Question

(a) If Angela has $100 to invest at 2.5% interest per year compounded continuously. How long will it take for Angela's investment to double? If she invested $1,000, how will the doubling time change? (b) Suppose April has access to an investment that will pay 10% interest compounded continuously. Which is better to be given s1000 now and invest or be given 51325 after 3 years without investing? (c) A child's grandparents are considering buying an $80,000 face-value, zero-coupon bond at her birth so that she will have enough money for her college education 17 years later. If they want a rate of return of 6% compounded continuously, what should they pay for the bond?

Answer 1

(a) If Angela has $100 to invest at 2.5% interest per year compounded continuously. How long will it take for Angela's investment to double? If she invested $1,000, how will the doubling time change?

When the invested amount is compounded continuously,

A = P e^(rt)

P : initial amount invested

A : amount at the end of t years

r : rate of interest ( expressed as a decimal )

t : number of years

Here, P = $100 , r = 2.5% = 0.025 , A = $200 ( as the investment has to double )

200 = 100 e^(0.025t)

2 = e^(0.025t)

0.025t = ln 2

t = ln2 / 0.025 = 27.7258 ≈ 27.73

t = 27.73 years

If P = $1000 , then A = $2000

So, 2000 = 1000 e^(0.025t)

2 = e^(0.025t)

On solving for t, the doubling time does not change.

The doubling time remains the same.