Question

Ruth has borrowed K2500.00 and is being charged interest at 7.8% interest, compounded weekly.
A. How much will she owe after 6 months?
B. If she repays K1000.00 after 6 months, how much will she owe after another 6 months?

Answer 1

The answer to A is K2599.35 and the answer to B is K1662.91

Answer 2

A. The formula for compound interest is given by A = P(1 + r/n)^(nt), where:
A = the future value of the investment/loan, including interest
P = the principal amount (initial loan amount) = K2500
r = the annual interest rate as a decimal = 7.8% = 0.078
n = the number of times that interest is compounded per year = 52 (weekly compounding)
t = the number of years the money is invested/borrowed for = 6 months = 0.5 years

Plugging in the values:
A = K2500(1 + 0.078/52)^(52*0.5)
A = K2500(1 + 0.0015)^26
A = K2500(1.0015)^26
A = K2500(1.039769)

A = K2594.422

After 6 months, Ruth will owe K2594.42

B. If Ruth repays K1000 after 6 months, she will still owe K2594.42 - K1000 = K1594.42.

To find out how much she will owe after another 6 months, we will use the same formula as in part A, but with a new value of P (K1594.42), and t = 0.5 years.

A = K1594.42(1 + 0.078/52)^(52*0.5)
A = K1594.42(1.0015)^26
A = K1594.42(1.039769)

A = K1658.31

After another 6 months, Ruth will owe K1658.31.

Answer 3

I apologize for the error in my calculations. Let's recalculate:

A. Using the compound interest formula:

A = P(1 + r/n)^(nt)
A = K2500(1 + 0.078/52)^(52*0.5)
A = K2500(1 + 0.0015)^26
A = K2500(1.0015)^26
A = K2500(1.040012)
A = K2600.03

After 6 months, Ruth will owe K2600.03.

B. If Ruth repays K1000 after 6 months, she will owe K2600.03 - K1000 = K1600.03.
Using the same formula:

A = K1600.03(1 + 0.078/52)^(52*0.5)
A = K1600.03(1.0015)^26
A = K1600.03(1.040012)
A = K1664.45

After another 6 months, Ruth will owe K1664.45.

Thank you for pointing out the error.