Rational Numbers — Assertion-Reason Type Questions

Assertion (A): The product (multiplication) of two rational numbers is $\dfrac{3}{8}$. If one of them is $-\dfrac{3}{20}$, then the other rational number is $-\dfrac{5}{2}$.

Reason (R): The product of two rational numbers $\dfrac{a}{b}$ and $\dfrac{c}{d}$ is $\dfrac{a \times c}{b \times d}$.

1. A is true, R is false.

2. A is false, R is true.

3. Both A and R are true.

4. Both A and R are false.

Answer

Let x be the other rational number.

$\Rightarrow -\dfrac{3}{20} \times x = \dfrac{3}{8} \\[1em] \Rightarrow x = \dfrac{3}{8} \div \left(-\dfrac{3}{20}\right) \\[1em] \Rightarrow x = \dfrac{3}{8} \times \left(-\dfrac{20}{3}\right) \\[1em] \Rightarrow x = \dfrac{3 \times (-20)}{8 \times 3} \\[1em] \Rightarrow x = \dfrac{-60}{24} \\[1em] \Rightarrow x = -\dfrac{5}{2}$

So, Assertion (A) is true. Reason (R) correctly states the rule for multiplication of rational numbers, so it is also true.

Hence, both A and R are true.

Hence, Option 3 is the correct option.

Assertion (A): On dividing the sum of $\dfrac{65}{12}\text { and }\dfrac{8}{3}$ by their difference, you get $\dfrac{97}{33}$.

Reason (R): If $\dfrac{p}{q} \text{ and } \dfrac{r}{s}$ are two rational numbers such that $\dfrac{r}{s} \neq 0$, then $\dfrac{p}{q} \div \dfrac{r}{s} = \dfrac{p}{q} \times \dfrac{s}{r}$.

1. A is true, R is false.

2. A is false, R is true.

3. Both A and R are true.

4. Both A and R are false.

Answer

By Division Method,

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

LCM of 12 and 3 is 2 × 2 × 3 = 12

Sum:

$\dfrac{65}{12} + \dfrac{8}{3} = \dfrac{65}{12} + \dfrac{8 \times 4}{3 \times 4} = \dfrac{65}{12} + \dfrac{32}{12} = \dfrac{97}{12}$

Difference:

$\dfrac{65}{12} - \dfrac{8}{3} = \dfrac{65}{12} - \dfrac{32}{12} = \dfrac{33}{12}$

Dividing the sum by the difference:

$\dfrac{97}{12} \div \dfrac{33}{12} = \dfrac{97}{12} \times \dfrac{12}{33} = \dfrac{97}{33}$

So, Assertion (A) is true. Reason (R) correctly states the rule for division of rational numbers, so it is also true.

Hence, both A and R are true.

Hence, Option 3 is the correct option.

Assertion (A): 1 and 0 are two co-prime integers, hence $\dfrac{1}{0}$ is rational.

Reason (R): Every integer is a rational number. Its converse is also true.

1. A is true, R is false.

2. A is false, R is true.

3. Both A and R are true.

4. Both A and R are false.

Answer

A rational number is of the form $\dfrac{p}{q}$ where $q \neq 0$. Since $\dfrac{1}{0}$ has denominator 0, it is not a rational number. So, Assertion (A) is false.

Every integer is a rational number, but its converse "every rational number is an integer" is not true (for example, $\dfrac{1}{2}$ is rational but not an integer). So, Reason (R) is also false.

Hence, both A and R are false.

Hence, Option 4 is the correct option.

Assertion (A): $\dfrac{34}{-51} \text { and }\dfrac{-2}{3}$ are equivalent rational numbers.

Reason (R): If $\dfrac{p}{q}$ is a rational number and n is a non-zero integer, then $\dfrac{p \times n}{q \times n} = \dfrac{p}{q}$

1. A is true, R is false.

2. A is false, R is true.

3. Both A and R are true.

4. Both A and R are false.

Answer

Reducing $\dfrac{34}{-51}$ to standard form:

HCF of 34 and 51 is 17.

$\dfrac{34}{-51} = \dfrac{34 \div 17}{-51 \div 17} = \dfrac{2}{-3} = \dfrac{-2}{3}$

So, $\dfrac{34}{-51} \text { and } \dfrac{-2}{3}$ are equivalent rational numbers and Assertion (A) is true. Reason (R) correctly states the property of equivalent rational numbers, so it is also true.

Hence, both A and R are true.

Hence, Option 3 is the correct option.