67}{99}$$ $$0.\overline{67} = \frac{67}{99}$$



Advertisement: