Banking — Case-Study Based Questions

Case study: Joseph has a recurring deposit account in a bank for two years at the rate of 8% per annum simple interest.

1. If at the time of maturity Joseph receives ₹2,000 as interest, then the monthly instalment is:
(a) ₹1,200
(b) ₹600
(c) ₹1,000
(d) ₹1,600

2.The total amount deposited in the bank is:
(a) ₹25,000
(b) ₹24,000
(c) ₹26,000
(d) ₹23,000

3.The amount Joseph receives on maturity is:
(a) ₹27,000
(b) ₹25,000
(c) ₹26,000
(d) ₹28,000

4. If the monthly instalment is ₹100 and the rate of interest is 8%, in how many months Joseph will receive ₹52 as interest?
(a) 18
(b) 30
(c) 12
(d) 6

Answer

1.Given,

I = 2000

R = 8%

n = 2 years = 24 months

$I = P \times \dfrac{n(n+1)}{2 \times 12} \times \dfrac{R}{100}\\[1em] \therefore I = P \times \dfrac{24 \times 25}{2 \times 12} \times \dfrac{8}{100}\\[1em] I = P \times \dfrac{600}{24} \times 0.08\\[1em] I = P \times 25 \times 0.08\\[1em] 2000 = P \times 2 \\[1em] P=\dfrac{2000}{2}\\[1em] P =₹ 1,000$

Hence, Option (c) is the correct option.

2. Given,

P = ₹1000

n = 24 months

Total deposit = P x n

Total deposit = 1000 x 24

Total deposit= ₹24,000

Hence, Option (b) is the correct option.

3. Given:

Total deposit = ₹24,000

Interest = ₹2,000

Maturity amount = Total Deposit + Interest

Maturity amount = 24000 + 2000

Maturity amount= ₹ 26,000

Hence, Option (c) is the correct option.

4. Given:

I = ₹52

P = ₹100

r = 8%

$I = P \times \dfrac{n(n+1)}{2 \times 12} \times \dfrac{R}{100} \\[1em] \therefore I = 100 \times \dfrac{n(n+1)}{2 \times 12} \times \dfrac{8}{100} \\[1em] 52 = \dfrac{n(n+1)}{24}\times 8 \\[1em] 52 = \dfrac{n(n+1)}{3} \\[1em] n(n+1) =52 \times 3 \\[1em] n(n+1) = 156 \\[1em] n^2+n-156 = 0 \\[1em] n^2+13n - 12n-156 = 0 \\[1em] n(n+13) - 12(n+13) = 0 \\[1em] (n+13)(n-12) = 0 \\[1em] n=-13 \text{ or } n=12$

Since the number of months cannot be negative.

∴ n = 12 months

Hence, Option (c) is the correct option.