Question
Imani invests $3000 in a bank that pays a nominal annual interest rate of 1.25% compounded monthly.
(a) Calculate the amount of money Imani will have in the bank at the end of 6 years. Give your answer correct to two decimal places.
(b) Calculate the number of months it takes until Imani has at least $3550 in the bank.
Imani uses the $3550 as a partial payment for a used car costing $22,000. For the remainder, she takes out a loan from a bank.
(c) Write down the amount of money that Imani takes out as a loan.
The loan is for 8 years and the nominal annual interest rate is 12.6% compounded monthly. Imani will pay the loan in fixed monthly installments at the end of each month.
(d) Calculate the amount, correct to the nearest dollar, that Imani will have to pay the bank each month.
▶️ Answer/Explanation
Detailed Solution
(a) Compound Interest Calculation:
Using either the finance app on a GDC or the compound interest formula:
\[ FV = 3000 \left( 1 + \frac{1.25}{1200} \right)^{72} \]
where:
- \( N = 72 \) (6 years × 12 months)
- \( I = 1.25\% \) annual interest
- \( PV = -3000 \) (initial investment)
- \( P/Y = 12 \) (payments per year)
- \( C/Y = 12 \) (compounding per year)
Calculating, we get:
\[ FV = 3233.53 \]
Conclusion: The future value is \$3233.53, correct to two decimal places.
(b) Time Required to Reach $3550:
Using either the finance app on GDC or the compound interest formula:
\[ 3550 = 3000 \left( 1 + \frac{1.25}{1200} \right)^N \]
Solving for \( N \):
\[ N = 162 \quad (\text{or } 161.686\ldots \text{ months}) \]
Conclusion: It takes approximately 162 months for Imani’s investment to reach at least \$3550.
(c) Loan Amount:
\[ \text{Loan} = 22000 – 3550 = 18450 \]
Conclusion: Imani takes out a loan of \$18,450.
(d) Monthly Loan Payments:
Using the loan payment formula:
\[ PMT = \frac{PV \times r}{1 – (1 + r)^{-N}} \]
where:
- \( N = 96 \) (8 years × 12 months)
- \( I = 12.6\% \) nominal annual interest
- \( PV = -18450 \) (loan amount)
- \( FV = 0 \) (loan fully paid)
- \( P/Y = 12 \), \( C/Y = 12 \) (monthly payments and compounding)
Calculating, we get:
\[ PMT = 306 \]
Conclusion: The monthly payment Imani has to make is \$306.
…………………………..Markscheme…………………………..
- (a) Correct use of compound interest formula and solving for future value.
- (b) Correct logarithmic approach to solve for \( N \).
- (c) Correct subtraction to determine loan amount.
- (d) Correct loan payment formula application and solving for monthly payment.
Question
Maan deposited $100,000 into a savings account with a nominal annual interest rate of \( I \)% compounded monthly. At the end of the eighth year, the amount in the account had increased to $150,000.
(a) Find the value of \( I \). [3]
Maan withdraws the $150,000 and places it in an annuity, earning a nominal annual interest rate of 6.1% compounded monthly. At the end of each month, Maan will receive a payment of $1000.
(b) Find the amount of money remaining in the annuity at the end of 10 years. Express your answer to the nearest dollar.
▶️Answer/Explanation
Detailed solution
(a) Finding the Interest Rate \( I \)
The compound interest formula is:
\[ FV = PV \left(1 + \frac{I}{100 \times 12}\right)^{12 \times t} \]
Substituting the given values:
\[ 150000 = 100000 \left(1 + \frac{I}{1200}\right)^{96} \]
Rearranging for \( I \):
\[ \left(1 + \frac{I}{1200}\right)^{96} = \frac{150000}{100000} \]
\[ \left(1 + \frac{I}{1200}\right)^{96} = 1.5 \]
Taking the natural logarithm on both sides:
\[ 96 \ln \left(1 + \frac{I}{1200}\right) = \ln(1.5) \]
\[ \ln \left(1 + \frac{I}{1200}\right) = \frac{\ln(1.5)}{96} \]
Solving for \( I \):
\[ 1 + \frac{I}{1200} = e^{\frac{\ln(1.5)}{96}} \]
\[ \frac{I}{1200} = e^{\frac{\ln(1.5)}{96}} – 1 \]
\[ I = 1200 \times (e^{\frac{\ln(1.5)}{96}} – 1) \]
\[ I = 5.08\% (5.07903\ldots) \]
So, the annual nominal interest rate is 5.08%.
(b) Finding the Amount Remaining in the Annuity After 10 Years
The future value of an annuity formula is:
\[ FV = PV(1 + r)^N – \frac{PMT}{r} \left( (1 + r)^N – 1 \right) \]
Given:
- \( N = 120 \) (10 years × 12 months)
- \( I = 6.1\% \) per year → \( r = \frac{6.1}{1200} \) per month
- \( PV = 150000 \)
- \( PMT = 1000 \)
Substituting the values:
\[ FV = 150000(1 + \frac{6.1}{1200})^{120} – \frac{1000}{\frac{6.1}{1200}} \left( (1 + \frac{6.1}{1200})^{120} – 1 \right) \]
Solving for \( FV \):
\[ FV = 110867 \]
So, the amount remaining in the annuity after 10 years is $110,867.
……………………………Markscheme……………………………….
(a)
$N=96$
$PV=\mp100000$
$FV=\pm150000$
$P/Y=12$
$C/Y=12$
OR
$150000=100000(1+\frac{I}{100\times12})^{12\times8}$
$I = 5.08 (5.07903…)$
(b)
$N = 120$
$I\% = 6.1$
$PV = \pm 150000$
$PMT = \pm 1000$
$P/Y = 12$
$C/Y = 12$
$FV = (\$) 110867$