U1 Question 1 AI HL Paper 3

Number and Algebra

• Finance • Loans and Mortgages • Compound Interest • Annuities • Amortization • Interpretation

1) Overview

A buyer wants to purchase a house and takes out a long-term mortgage from a bank. To understand the true cost of borrowing, we need to model the down payment, the monthly interest rate, the fixed monthly repayment, and the effect of paying extra each month.

This is a classic IB Applications and Interpretation finance investigation. It uses the annuity formula for loan repayments, then builds further to find the total paid, the number of payments under a faster repayment plan, the final smaller payment, and the total saving. Extra parts have been added to make the structure more complete and easier for students to follow independently.

2) Key Concepts Related to the Question

You do not need the following formulas for IB Applications and Interpretation, but it is useful to know them.

A down payment is paid immediately, so the loan principal is the house price minus the down payment.
If the nominal annual interest rate is compounded monthly, then the monthly interest rate is

$i=\frac{r}{12}$

The present value of a loan repaid by fixed monthly payments is modelled by an annuity formula.
If the principal is $L$ , the monthly payment is $P$ , the monthly interest rate is $i$ , and the number of monthly payments is $n$ , then

$L=P\cdot \frac{1-(1+i)^{-n}}{i}$

Rearranging gives the monthly payment formula

$P=L\cdot \frac{i}{1-(1+i)^{-n}}$

Paying more each month reduces both the repayment time and the total interest paid.
In the final month of a loan, the payment is usually smaller because only the remaining balance plus one month of interest must be paid.

TI-84 TVM Solver Guide

You can solve the main finance parts of this question quickly using the TVM Solver in the TI-84 Finance app. The TVM Solver solves for one variable when the others are known. The main variables are:

N = total number of payments
I% = annual nominal interest rate (as a percentage)
PV = present value, or loan amount
PMT = payment each period
FV = future value
P/Y = payments per year
C/Y = compounding periods per year

On the TI-84 Plus CE Finance app, the TVM Solver is opened from the Finance app, and you enter four TVM values, set P/Y and C/Y, choose END or BEGIN, then move to the unknown variable and press ALPHA then ENTER to solve. TI also notes that cash inflows should be entered as positive values and cash outflows as negative values.

How to open the TVM Solver

Press APPS, choose Finance, then select TVM Solver. On TI’s guide, this appears as entering the Finance app first and then choosing 1: TVM Solver.

Using TVM Solver for part (b)(i): monthly payment

For Talia’s mortgage:

$PV=393600,\qquad N=300,\qquad I\%=6.6,\qquad FV=0$

Because payments are monthly and interest is compounded monthly, enter

$P/Y=12,\qquad C/Y=12$

Since payments are made at the end of each month, choose END. This matches TI’s description that END means payments are made at the end of each period.

Enter the loan as money received, so use

$PV=393600$

and let the calculator return a negative payment, since the monthly repayment is money paid out. TI’s finance guide uses this cash-flow sign convention.

So enter:

N = 300
I% = 6.6
PV = 393600
PMT = leave blank or put any value, since this is what you are solving for
FV = 0
P/Y = 12
C/Y = 12
PMT: END

Then move the cursor to PMT and press ALPHA then ENTER to solve. The calculator will return approximately

$PMT\approx -2660.38$

The negative sign means money is leaving Talia. So the monthly payment is

$\$2660.38$

Using TVM Solver for part (d)(i): number of payments if PMT = $3200

Now keep the same loan information, but change the payment:

I% = 6.6
PV = 393600
PMT = -3200
FV = 0
P/Y = 12
C/Y = 12
PMT: END

Move to N and press ALPHA then ENTER.
The calculator gives approximately

$N\approx 205.9$

So Talia needs

$206$

payments in total.

Helpful exam note

For loan questions, it is usually best to enter:

PV as positive, because it is money borrowed,
PMT as negative, because repayments are money paid out,
FV = 0, because the loan should be fully paid off at the end.

3) Question and Fully Worked Solutions

Talia wants to buy a house costing

$\$480\,000$

To qualify for a mortgage, she must make an initial down payment equal to

$18\%$

of the house price.

The bank offers her a

$25\text{-year}$

loan for the remaining balance at a nominal interest rate of

$6.6\%\text{ per annum}$

compounded monthly.
She will repay the loan in fixed monthly payments made at the end of each month.

(a)(i) Find the down payment.

Solution

The down payment is

$18\%\text{ of }480000$

$=0.18(480000)$

$=86400$

Answer:

$\$86\,400$

(a)(ii) Find the original amount of the loan after the down payment is paid.

Solution

The loan principal is

$480000-86400$

$=393600$

Answer:

$\$393\,600$

(a)(iii) Find the monthly interest rate.

Solution

The nominal annual rate is

$6.6\%=0.066$

So the monthly rate is

$i=\frac{0.066}{12}=0.0055$

Answer:

$i=0.0055$

which is 0.55% per month.

(a)(iv) State the total number of monthly payments in the original loan plan.

Solution

A 25-year loan with monthly payments has

$25\times 12=300$

payments.

Answer:

$300$

monthly payments.

