DP Maths AA SL - Functions - Modeling with Functions

DP Maths AA SL - Functions - Modeling with Functions - Full notes

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DP Maths AA SL | Modeling with Functions

Modeling with Functions

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1.4 Financial Applications

DP MATHEMATICS AA · TOPIC 1: NUMBER AND ALGEBRA

1.4

Financial Applications

Full notes for SL · Syllabus SL 1.4

Contents

1	Simple and compound growth
2	Compound interest
3	Compounding periods
4	Finding the time, rate or principal
5	Depreciation
6	Inflation and real value
7	Using the financial solver (TVM)
8	Loans and annuities
9	Compound interest as a geometric sequence
10	Compounding frequency and nominal rates
11	Effective annual rate
12	Depreciation
13	Inflation and real value
14	Loans and repayments
15	Mortgages and the effect of the term
16	Savings plans and annuities
17	Finding the number of payments or the rate
18	Investment decisions
19	Mixed review: worked problems
20	Additional exam-style practice
21	Exam-style questions with markscheme
22	Common misconceptions and links
23	Check your understanding
24	Summary and topic map
25	Answers to practice exercises

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and work through every worked example before attempting the Quick check.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
distinguish simple and compound growth and link them to arithmetic and geometric sequences
calculate compound interest with annual and more frequent compounding
find the time, rate or principal needed to reach a target value
calculate depreciation using a geometric model
adjust for inflation to find real values
use the financial (TVM) solver correctly, including the sign convention
calculate repayments and totals for loans and regular savings

This booklet covers SL 1.4, which is common to SL and HL.

Guiding questions

How can sequences model money growing or losing value over time? Why does the frequency of compounding matter? How do banks calculate loan repayments?

How this topic is examined

Financial applications are usually tested on Paper 2, where a GDC with a TVM solver is available. Questions ask for final values, interest rates, the number of years needed, comparisons between investment options, and loan repayments.

Show the formula or list every TVM input (N, I%, PV, PMT, FV, P/Y, C/Y) you used: without this, a wrong answer earns no method marks. Give money answers to two decimal places.

Command term	What the examiner wants
Write down	an answer with little or no working; it should be obtainable by inspection
Find / Calculate / Determine	obtain the answer showing relevant working
Solve	find the values of the unknowns, using algebra or technology as allowed
Show that	obtain the given result; every step must be shown and the result must not be assumed
Hence / Hence or otherwise	use the previous result (or another valid method)
Sketch	a diagram showing the general shape and key features, labelled
Prove	a formal, rigorous argument from accepted facts
Justify / Explain	give valid reasons or evidence

How to use these notes

Each section teaches one part of the syllabus. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section
Why it works	a derivation or proof of a key result, so you understand where formulas come from
Key point	a result or idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
GDC tip	how to use your graphic display calculator efficiently on Paper 2
In the real world	where the mathematics is used outside the classroom
Worked example	a full solution with IB-style marks (M1 method, A1 answer, R1 reasoning, AG answer given); examples are graded basic, standard and exam level
Practice exercise	graded questions for each section; full answers are at the end of the booklet
Quick check	short questions to check the key ideas

1 Simple and compound growth

Interest on interest.

Think about it

You invest $1000 at 5% per year. Will you have the same amount after 10 years whether interest is added to the balance or paid out to you each year?

With simple interest, interest is calculated on the original amount (the principal) only. $1000 at 5% earns $50 every year, so the balance after n years is 1000 + 50n: an arithmetic sequence. After 10 years it is $1500.

With compound interest, interest is added to the balance at the end of each period and then earns interest itself. The balance is multiplied by 1.05 each year: a geometric sequence. After 10 years it is 1000 × 1.0510 = $1628.89. The extra $128.89 is 'interest on interest'.

Over long periods, compound growth outstrips simple growth dramatically: after 40 years at 5%, simple interest gives $3000 while compound interest gives $7039.99.

Figure 1. Compound growth overtakes simple growth.

