In: Accounting
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.
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)
- 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 37th 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 37th 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.
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 savings 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, 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 |
Weekly savings 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 =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 |
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