Rational Numbers — Multiple Choice Questions

Additive inverse of $\dfrac{-2}{-5}$ is

1. $\dfrac{2}{5}$

2. $\dfrac{5}{2}$

3. $\dfrac{2}{-5}$

4. $\dfrac{5}{-2}$

Answer

First express the rational number with a positive denominator, then change its sign to get the additive inverse.

$\Rightarrow \dfrac{-2}{-5} = \dfrac{-2 \times (-1)}{-5 \times (-1)} = \dfrac{2}{5} \\[1em] \Rightarrow \text{Additive inverse} = -\left(\dfrac{2}{5}\right) \\[1em] \Rightarrow \dfrac{-2}{5} = \dfrac{2}{-5}$

Hence, Option 3 is the correct option.

Multiplicative inverse of $\dfrac{-3}{7}$ is

1. $\dfrac{7}{3}$

2. $\dfrac{-7}{3}$

3. $\dfrac{3}{7}$

4. None of these

Answer

To find the multiplicative inverse of a non-zero rational number, interchange its numerator and denominator.

$\Rightarrow \text{Multiplicative inverse of } \dfrac{-3}{7} = \dfrac{7}{-3} \\[1em] \Rightarrow \dfrac{-7}{3} \\[1em] [\text{Check : } \dfrac{-3}{7} \times \dfrac{-7}{3} = 1]$

Hence, Option 2 is the correct option.

Sum of a rational number and its additive inverse is

1. 1

2. 0

3. -1

4. None of these

Answer

The additive inverse of a rational number is obtained by changing its sign.

For every rational number $\dfrac{p}{q}$, we have $\dfrac{p}{q} + \left(-\dfrac{p}{q}\right) = 0$

Hence, Option 2 is the correct option.

Rational numbers are not closed under

1. addition

2. subtraction

3. multiplication

4. division

Answer

Rational numbers are closed under addition, subtraction and multiplication.

However, division of a rational number by zero is not defined. For example, $\dfrac{4}{5} \div 0$ is not defined.

Hence, Option 4 is the correct option.

$0 \div \dfrac{2}{3}$ is equal to

1. $\dfrac{2}{3}$

2. $\dfrac{3}{2}$

3. 0

4. not defined

Answer

Zero divided by any non-zero rational number is always zero.

$\Rightarrow 0 \div \dfrac{2}{3} \\[1em] \Rightarrow 0 \times \dfrac{3}{2} \\[1em] \Rightarrow 0$

Hence, Option 3 is the correct option.

$\dfrac{2}{3} \div 0$ is equal to

1. $\dfrac{2}{3}$

2. $\dfrac{3}{2}$

3. 0

4. not defined

Answer

Divisibility of a rational number by zero is not defined, i.e. $\dfrac{p}{q} \div 0$ does not exist as 0 has no reciprocal.

Hence, Option 4 is the correct option.

$\dfrac{p}{q} + \left(\dfrac{r}{s} + \dfrac{t}{u}\right) = \left(\dfrac{p}{q} + \dfrac{r}{s}\right) + \dfrac{t}{u}$ is called

1. commutative property

2. associative property

3. distributive property

4. None of these

Answer

Here the grouping of the three rational numbers has been changed but the sum remains the same.

If $\dfrac{p}{q}, \dfrac{r}{s} \text{ and } \dfrac{t}{u}$ are any three rational numbers,

Then $\dfrac{p}{q} + \left(\dfrac{r}{s} + \dfrac{t}{u}\right) = \left(\dfrac{p}{q} + \dfrac{r}{s}\right) + \dfrac{t}{u}$

Hence, Option 2 is the correct option.

Multiplication of a non-zero rational number and its reciprocal is

1. 0

2. 1

3. -1

4. none of these

Answer

The reciprocal of a non-zero rational number is obtained by interchanging its numerator and denominator.

For every non-zero rational number $\dfrac{p}{q}$, we have $\dfrac{p}{q} \times \dfrac{q}{p} = 1$

Hence, Option 2 is the correct option.

Product of rational number $\dfrac{-2}{5}$ and its additive inverse is

1. 0

2. 1

3. $\dfrac{-4}{25}$

4. $\dfrac{-5}{2}$

Answer

First find the additive inverse by changing the sign, then multiply.

$\Rightarrow \text{Additive inverse of } \dfrac{-2}{5} = \dfrac{2}{5} \\[1em] \Rightarrow \dfrac{-2}{5} \times \dfrac{2}{5} \\[1em] \Rightarrow \dfrac{(-2) \times 2}{5 \times 5} \\[1em] \Rightarrow \dfrac{-4}{25}$

Hence, Option 3 is the correct option.

