Rational Numbers — Multiple Choice Questions

The rational number equivalent to $\dfrac{-2}{3}$ and with numerator -10 is:

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

2. $\dfrac{-10}{30}$

3. $\dfrac{-10}{15}$

4. $\dfrac{10}{15}$

Answer

To get numerator -10 from -2, we multiply by 5.

$\dfrac{-2}{3} = \dfrac{-2 \times 5}{3 \times 5} = \dfrac{-10}{15}$

Hence, Option 3 is the correct option.

The largest rational number out of $\dfrac{-5}{6}, \dfrac{-19}{24} \text{ and } \dfrac{37}{-12}$ is:

1. $-\dfrac{5}{6}$

2. $\dfrac{-19}{24}$

3. $\dfrac{37}{-12}$

4. none of these

Answer

By Division Method,

$\begin{array}{l|r} 2 & 6, 24, 12 \\ \hline 2 & 3, 12, 6 \\ \hline 2 & 3, 6, 3 \\ \hline 3 & 3, 3, 3 \\ \hline & 1, 1, 1 \end{array}$

LCM of 6, 24 and 12 is 2 × 2 × 2 × 3 = 24

Converting each rational number to denominator 24:

$\Rightarrow \dfrac{-5}{6} = \dfrac{-5 \times 4}{6 \times 4} = \dfrac{-20}{24} \\[1em] \dfrac{-19}{24} = \dfrac{-19}{24} \\[1em] \dfrac{37}{-12} = \dfrac{-37}{12} = \dfrac{-37 \times 2}{12 \times 2} = \dfrac{-74}{24}$

Since $-74 \lt -20 \lt -19$, the largest rational number is $\dfrac{-19}{24}$.

Hence, Option 2 is the correct option.

$\dfrac{-88}{-136}$ in standard form is:

1. $\dfrac{88}{-13}$

2. $\dfrac{-88}{136}$

3. $\dfrac{11}{17}$

4. $-\dfrac{11}{17}$

Answer

$\dfrac{-88}{-136} = \dfrac{88}{136}$

HCF of 88 and 136 is 8.

$\dfrac{88}{136} = \dfrac{88 \div 8}{136 \div 8} = \dfrac{11}{17}$

Hence, Option 3 is the correct option.

Which of the following is true for the rational numbers $-\dfrac{3}{5}$ and $-\dfrac{1}{3}$?

1. $-\dfrac{3}{5} \gt -\dfrac{1}{3}$

2. $-\dfrac{1}{3} \gt -\dfrac{3}{5}$

3. none of these

Answer

By Division Method,

$\begin{array}{l|r} 3 & 5, 3 \\ \hline 5 & 5, 1 \\ \hline & 1, 1 \end{array}$

LCM of 5 and 3 is 3 × 5 = 15

$\Rightarrow -\dfrac{3}{5} = \dfrac{-3 \times 3}{5 \times 3} = \dfrac{-9}{15} \\[1em] \Rightarrow -\dfrac{1}{3} = \dfrac{-1 \times 5}{3 \times 5} = \dfrac{-5}{15}$

Since -5 > -9, we have $\dfrac{-5}{15} \gt \dfrac{-9}{15}$.

Hence, $-\dfrac{1}{3} \gt -\dfrac{3}{5}$

Hence, Option 2 is the correct option.

According to the number line given above, the values of A and B are:

1. $A = -\dfrac{5}{4} \text{ and } B = \dfrac{7}{3}$

2. $A = -\dfrac{5}{4} \text{ and } B = -\dfrac{7}{3}$

3. $A = -\dfrac{7}{4} \text{ and } B = \dfrac{7}{3}$

4. $A = \dfrac{7}{4} \text{ and } B = -\dfrac{7}{3}$

Answer

From the number line, A lies between -2 and -1 at the point $-\dfrac{7}{4}$, and B lies between 2 and 3 at the point $\dfrac{7}{3}$.

Hence, $A = -\dfrac{7}{4} \text{ and } B = \dfrac{7}{3}$

Hence, Option 3 is the correct option.

