Compound Interest — Assertion-Reason Type Questions

Assertion (A): ₹ 1,00,000 amounts to ₹ 1,21,000 in 1 year and to ₹1,46,410 in 2 years, compounded annually.

Reason (R): When the interest is compounded annually, then, we have A = $P \times \Big(1 + \dfrac{R}{100} \Big)^n$

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

Given,

P = ₹ 1,00,000

A = ₹ 1,21,000

n = 1 year

By formula,

A = $P \times \Big(1 + \dfrac{r}{100}\Big)^n$

Substituting the values in formula,

$\Rightarrow 121000 = 100000 \times \Big(1 + \dfrac{r}{100}\Big)^1 \\[1em] \Rightarrow \dfrac{121000}{100000} = 1 + \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{r}{100} = 1.21 - 1 \\[1em] \Rightarrow r = 0.21 \times 100 \\[1em] \Rightarrow r = 21$

Lets calculate the amount for 2 years with this rate of interest.

By formula,

A = $P \times \Big(1 + \dfrac{r}{100}\Big)^n$

Substituting the values in formula,

$\Rightarrow A = 100000 \times \Big(1 + \dfrac{21}{100}\Big)^2 \\[1em] \Rightarrow A = 100000 \times \Big(\dfrac{100 + 21}{100}\Big)^2 \\[1em] \Rightarrow A = 100000 \times \Big(\dfrac{121}{100}\Big)^2 \\[1em] \Rightarrow A = 100000 \times (1.21)^2 \\[1em] \Rightarrow A = 100000 \times 1.4641 \\[1em] \Rightarrow A = 146410.$

Assertion (A) is true.

Formula,

A = $P \times \Big(1 + \dfrac{R}{100} \Big)^n$

Reason is true, as it is the formula to calculate the amount for compound interest when the interest is compounded annually.

Hence, option 3 is correct option.

Assertion (A): The present population of a town decreases at the rate of 5% p.a. due to migration. Its present population is 21,600 and 2 years ago it was 24,000.

Reason (R): Let there be an increase in population at R% p.a. Then population after n years is given by $P \times \Big(1 - \dfrac{R}{100} \Big)^n$, where P is the present population.

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

Given,

P (two years ago) = 24,000

R = 5% p.a.

n = 2 year

By formula,

Population after n years = $P\Big(1 - \dfrac{r}{100}\Big)^n$

Substituting the values in formula,

$\text{Present population }=24000 \times \Big(1 - \dfrac{5}{100}\Big)^2 \\[1em] =24000 \times \Big(\dfrac{100 - 5}{100}\Big)^2 \\[1em] =24000 \times \Big(\dfrac{95}{100}\Big)^2 \\[1em] =24000 \times (0.95)^2\\[1em] =24000 \times 0.9025\\[1em] =21,660$

Assertion is False

Reason is also false as we do not use that formula to calculate the growth in population. We use,

Population growth after n years at R% with initial Population as P is calulated using formula : $P \times \Big(1 + \dfrac{R}{100}\Big)^n$

Hence, option 4 is correct option.