Practice exercise 1

1. Find the simple interest on $2400 at 3.5% per year for 6 years.

2. Find the value of $2400 at 3.5% compound interest per year after 6 years.

3. Which sequence type models a balance with simple interest, and why?

2 Compound interest

A geometric sequence of balances.

If a present value PV is invested at an annual interest rate of r% compounded once a year, the future value after n years is:

This is the nth term of a geometric sequence starting at PV (with u0 = PV) and ratio 1 + r/100. The interest earned is FV − PV.

WORKED EXAMPLE 1 Future value (basic)

Find the value of $4500 invested for 8 years at 3.2% per annum compounded annually, and the interest earned.

Solution

FV = 4500 × 1.0328M1

= $5789.62A1

Interest = 5789.62 − 4500 = $1289.62A1

WORKED EXAMPLE 2 Present value (standard)

How much must be invested now at 4% compounded annually to have $10 000 in 7 years?

Solution

PV × 1.047 = 10 000M1

PV = 10 000 ÷ 1.047M1

= $7599.18A1

Practice exercise 2

1. Find the value of $600 after 3 years at 10% compound interest.

2. Find the value of $15 000 after 12 years at 2.75% compounded annually.

3. How much must be invested now at 6% to have $5000 in 10 years?

3 Compounding periods

More frequent compounding, slightly more interest.

Interest is often compounded more than once a year: half-yearly (k = 2), quarterly (k = 4), monthly (k = 12) or daily (k = 365). The annual rate r% is divided equally between periods, so the rate per period is r/k %, and the number of periods is kn:

WORKED EXAMPLE 3 Monthly compounding (basic)

Find the value of $5000 after 6 years at 4% per annum compounded monthly.

Solution

FV = 5000 × (1 + 4/1200)72M1

= $6353.71A1

Compounding	k	Value of $1000 after 1 year at 12%
annually	1	$1120.00
half-yearly	2	$1123.60
quarterly	4	$1125.51
monthly	12	$1126.83
daily	365	$1127.47

The more frequent the compounding, the larger the final value, but the gains become very small. As k → ∞ (continuous compounding), the factor approaches er/100: $1000 at 12% compounded continuously gives 1000e0.12 = $1127.50.

Effective annual rate. To compare offers, compute the equivalent annual rate: (1 + r/(100k))k − 1. 12% compounded monthly is equivalent to 12.68% per year.

WORKED EXAMPLE 4 Comparing offers (exam level)

Bank A offers 5.1% compounded annually; bank B offers 5% compounded monthly. Which gives more after one year?

Solution

A: 1.051 per yearA1

B: (1 + 5/1200)12 = 1.0512M1 A1

Bank B is slightly better (5.12% effective)R1

Common mistake

Using r instead of r/k for each period, or n instead of kn for the number of periods.

Practice exercise 3

1. Find the value of $3000 after 2 years at 6% compounded quarterly.

2. Find the effective annual rate for 8% compounded quarterly.

3. $2000 is invested at 3% compounded half-yearly. Find its value after 5 years.

4. Which is better: 6% compounded annually or 5.85% compounded monthly?

4 Finding the time, rate or principal

Solve with logarithms or technology.

If the unknown is n, use logarithms (1.5) or the TVM solver. If the unknown is the rate, take the nth root. If the unknown is the principal, divide.

WORKED EXAMPLE 5 Finding the time (standard)

How many complete years are needed for $2000 to grow to more than $3000 at 4.5% per annum compounded annually?

Solution

2000 × 1.045n > 3000, so 1.045n > 1.5M1

n > ln 1.5 ÷ ln 1.045 = 9.21A1

10 yearsA1

WORKED EXAMPLE 6 Finding the rate (standard)

An investment of $8000 grows to $10 000 in 5 years with annual compounding. Find the interest rate.

Solution

8000(1 + r/100)5 = 10 000M1

1 + r/100 = 1.251/5 = 1.0456A1

r = 4.56% (3 s.f.)A1

WORKED EXAMPLE 7 Time with monthly compounding (exam level)

How many months will it take $5000 to reach $7000 at 3.6% per annum compounded monthly?