The rational number between $-\dfrac{1}{3} \text { and }\dfrac{1}{3}$ is:

1. 1

2. -1

3. 0

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

Answer

A rational number between $-\dfrac{1}{3} \text { and } \dfrac{1}{3}$ is their mean.

$\dfrac{1}{2}\left(-\dfrac{1}{3} + \dfrac{1}{3}\right) = \dfrac{1}{2} \times 0 = 0$

Hence, 0 lies between $-\dfrac{1}{3} \text { and }\dfrac{1}{3}$.

Hence, Option 3 is the correct option.

A boy walks $\dfrac{2}{3}$ km from a place P towards north and then from there $\dfrac{5}{6}$ km towards south. The position of the boy from the place P is:

1. $\dfrac{1}{6}$ km towards north

2. $\dfrac{1}{6}$ km towards south

3. $\dfrac{9}{6}$ km towards north

4. $\dfrac{9}{6}$ km towards south

Answer

Taking the direction towards north as positive and south as negative:

By Division Method,

$\begin{array}{l|r} 2 & 3, 6 \\ \hline 3 & 3, 3 \\ \hline & 1, 1 \end{array}$

LCM of 3 and 6 is 2 × 3 = 6

$\Rightarrow \dfrac{2}{3} - \dfrac{5}{6}\\[1em] = \dfrac{2 \times 2}{3 \times 2} - \dfrac{5}{6} \\[1em] = \dfrac{4}{6} - \dfrac{5}{6} = \dfrac{4 - 5}{6}\\[1em] = \dfrac{-1}{6}$

The negative sign shows the boy is $\dfrac{1}{6}$ km towards south of P.

Hence, Option 2 is the correct option.

The rational number that should be added to $-\dfrac{3}{7} \text { to get } -\dfrac{5}{7}$ is:

1. $\dfrac{8}{7}$

2. $-\dfrac{8}{7}$

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

4. $\dfrac{2}{7}$

Answer

Let x be added to $-\dfrac{3}{7}$.

$\Rightarrow -\dfrac{3}{7} + x = -\dfrac{5}{7} \\[1em] \Rightarrow x = -\dfrac{5}{7} - \left(-\dfrac{3}{7}\right) \\[1em] \Rightarrow x = \dfrac{-5}{7} + \dfrac{3}{7} \\[1em] \Rightarrow x = \dfrac{-5 + 3}{7} \\[1em] \Rightarrow x = \dfrac{-2}{7}$

Hence, Option 3 is the correct option.

The rational number that should be subtracted from $-\dfrac{2}{7}\text { to get }-\dfrac{5}{7}$ is:

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

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

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

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

Answer

Let x be subtracted from $-\dfrac{2}{7}$.

$\Rightarrow -\dfrac{2}{7} - x = -\dfrac{5}{7} \\[1em] \Rightarrow x = -\dfrac{2}{7} - \left(-\dfrac{5}{7}\right) \\[1em] \Rightarrow x = \dfrac{-2}{7} + \dfrac{5}{7} \\[1em] \Rightarrow x = \dfrac{-2 + 5}{7} \\[1em] \Rightarrow x = \dfrac{3}{7}$

Hence, Option 1 is the correct option.

The sum of two rational numbers is 2. If one of them is $-\dfrac{1}{2}$; the other is:

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

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

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

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

Answer

Let x be the other number.

$\Rightarrow -\dfrac{1}{2} + x = 2 \\[1em] \Rightarrow x = 2 - \left(-\dfrac{1}{2}\right) \\[1em] \Rightarrow x = \dfrac{2}{1} + \dfrac{1}{2} \\[1em] \Rightarrow x = \dfrac{2 \times 2}{1 \times 2} + \dfrac{1}{2} \\[1em] \Rightarrow x = \dfrac{4}{2} + \dfrac{1}{2} \\[1em] \Rightarrow x = \dfrac{5}{2}$

Hence, Option 1 is the correct option.