(b)(i) Calculate Talia’s monthly payment for this loan, to two decimal places.

Solution

Use

$P=L\cdot \frac{i}{1-(1+i)^{-n}}$

with

$L=393600,\qquad i=0.0055,\qquad n=300$

$P=393600\cdot \frac{0.0055}{1-(1.0055)^{-300}}$

$P\approx 2660.38$

Answer:

$\$2660.38$

per month.

(b)(ii) Explain why the monthly payment is greater than simply dividing the loan amount by 300.

Solution

If there were no interest, the payment would just be the loan divided equally over 300 months. But here, interest is charged every month on the remaining balance, so the payment must cover both principal and interest.

Answer: The monthly payment is greater because it must pay off both the borrowed amount and the monthly interest.

(c)(i) Using your answer to part (b)(i), calculate the total amount Talia will pay over the life of the loan. Do not include the down payment.

Solution

She makes

$300$

payments of

$\$2660.38$

So the total paid is

$300(2660.38)=798114$

Answer: The total amount paid on the loan is approximately

$\$798\,114$

(c)(ii) Hence find the total interest paid over the life of the loan.

Solution

Interest paid is total loan repayment minus the original loan:

$798114-393600=404514$

Answer: The total interest paid is approximately

$\$404\,514$

Talia decides to repay the loan faster by increasing her monthly payment to

$\$3200$

per month.

(d)(i) Find the total number of monthly payments she will need to make to pay off the loan.

Solution

Use the annuity model

$L=P\cdot \frac{1-(1+i)^{-n}}{i}$

with

$L=393600,\qquad P=3200,\qquad i=0.0055$

$393600=3200\cdot \frac{1-(1.0055)^{-n}}{0.0055}$

Multiply through:

$393600\cdot \frac{0.0055}{3200}=1-(1.0055)^{-n}$

$0.6765=1-(1.0055)^{-n}$

$(1.0055)^{-n}=0.3235$

Take logarithms:

$-n\ln(1.0055)=\ln(0.3235)$

$n=\frac{-\ln(0.3235)}{\ln(1.0055)}$

$n\approx 205.9$

So she will need

$206$

monthly payments, with the final one smaller than $3200.

Answer: She will need 206 monthly payments.

(d)(ii) Explain why the answer from part (d)(i) must be rounded up.

Solution

A value like 205.9 means that 205 full payments are not enough to finish the loan. She must make one more payment to clear the remaining balance.

Answer: It must be rounded up because a loan cannot be cleared with only part of a monthly payment count; one more full payment month is needed.

(e)(i) Find the remaining balance immediately after the 205th payment.

Solution

The balance after 205 payments is the present value of the one remaining payment month, or equivalently the unpaid balance after amortization:

$B_{205}=L(1+i)^{205}-P\cdot \frac{(1+i)^{205}-1}{i}$

Substitute the values:

$B_{205}=393600(1.0055)^{205}-3200\cdot \frac{(1.0055)^{205}-1}{0.0055}$

$B_{205}\approx 1546.52$

Answer: The remaining balance after the 205th payment is approximately

$\$1546.52$

(e)(ii) Determine the amount of Talia’s final payment, to two decimal places.

Solution

Before the final payment is made, one more month of interest is added:

$\text{Final payment}=1546.52(1.0055)$

$\approx 1555.03$

Answer: The final payment is approximately

$\$1555.03$

(e)(iii) Find the total amount Talia pays under the faster repayment plan.

Solution

She makes

$205$

payments of

$\$3200$

and then a final payment of

$\$1555.03$

So the total is

$205(3200)+1555.03$

$=656000+1555.03$

$=657555.03$

Answer: The total amount paid is approximately

$\$657\,555.03$

(f)(i) Hence determine the total amount Talia saves, to the nearest dollar, by making the higher monthly payments.

Solution

Original total paid:

$\$798114$

New total paid:

$\$657555.03$

So the saving is

$798114-657555.03=140558.97$

Answer: The total saving is approximately

$\$140\,559$

(f)(ii) Find how many years and months earlier the loan is repaid under the faster plan.

Solution

Original duration:

$300\text{ months}$

New duration:

$206\text{ months}$

Difference:

$300-206=94\text{ months}$

Convert to years and months:

$94=7\times 12+10$

Answer: The loan is repaid 7 years and 10 months earlier.

(g) State one financial reason, other than the total saving, why the faster repayment plan could still be difficult for some borrowers.

Solution

Although the faster plan reduces interest overall, it requires a much larger monthly commitment, which may not be affordable within a normal household budget.

Answer: One reason is that the higher monthly payment may place too much pressure on monthly cash flow.

Common Mistakes

Calculating the down payment correctly but forgetting to subtract it from the house price before finding the loan amount.
Using the annual interest rate directly instead of converting to a monthly rate.
Using 25 instead of 300 monthly payments.
Rounding the monthly repayment too early and then carrying that rounded value through all later calculations without care.
Rounding the number of payments down instead of up.
Assuming the final payment is another full $3200 payment.
Comparing totals incorrectly by including the down payment in one case but not the other.

