Question

1. Calculate the amount of money that a person must have in a bank today (the beginning of the year) to be able to withdraw $375 at the end of each year for the next 10 years if the bank pays interest compounded yearly at j₁ = 5.8% pa. Give your answer in dollars and cents to the nearest cent.

Account balance = $

2. Calculate the simple interest rate pa that must be earned for $60,000 invested on 29 October 2019 to be worth $60,561.01 on 9 January 2020. Give your answer as a percentage per annum to 2 decimal places. This days between dates calculator may assist you.

r =  % pa

3. Calculate the accumulated value (S) that payments of $10 per quarter (paid at the end of the month) will accumulate to after 18 years if interest is paid at a rate of 1% pa compounded quarterly. Give your answer in dollars and cents to the nearest cent.

S = $

4. Calculate the amount of money you should invest now, in an account earning 8.9% pa simple interest, in order to have $6,900 after 3 months. Give your answer in dollars and cents to the nearest cent.

Amounted invested = $

5. Find the nominal annual rate of interest convertible daily (j₃₆₅) that is equivalent to 6% pa effective. Give your answer as a percentage per annum to 3 decimal places.

j₃₆₅ = % pa

6. Calculate the discounted (present) value (P) at 5.69% pa simple interest of a payment of $73,000 due at the end of 14 months. Give your answer in dollars and cents to the nearest cent.

P = $

7. If $30,000 is paid at the end of each year for 11 years, calculate the equivalent single payment now (P) if interest is 15% pa effective. Give your answer in dollars and cents to the nearest cent.

P = $

8. An amount of $10,000 is invested on 14 April 2019 at 13% pa compounded quarterly. Calculate the interest (I) earned between 14 April 2022 and 14 April 2025. Give your answer in dollars and cents to the nearest cent.

I = $

9. Calculate the present value on 3 August 2019 of $13,500 due on 3 November 2019 at a simple interest rate of 4% pa. Give your answer in dollars and cents to the nearest cent. This days between dates calculator may assist you.

P = $

10. Neddy invests $3,061 at 11% pa simple interest and this investment grows over time to $3,274. Calculate the time period (t) over which Neddy made the investment. Give your answer in days rounded to the nearest day.

t =  days

Answer 1

1). PMT = 375; N = 10; I = 5.8%, solving for PV : PV = $2,786.4 (Answer)

2). Principal = 60,000; Total = 60,561.06

Interest earned = total - principal = 561.06

Simple interest rate = interest earned/principal = 561.06/60,000 = 0.94%

Time period from 29 Oct 2019 to Jan 2020 = 72 days

Simple interest rate is 0.94% for 72 days.

Number of times, it gets compounded in a year (365 days) = 365/72 = 5.0694

Annualized simple interest = (1+0.94%)^5.0694 -1 = 4.83% (Answer)

3). PMT = 10; I = 1%; n = 18*12 = 216, solve for FV. FV = $7,578.6 (Answer)

4). Interest rate p.a. = 8.9%

Interest rate for a quarter = 8.9%/4 = 2.23%

Let amount invested be P

Then, P + 2.23%P = 6,900

1.0223P = 6,900

P = $6,749.8 (Answer)

5). If the daily interest rate is r then

(1+r)^365 = (1+6%)^1

Solving for r, we get r = 0.0160% (Answer)

6). 5.69% p.a. for 14 months is (1+5.69%)^(14/12) -1 = 6.67%

If present value is P then

P + 6.67%P = 73,000

P = $68,435.8 (Answer)

7). PMT = 30,000; I = 15%; N = 11, solve for PV.

PV = $157,011.4 (Answer)