Fractions — Assertion-Reason Type Questions

Assertion (A): Let us consider the product of a proper fraction $\dfrac{3}{4}$ and a mixed fraction $1\dfrac{1}{5}$. The product is $\dfrac{9}{10} \text{ and } \dfrac{3}{4} \lt \dfrac{9}{10} \lt 1\dfrac{1}{5}$.

Reason (R): The product of a proper fraction and a mixed fraction is less than the proper fraction but greater than the improper fraction.

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

The product of $\dfrac{3}{4} \text{ and } 1\dfrac{1}{5}$ is :

$\dfrac{3}{4} \times 1\dfrac{1}{5} \\[1em] = \dfrac{3}{4} \times \dfrac{6}{5} \\[1em] = \dfrac{18}{20} \\[1em] = \dfrac{9}{10}$

Checking the order:

$\dfrac{3}{4} = 0.75$, $\dfrac{9}{10} = 0.9 \text{ and } 1\dfrac{1}{5} = 1.2,$

$\text{ so }\dfrac{3}{4} \lt \dfrac{9}{10} \lt 1\dfrac{1}{5}$.

Thus, Assertion (A) is true.

The Reason states the product is less than the proper fraction but greater than the improper fraction. In fact the product is greater than the proper fraction but less than the improper (mixed) fraction, so the Reason is false.

Thus, Reason (R) is false.

∴ A is true, R is false.

Hence, Option 1 is the correct option.

Assertion (A): The product of two improper fractions $1\dfrac{1}{3} \text{ and } 2\dfrac{1}{2}$ is greater than both $1\dfrac{1}{3} \text{ and } 2\dfrac{1}{2}$.

Reason (R): The product of two improper positive fractions is greater than each of the improper fractions multiplied together.

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

The product of $1\dfrac{1}{3} \text{ and } 2\dfrac{1}{2}$ is :

$\Rightarrow 1\dfrac{1}{3} \times 2\dfrac{1}{2} \\[1em] = \dfrac{4}{3} \times \dfrac{5}{2}\\[1em] = \dfrac{20}{6} \\[1em] = \dfrac{10}{3}\\[1em] = 3\dfrac{1}{3}$

Since $3\dfrac{1}{3}$ is greater than both $1\dfrac{1}{3} \text{ and } 2\dfrac{1}{2}$,

Thus, Assertion (A) is true.

We know that,

The product of two improper positive fractions is greater than each of the improper fractions multiplied together.

Thus, Reason (R) is true.

∴ Both A and R are true.

Hence, Option 3 is the correct option.

Assertion (A): The fraction $\dfrac{32}{56}$ can be reduced to $\dfrac{4}{7}$.

Reason (R): Two fractions $\dfrac{p}{q} \text{ and } \dfrac{r}{s}$ are said to be equivalent only when ps = rq.

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

For the Assertion, HCF of 32 and 56 is 8.

$\dfrac{32}{56} = \dfrac{32 \div 8}{56 \div 8} = \dfrac{4}{7}$

Thus, Assertion (A) is true.

The Reason states the correct condition for two fractions to be equivalent, namely ps = rq.

Checking: $32 \times 7 = 224 \text{ and } 4 \times 56 = 224$, so the condition holds.

Thus, Reason (R) is true.

∴ Both A and R are true.

Hence, Option 3 is the correct option.