If the cost of 1 m cloth is ₹ $20\dfrac{5}{8}$, then the cost of $2\dfrac{2}{7}$ m is:

1. ₹ $\left(20\dfrac{5}{8} + 2\dfrac{2}{7}\right)$

2. ₹ $\left(20\dfrac{5}{8} \div 2\dfrac{2}{7}\right)$

3. ₹ $\left(20\dfrac{5}{8} \times 2\dfrac{2}{7}\right)$

4. none of these

Answer

The cost of $2\dfrac{2}{7}$ m of cloth is obtained by multiplying the cost of 1 m by the length.

$\text{Cost} = 20\dfrac{5}{8} \times 2\dfrac{2}{7}$

Hence, Option 3 is the correct option.

The product of two rational numbers is -1. If one of them is $\dfrac{5}{6}$, then the other rational number is:

1. $\dfrac{6}{5}$

2. $-\dfrac{6}{5}$

3. $-\dfrac{5}{6}$

4. none of these

Answer

Let x be the other rational number.

$\Rightarrow \dfrac{5}{6} \times x = -1 \\[1em] \Rightarrow x = -1 \div \dfrac{5}{6} \\[1em] \Rightarrow x = -1 \times \dfrac{6}{5} \\[1em] \Rightarrow x = -\dfrac{6}{5}$

Hence, Option 2 is the correct option.

The distance covered in $2\dfrac{2}{9}$ hours at the speed of $5\dfrac{2}{5}$ km per hour is:

1. $\left(\dfrac{27}{5} + \dfrac{20}{9}\right)$ km

2. $\left(\dfrac{27}{5} \div \dfrac{20}{9}\right)$ km

3. $\left(\dfrac{20}{9} \div \dfrac{27}{5}\right)$ km

4. none of these

Answer

Distance = Speed × Time

$\text{Distance} = 5\dfrac{2}{5} \times 2\dfrac{2}{9} = \dfrac{27}{5} \times \dfrac{20}{9}$

The distance is obtained by multiplication, which is not given in options 1, 2 or 3.

Hence, Option 4 is the correct option.

$2\dfrac{5}{8}$ divided by $1\dfrac{1}{6}$ gives:

1. $\dfrac{4}{9}$

2. $3\dfrac{19}{24}$

3. $2\dfrac{1}{4}$

4. none of these

Answer

$\Rightarrow 2\dfrac{5}{8} \div 1\dfrac{1}{6} = \dfrac{21}{8} \div \dfrac{7}{6} \\[1em] = \dfrac{21}{8} \times \dfrac{6}{7} \\[1em] = \dfrac{21 \times 6}{8 \times 7} \\[1em] = \dfrac{126}{56} \\[1em] = \dfrac{9}{4} \\[1em] = 2\dfrac{1}{4}$

Hence, Option 3 is the correct option.

$-\dfrac{1}{3}, \dfrac{-5}{9} \text{ and }\dfrac{-4}{3}$ in ascending order are:

1. $-\dfrac{1}{3} \lt \dfrac{-5}{9} \lt \dfrac{-4}{3}$

2. $-\dfrac{1}{3} \lt \dfrac{-4}{3} \lt \dfrac{-5}{9}$

3. $\dfrac{-4}{3} \lt \dfrac{-5}{9} \lt -\dfrac{1}{3}$

4. $\dfrac{-5}{9} \lt \dfrac{-4}{3} \lt -\dfrac{1}{3}$

Answer

By Division Method,

$\begin{array}{l|r} 3 & 3, 9, 3 \\ \hline 3 & 1, 3, 1 \\ \hline & 1, 1, 1 \end{array}$

LCM of 3 and 9 = 3 × 3 = 9

Converting each rational number to denominator 9:

$\Rightarrow -\dfrac{1}{3} = \dfrac{-1 \times 3}{3 \times 3} = \dfrac{-3}{9} \\[1em] \Rightarrow \dfrac{-5}{9} = \dfrac{-5}{9} \\[1em] \Rightarrow \dfrac{-4}{3} = \dfrac{-4 \times 3}{3 \times 3} = \dfrac{-12}{9}$

Comparing the numerators: -12 < -5 < -3

Hence, $\dfrac{-4}{3} \lt \dfrac{-5}{9} \lt -\dfrac{1}{3}$

Hence, Option 3 is the correct option.