Banking — Assertion-Reason Type Questions

Assertion (A) : Sunidhi deposits ₹1,600 per month in a bank for $1\dfrac{1}{2}$ years in a recurring deposit account at 10% p.a. She gets ₹31,080 on maturity.

Reason (R): Maturity value is given by MV = (P x n) - S.I.

1. Both A and R are true, and R is the correct explanation of A.

2. Both A and R are true, but R is not the correct explanation of A.

3. A is true, but R is false.

4. A is false, but R is true.

Answer

According to Assertion:

Given,

P = ₹1,600

n = $1\dfrac{1}{2}$ years = 18 months

r = 10%

I = $P \times \dfrac{n(n+1)}{2 \times 12} \times \dfrac{r}{100}$

$\therefore I = 1600\times \dfrac{18\times 19}{2 \times 12} \times \dfrac{10}{100} \\[1em] I = 1600 \times \dfrac{342}{24} \times 0.1 \\[1em] I = 1600 \times 14.25 \times 0.1 \\[1em] I = ₹2,280$

Sum deposited = ₹1,600 x 18 = ₹28,800

Maturity value = Sum deposited + Interest = ₹28,800 + ₹2,280 = ₹31,080

So, Assertion (A) is true.

According to Reason:

Maturity value is given by MV = (P x n) - S.I.

But,

Maturity value = Sum deposited + Interest

Sum deposited = P × n

Maturity value = (P × n) + Interest

So, Reason (R) is false.

Hence, Option 3 is the correct option.

Assertion (A): Pawandeep opened a recurring deposit account in a bank for a period of 2 years. If the bank pays interest at the rate of 6% p.a. and the monthly instalment is ₹1,000, then the maturity amount is ₹25,000.

Reason (R): For a recurring deposit account, we compute the interest using the following formula:

$S.I. = P \times \dfrac{n(n+1)}{2} \times \dfrac{1}{12} \times \dfrac{R}{100}$

1. Both A and R are true, and R is the correct explanation of A.

2. Both A and R are true, but R is not the correct explanation of A.

3. A is true, but R is false.

4. A is false, but R is true.

Answer

According to Assertion:

The maturity amount is ₹25,000.

Given,

P = ₹1,000

n = 2 years = 24 months

r = 6%

I = $P \times \dfrac{n(n+1)}{2 \times 12} \times \dfrac{r}{100}$

Substituting values we get :

$\therefore I = 1000\times \dfrac{24\times 25}{2 \times 12} \times \dfrac{6}{100} \\[1em] I = 1000 \times \dfrac{600}{24} \times 0.06\\[1em] I = 1000 \times 25 \times 0.06 \\[1em] I = ₹1500$

Sum deposited = ₹1,000 x 24 = ₹24,000

Maturity value = Sum deposited + Interest = ₹24,000 + ₹1,500 = ₹25,500

The given Maturity amount = ₹25,500

So, Assertion(A) is false.

For a recurring deposit account, we compute the interest using the following formula:

I = $P \times \dfrac{n(n+1)}{2 \times 12} \times \dfrac{r}{100}$

So, Reason (R) is true.

Hence, Option 4 is the correct option.