One Additional Practice Question

Marcus wants to buy an apartment costing

$\$360\,000$

He must make a down payment of

$15\%$

of the purchase price.
The bank offers a

$20\text{-year}$

loan at a nominal interest rate of

$5.4\%\text{ per annum}$

compounded monthly.
He makes fixed monthly repayments at the end of each month.
Later, he increases his monthly payment to

$\$2500$

to clear the loan faster.

Answer the following:

(a) Find the down payment and the loan principal.
(b) Find the monthly interest rate and the total number of monthly payments in the original plan.
(c) Calculate the original monthly payment.
(d) Find the total amount paid over the life of the original loan.
(e) Find the total amount of interest paid in the original plan.
(f) Determine how many monthly payments are needed if Marcus pays $2500 each month.
(g) Find the amount of the final payment.
(h) Find the total paid under the faster plan.
(i) Find the total saving and how much earlier the loan is repaid.

Show answers

(a)

$\text{Down payment}=0.15(360000)=54000$

$\text{Loan principal}=360000-54000=306000$

(b)

$i=\frac{0.054}{12}=0.0045$

$n=20\times 12=240$

(c)

$P=306000\cdot \frac{0.0045}{1-(1.0045)^{-240}} \approx 2088.52$

(d)

$240(2088.52)=501244.80$

(e)

$501244.80-306000=195244.80$

(f)

Solve

$306000=2500\cdot \frac{1-(1.0045)^{-n}}{0.0045}$

to obtain

$n\approx 163.1$

$164$

payments are needed.

(g)

Using the remaining balance after 163 full payments gives a final payment of approximately

$\$419.46$

(h)

$163(2500)+419.46=407919.46$

(i)

Saving:

$501244.80-407919.46=93325.34$

So the saving is about

$\$93\,325$

Time saved:

$240-164=76\text{ months}=6\text{ years }4\text{ months}$

Number and Algebra

1) Overview

2) Key Concepts Related to the Question

TI-84 TVM Solver Guide

How to open the TVM Solver

Using TVM Solver for part (b)(i): monthly payment

Using TVM Solver for part (d)(i): number of payments if PMT = $3200

Helpful exam note

3) Question and Fully Worked Solutions

(a)(i) Find the down payment.

(a)(ii) Find the original amount of the loan after the down payment is paid.

(a)(iii) Find the monthly interest rate.

(a)(iv) State the total number of monthly payments in the original loan plan.

(b)(i) Calculate Talia’s monthly payment for this loan, to two decimal places.

(b)(ii) Explain why the monthly payment is greater than simply dividing the loan amount by 300.

(c)(i) Using your answer to part (b)(i), calculate the total amount Talia will pay over the life of the loan. Do not include the down payment.

(c)(ii) Hence find the total interest paid over the life of the loan.

(d)(i) Find the total number of monthly payments she will need to make to pay off the loan.

(d)(ii) Explain why the answer from part (d)(i) must be rounded up.

(e)(i) Find the remaining balance immediately after the 205th payment.

(e)(ii) Determine the amount of Talia’s final payment, to two decimal places.

(e)(iii) Find the total amount Talia pays under the faster repayment plan.

(f)(i) Hence determine the total amount Talia saves, to the nearest dollar, by making the higher monthly payments.

(f)(ii) Find how many years and months earlier the loan is repaid under the faster plan.

(g) State one financial reason, other than the total saving, why the faster repayment plan could still be difficult for some borrowers.

Common Mistakes

One Additional Practice Question

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IB Mathematics: Applications and Interpretation HL Paper 3 Exam Practice Questions

U1 Question 1 AI HL Paper 3

Number and Algebra

1) Overview

2) Key Concepts Related to the Question

TI-84 TVM Solver Guide

How to open the TVM Solver

Using TVM Solver for part (b)(i): monthly payment

Using TVM Solver for part (d)(i): number of payments if PMT = $3200

Helpful exam note

3) Question and Fully Worked Solutions

(a)(i) Find the down payment.

(a)(ii) Find the original amount of the loan after the down payment is paid.

(a)(iii) Find the monthly interest rate.

(a)(iv) State the total number of monthly payments in the original loan plan.

(b)(i) Calculate Talia’s monthly payment for this loan, to two decimal places.

(b)(ii) Explain why the monthly payment is greater than simply dividing the loan amount by 300.

(c)(i) Using your answer to part (b)(i), calculate the total amount Talia will pay over the life of the loan. Do not include the down payment.

(c)(ii) Hence find the total interest paid over the life of the loan.

(d)(i) Find the total number of monthly payments she will need to make to pay off the loan.

(d)(ii) Explain why the answer from part (d)(i) must be rounded up.

(e)(i) Find the remaining balance immediately after the 205th payment.

(e)(ii) Determine the amount of Talia’s final payment, to two decimal places.

(e)(iii) Find the total amount Talia pays under the faster repayment plan.

(f)(i) Hence determine the total amount Talia saves, to the nearest dollar, by making the higher monthly payments.

(f)(ii) Find how many years and months earlier the loan is repaid under the faster plan.

(g) State one financial reason, other than the total saving, why the faster repayment plan could still be difficult for some borrowers.

Common Mistakes

One Additional Practice Question

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