Question

How much interest is earned in the third year on a $1,000 deposit that earns 10% interest compounded annually? (2 points) I. 2. If you seek to be a millionaire (savings- $1,000,000) when you retire in 45 years, how much must you deposit at the end of each year if you earn 6% compounded annually? (2 points) rate of interest. The loan will have annual ment will be $10,000. The other three 3. You would like to loan me $100,000 for five years at a 6% payments, but to make it attractive to me, the first and last pay payments will be equal. What do those payments need tobein rdert pro ey urg% return? (6 points)

Answer 1

1. Amount invested = $1000

Interest rate = 10%

Value of Investment after 1 year = 1000 * 1.1 = 1100

Value of Investment after 2 years = 1100 * 1.1 = 1210

Value of Investment after 3 years = 1210 * 1.1 = 1331

Hence, interest earned in Year 3 = 1331 - 1210 = $121

2. Let amount deposited be P

Interest Rate = r = 6%

Time frame = n = 45 years

Value of investment in 45 years = P[(1+r)ⁿ -1]/r

This will be equal to $1000000

=> P(1.06⁴⁵-1)/0.06 = 1000000

=> P = $4700.50

Hence $4700.50 should be invested for 45 years to become a millionaire

3. Let the payments for 3 years be P

rate of return required = r = 8%

CF1 = 10000
CF2 = P
CF3 = P
CF4 = P
CF5 = 10000

Present Value of Payments = 10000/(1+r) + P/(1+r)² + P/(1+r)³ + P/(1+r)⁴ + 10000/(1+r)⁵

This is equal to $100000

=> 10000/1.08 + P/1.08² + P/1.08³ + P/1.08⁴ + 10000/1.08⁵ = 100000

=> P = 83934.91/2.39 = $35119.21

Hence Payments for 3 years = $35,119.21