Neha borrowed Rs. 24000 from the State Bank of India to buy a scooter.

1.	Neha borrowed Rs. 24000 from the State Bank of India to buy a scooter. If the rate of interest be 10% per annum compounded annually, what payment will she have to make after 2 years 3 months?
Answer» Present value = Rs.24000 Interest rate = 10 % per annum Time = 2 years 3 month = (2 + 1/4) years = \(2\frac{1}{4}\) years. Amount (A) = P (1 + R/100)ⁿ× [1 + (1/4 × R)/100] [Where, P = Present value R = Annual interest rate n = Time in years] ∴ A = 24000 (1 + 10/100)² × [1 + (1/4 × 10)/100] ⇒ A = 24000 (1 + 1/10)² × [1 + 1/40] ⇒ A = 24000 (11/10)² × [41/40] ⇒ A = 24000 × 121/100 × 41/40 ⇒ A = 24 × 121 × 41/4 ⇒ A = 6 × 121 × 41 ⇒ A = 29766 ∴ Amount = Rs.29766 ∴ Neha should pay Rs. 29766 to the bank after 2 years 3 months.

Neha borrowed Rs. 24000 from the State Bank of India to buy a scooter. If the rate of interest be 10% per annum compounded annually, what payment will she have to make after 2 years 3 months?

Answer»

Present value = Rs.24000

Interest rate = 10 % per annum

Time = 2 years 3 month = (2 + 1/4) years = \(2\frac{1}{4}\) years.

Amount (A) = P (1 + R/100)ⁿ× [1 + (1/4 × R)/100]

[Where, P = Present value

R = Annual interest rate

n = Time in years]

∴ A = 24000 (1 + 10/100)² × [1 + (1/4 × 10)/100]

⇒ A = 24000 (1 + 1/10)² × [1 + 1/40]

⇒ A = 24000 (11/10)² × [41/40]

⇒ A = 24000 × 121/100 × 41/40

⇒ A = 24 × 121 × 41/4

⇒ A = 6 × 121 × 41

⇒ A = 29766

∴ Amount = Rs.29766

∴ Neha should pay Rs. 29766 to the bank after 2 years 3 months.