Question

This question is worth 10 marks in total. This is a written calculation question, and you should perform the necessary calculations/working on paper to later be scanned and uploaded. Start a new page for this question. For dollar amounts, give your answer to the nearest cent. For interest rates, give our answer as a percentage rounded to 2 decimal places.

If any parts of the question use values from earlier parts, use the EXACT values from earlier parts.

QUESTION START

You want to buy a car which will cost you $10,000. You do not have sufficient funds to purchase the car. You do not expect the price of the car to change in the foreseeable future. You can either save money or borrow money to buy the car.

• Plan 1: You decide to open a bank account and start saving money. You will purchase the car when you have sufficient savings. The nominal interest rate for the bank account is 6% per annum compounded monthly.

a) You will make regular deposits in your bank account at the start of each month for the next 2.5 years. Calculate the minimum required monthly savings to be deposited into the bank such that you would have sufficient funds to purchase the car in 2.5 years. (1 mark)

b) You will make regular deposits in your bank account at the start of each week for the next 2.5 years. Calculate the minimum required weekly savings to be deposited into the bank such that you would have sufficient funds to purchase the car in 2.5 years.

c) You will make regular deposits of $2,000 at the end of each year. Calculate how long will it take for you to have sufficient funds to purchase the car. (1 mark)

• Plan 2: You decide to borrow $13,000 from the bank and purchase the car now, as well as cover some other expenses. The bank offers two options for the structure of the repayments.

- Option 1: The first repayment will not start until you graduate from university. Therefore, no month-end-instalments will be made for the first 36 months. Then, commencing at the end of the 37^th month, a total of 30 month-end-instalments of $X will be made over the life of the loan. The nominal interest rate is 6% per annum compounded monthly.

d) Calculate X. (2 mark)

e) Your parents agree to help you repay the loan by contributing a lump sum of $1,800 when you successfully graduate from university. Calculate the new value of X. (1 mark)

- Option 2: For the first 36 months (while you are still studying), you will be making month-end-instalments of $Y. Then, commencing at the end of the 37^th month (when you graduate from university), you will double the amount of monthly repayment for the remaining 30 month-end-instalments. The nominal interest rate is 6% per annum compounded monthly.

f) Calculate the value of Y.

Answer 1

PLAN 1

a) Monthly interest rate = 6%/12 = 0.5% =0.005

No. of deposits = 2.5*12= 30

If the deposits per month are $A, future value of deposits = value of car

=> A*1.005^30+....+ A*1.005 = 10000

=> A/0.005*(1.005^30-1)*1.005 = 10000

=> 32.44142 *A =10000

=> A = $308.25

Monthly deposits required are $308.25 per month at the start of the month for 2.5 years

b) Interest rate per week (r) is given by (1+r)^52 = (1+0.005)^12 = > r = 0.00115163

No. of deposits = 2.5*52= 130

If the deposits per week are $A, future value of deposits = value of car

=> A*1.00115163^130+....+ A*1.00115163 = 10000

=> A/0.00115163*(1.00115163^130-1)*1.00115163 = 10000

=> 140.3102 *A =10000

=> A = $71.27

Deposits required are $71.27 per week at the start of the week for 2.5 years

c) Effective interest rate per year = (1+0.06/12)^12-1 = 0.061678

If n is the no of years required , then

2000/0.061678*(1.061678^n-1) = 10000

1.061678^n= 1.308389

Taking natural log of both sides

n = 4.491135

So, it will take approximately 4.5 years to accumulate the required amount to buy the car.

PLAN 2

OPTION 1

d) Monthly interest rate = 6%/12 = 0.5% =0.005

No. of deposits = 30

The present value of 30 payments must equal the loan amount of $13000

So, X/1.005^37+ .... +X/1.005^66 = 13000

=> 1/1.005^36* X/0.005*(1-1/1.005^30) = 13000

=> 23.22596 *X = 13000

=> X =$559.72

e) If the deposit of $1800 is made to bank at the end of 36th month by parents,

X/1.005^37+ .... +X/1.005^66 + 1800/1.005^36 = 13000

=> 1/1.005^36* X/0.005*(1-1/1.005^30) + 1504.16= 13000

=> 23.22596 *X = 11495.84

=> X =$494.96

PLAN 2

f) In this case

(Y/1.005+ Y/1.005^2+...+Y/1.005^36)+ (2Y/1.005^37+...+2Y/1.005^66) = 13000

=> Y/0.005*(1-1/1.005^36) +  1/1.005^36* 2Y/0.005*(1-1/1.005^30) = 13000

=> 79.32294*Y = 13000

=> Y =$163.89