Rational & Irrational Numbers — Competency Focused Questions

The sum of all rational numbers between 0 and 0.1 is :

1. finite

2. infinite

3. can't say anything

4. none of these

Answer

Between any two real numbers like 0 and 0.1, there are infinitely many rational numbers. So the sum is infinite.

Hence, Option 2 is the correct option.

Four rational numbers p, q, r and s are such that q is the reciprocal of p and s is the reciprocal of r. The value of the expression $\Big([p + \dfrac{1}{q}] ÷ [r + \dfrac{1}{s}]\Big) ÷ \Big([s + \dfrac{1}{r}] ÷ [q + \dfrac{1}{p}]\Big)$ is equal to:

1. 1

2. 0

3. pr

4. $\dfrac{s}{q}$

Answer

Given,

$\Rightarrow q = \dfrac{1}{p} \\[1em] \Rightarrow \dfrac{1}{q} = p \\[1em] \Rightarrow s = \dfrac{1}{r} \\[1em] \Rightarrow \dfrac{1}{s} = r.$

Substituting value from above in given equation,

$\Rightarrow \Big([p + p] ÷ [r + r]\Big)÷ \Big([s + s] ÷ [q + q]\Big) \\[1em] \Rightarrow \Big(\dfrac{2p}{2r}\Big) ÷ \Big(\dfrac{2s}{2q}\Big) \\[1em] \Rightarrow \dfrac{p}{r} ÷ \dfrac{s}{q} \\[1em] \Rightarrow \Big(\dfrac{p}{r}\Big) \times \Big(\dfrac{q}{s}\Big) \\[1em] \Rightarrow \dfrac{p}{r} \times q \times \dfrac{1}{s} \\[1em] \Rightarrow \dfrac{p}{r} \times \dfrac{1}{p} \times r \\[1em] \Rightarrow 1.$

Hence, Option 1 is the correct option.

$0.6 + 0.\overline{7} + 0.4\overline{7}$ is equal to s:

1. $\dfrac{155}{90}$

2. $\dfrac{147}{90}$

3. $\dfrac{167}{90}$

4. none of these

Answer

Convert repeating decimals to fraction,

Let, x = $0.\overline{7}$

⇒ x = 0.7777....     .......(1)

⇒ 10x = 7.777...     .......(2)

Subtracting equation (1) from (2), we get :

⇒ 10x - x = 7.777..... - 0.777.....

⇒ 9x = 7

⇒ x = $\dfrac{7}{9}$

Let, y = $0.4\overline{7}$

⇒ 10y = 4.7777....     .......(3)

⇒ 100y = 47.777...     .......(4)

Subtracting equation (3) from (4), we get :

⇒ 100y - 10y = 47.777..... - 4.777.....

⇒ 90y = 43

⇒ y = $\dfrac{43}{90}$

$\Rightarrow 0.6 + 0.\overline{7} + 0.4\overline{7} = 0.6 + \dfrac{7}{9} + \dfrac{43}{90} \\[1em] = \dfrac{6}{10} + \dfrac{7}{9} + \dfrac{43}{90} \\[1em] = \dfrac{54 + 70 + 43}{90} \\[1em] = \dfrac{167}{90}.$

Hence, Option 3 is the correct option.

$\sqrt[4]{\sqrt[3]{3^2}}$ can be expressed as :

1. $3^6$

2. $6^{1/3}$

3. $3^{1/12}$

4. $3^{1/6}$

Answer

We know that,

$\Rightarrow \sqrt[n]{x} = x^\dfrac{1}{n}$

Solving,

$\Rightarrow \sqrt[4]{\sqrt[3]{3^2}} \\[1em] \Rightarrow \sqrt[4]{3^\dfrac{2}{3}} \\[1em] \Rightarrow 3^\dfrac{2}{3 \times 4} \\[1em] \Rightarrow 3^\dfrac{2}{12} \\[1em] = {3^\dfrac{1}{6}}$

Hence, Option 4 is the correct option.

$1.\overline{9} - 1.9$ is equal to :

1. 0

2. 1

3. 0.09

4. 0.1

Answer

Let x = $1.\overline{9}$

⇒ x = 1.999...     .......(1)

⇒ 10x = 19.999...     .......(2)

Subtracting equation (1) from (2), we get :

⇒ 10x - x = 19.999..... - 1.999....

⇒ 9x = 18

⇒ x = $\dfrac{18}{9}$ = 2.

⇒ $1.\overline{9} - 1.9$

⇒ 2 - 1.9

⇒ 0.1

Hence, Option 4 is the correct option.

When written in decimal form, which of the following will be a non-terminating and non-repeating number?

1. $1^{1/9}$

2. $2^{1/9}$

3. $2^{-9}$

4. $9^{1/2}$

Answer

A non-terminating and non-repeating decimal is the definition of an irrational number.

Since 2 is not a perfect 9th power of any rational number, its 9th root will be an irrational number.

Hence, Option 2 is the correct option.

Observe the values of a, b, c given in the table. If we choose numbers a, b and c from rows a, b and c respectively, what is the maximum possible value of $\dfrac{c - b}{a}$?

Answer

We get the maximum possible value of $\dfrac{c - b}{a}$, if c is the largest and a, b are the smallest values.

$\Rightarrow \dfrac{c - b}{a} \\[1em] \Rightarrow \dfrac{25 - 3}{2} \\[1em] \Rightarrow \dfrac{22}{2} \\[1em] \Rightarrow 11.$

Hence, the maximum possible value of $\dfrac{c - b}{a}$ = 11.

The portion of a number line between 0 and 4 has been divided into 16 equal parts. Highlight the portion of the number line in which the reciprocal of any rational number is greater than the number itself.

Answer

As, the complete line is divided in into 16 equal parts. Thus, A = 1, B = 2, C = 3.

Rational number less than 1 and greater than 0, will be the interval in which reciprocal of any rational number is greater than the number itself.