Question

N$1000 is placed in an account at 4% compounded annually for 2 years. It is then withdrawn at the end of the two years and placed in another bank at the rate of 5% compounded semiannually for 4 years. What is the balance in the second account after the 4 years?

Answer 1

The balance is N$1303.72

Answer 2

The balance in the second account after 4 years is N$1303.72.

Explanation:

To solve this problem, we need to use the compound interest formula:

A = P(1 + r/n)^(n*t)

where:

A = final amount

P = principal (initial amount)

r = annual interest rate (as a decimal)

n = number of times compounded per year

t = time (in years)

For the first account, we have:

P = 1000

r = 0.04

n = 1 (compounded annually)

t = 2

So, after 2 years, the balance in the first account is:

A1 = 1000(1 + 0.04/1)^(1*2) = 1081.60

This amount is then withdrawn and placed in the second account, where it earns interest at a rate of 5% compounded semiannually. This means that the interest rate per compounding period is:

r = 0.05/2 = 0.025

And the number of compounding periods over 4 years is:

n = 2*4 = 8

So, after 4 years in the second account, the balance is:

A2 = 1081.60(1 + 0.025/1)^(1*8) = 1303.72

Therefore, the balance in the second account after 4 years is N$1303.72.