Sum of rational number $\dfrac{4}{7}$ and its reciprocal is

1. $\dfrac{28}{65}$

2. $\dfrac{65}{28}$

3. $\dfrac{-28}{65}$

4. $\dfrac{-65}{28}$

Answer

First find the reciprocal by interchanging the numerator and denominator, then add.

$\Rightarrow \text{Reciprocal of } \dfrac{4}{7} = \dfrac{7}{4} \\[1em] \Rightarrow \dfrac{4}{7} + \dfrac{7}{4} \\[1em] [\text{ LCM of } 7 \text{ and } 4 = 28] \\[1em] \Rightarrow \dfrac{4 \times 4 + 7 \times 7}{28} \\[1em] \Rightarrow \dfrac{16 + 49}{28} \\[1em] \Rightarrow \dfrac{65}{28}$

Hence, Option 2 is the correct option.

Sum of two rational numbers is 0, if one of them is $\dfrac{-4}{5}$, then other is

1. $\dfrac{5}{4}$

2. $\dfrac{4}{5}$

3. $\dfrac{-5}{4}$

4. $\dfrac{-4}{5}$

Answer

As the sum of the two rational numbers is 0, the other rational number is the additive inverse of $\dfrac{-4}{5}$.

$\Rightarrow \text{Additive inverse of } \dfrac{-4}{5} = \dfrac{4}{5} \\[1em] [\text{Check : } \dfrac{-4}{5} + \dfrac{4}{5} = 0]$

Hence, Option 2 is the correct option.

Product of two rational numbers is 1, if one of them is $\dfrac{10}{3}$, then other is

1. $\dfrac{3}{10}$

2. $\dfrac{-3}{10}$

3. $\dfrac{10}{3}$

4. None of these

Answer

As the product of the two rational numbers is 1, the other rational number is the reciprocal of $\dfrac{10}{3}$.

$\Rightarrow \text{Reciprocal of } \dfrac{10}{3} = \dfrac{3}{10} \\[1em] [\text{Check : } \dfrac{10}{3} \times \dfrac{3}{10} = 1]$

Hence, Option 1 is the correct option.

Rational number represented by the point P on the number line is

1. $\dfrac{-5}{7}$

2. $\dfrac{-3}{7}$

3. $\dfrac{-5}{8}$

4. $\dfrac{-4}{8}$

Answer

The line segment between -1 and 0 has been divided into seven equal parts, so each division represents $\dfrac{1}{7}$.

The point P is the fourth point of division to the right of -1.

$\Rightarrow P = -1 + \dfrac{4}{7} \\[1em] \Rightarrow \dfrac{-7 + 4}{7} \\[1em] \Rightarrow \dfrac{-3}{7}$

Hence, Option 2 is the correct option.

What should be subtracted from $\dfrac{-5}{3}$ to get $\dfrac{-2}{7}$?

1. $\dfrac{29}{21}$

2. $\dfrac{-21}{29}$

3. $\dfrac{-29}{21}$

4. $\dfrac{21}{29}$

Answer

Let the required number be $x$.

According to the given condition,

$\Rightarrow \dfrac{-5}{3} - x = \dfrac{-2}{7} \\[1em] \Rightarrow x = \dfrac{-5}{3} - \left(\dfrac{-2}{7}\right) \\[1em] \Rightarrow x = \dfrac{-5}{3} + \dfrac{2}{7} \\[1em] [\text{ LCM of } 3 \text{ and } 7 = 21] \\[1em] \Rightarrow x = \dfrac{(-5) \times 7 + 2 \times 3}{21} \\[1em] \Rightarrow x = \dfrac{-35 + 6}{21} \\[1em] \Rightarrow x = \dfrac{-29}{21}$

Hence, Option 3 is the correct option.

Reciprocal of a negative number is

1. positive

2. negative

3. cannot say

4. does not exist

Answer

The reciprocal of a non-zero rational number is obtained by interchanging its numerator and denominator, so the sign of the rational number remains unchanged.

For example, the reciprocal of $\dfrac{-3}{5} \text{ is } \dfrac{5}{-3} = \dfrac{-5}{3}$, which is negative.

Hence, Option 2 is the correct option.

Between two given rational numbers, we can find

1. one and only one rational number

2. only two rational numbers

3. only ten rational numbers

4. infinitely many rational numbers

Answer

Between any two different rational numbers, we can always insert their mean, and repeating this process we can insert as many rational numbers as we want.

Hence, Option 4 is the correct option.

If x, y are rational numbers (x < y), then $\dfrac{x + y}{2}$ is a rational number

1. between x and y

2. less than x and y both

3. greater than x and y both

4. less than x but greater than y

Answer

If $x \text{ and } y$ are rational numbers such that $x$ < $y$, then the mean of $x \text{ and } y$ always lies between them, i.e. $x$ < $\dfrac{x + y}{2}$ < $y$

Hence, Option 1 is the correct option.