Solution

5000(1.003)m = 7000M1

m = ln 1.4 ÷ ln 1.003 = 112.3A1

113 monthsA1

Practice exercise 4

1. Find n if 1.06n = 2 (to 3 s.f.).

2. How long does it take money to double at 8% compounded annually?

3. Find the annual rate that turns $2500 into $4000 in 9 years.

4. How many quarters for $1000 to exceed $1500 at 5% compounded quarterly?

5 Depreciation

Losing value by a fixed percentage.

Depreciation is the loss in value of an asset (such as a car or machinery) over time. If the value decreases by r% each year, the value after n years is a geometric sequence with ratio 1 − r/100:

WORKED EXAMPLE 8 Depreciation (standard)

A machine bought for $45 000 depreciates by 18% per year. (a) Find its value after 4 years. (b) After how many complete years will its value first fall below $10 000?

Solution

(a) 45 000 × 0.824 = $20 345.48A1

(b) 45 000 × 0.82n < 10 000, n > ln(2/9) ÷ ln 0.82 = 7.58M1

8 yearsA1

WORKED EXAMPLE 9 Finding a depreciation rate (exam level)

A car bought for $32 000 is worth $14 000 after 6 years. Find the annual rate of depreciation.

Solution

32 000(1 − r/100)6 = 14 000M1

1 − r/100 = (14/32)1/6 = 0.8711A1

r = 12.9%A1

Practice exercise 5

1. A phone worth $1200 depreciates by 30% per year. Find its value after 3 years.

2. A boat loses 8% of its value per year. After how many years is it worth less than half its original value?

3. A $20 000 car is worth $9000 after 5 years. Find the rate of depreciation.

6 Inflation and real value

Money loses purchasing power.

Inflation is the general increase in prices over time. It reduces the purchasing power of money: if prices rise 3% per year, $100 buys less each year.

The real value of an investment measures what it can buy in today's prices. A simple approach used in the AA course: the real interest rate ≈ nominal interest rate − inflation rate. If savings earn 5% and inflation is 2%, real growth is about 3% per year. (A more precise version divides the growth factors: 1.05/1.02 = 1.0294, a real rate of 2.94%.)

To find what amount will be needed in future to buy the same goods, grow today's price at the inflation rate.

WORKED EXAMPLE 10 Real value (standard)

$10 000 is invested for 5 years at 4.5% per annum compounded annually. Inflation is 2.5% per year. Estimate the real value of the investment in today's money.

Solution

Real rate ≈ 4.5 − 2.5 = 2%M1

Real value ≈ 10 000 × 1.025 = $11 040.81A1

WORKED EXAMPLE 11 Future cost (standard)

A car costs $25 000 today. If inflation averages 3.2% per year, estimate its cost in 8 years.

Solution

25 000 × 1.0328M1

= $32 164.56A1

Practice exercise 6

1. Nominal rate 6%, inflation 3.5%. Approximate real rate?

2. $5000 is invested at 3% for 10 years with inflation 2%. Estimate its real value.

3. A coffee costs $4.50 today. Estimate its price in 15 years with 2.8% inflation.

7 Using the financial solver (TVM)

Let the GDC do the arithmetic, but enter it correctly.

On Paper 2, the time value of money (TVM) solver can find any one unknown. Inputs: N = total number of compounding periods (years × C/Y); I% = annual interest rate; PV = present value; PMT = regular payment (0 if none); FV = future value; P/Y = payments per year; C/Y = compounding periods per year.

Sign convention. Money leaving you (invested, deposited, repaid) is negative; money coming to you (received, withdrawn, borrowed) is positive. So an investment of $5000 is entered as PV = −5000 and FV comes out positive.

Input	Meaning	Example ($5000, 6 years, 4%, monthly)
N	number of periods	72
I%	annual rate	4
PV	present value (negative if invested)	−5000
PMT	regular payment	0
FV	future value	solve: 6353.71
P/Y, C/Y	periods per year	12, 12

GDC tip

If the solver gives an error, check the signs: PV and FV must have opposite signs. Also check that N is in periods, not years, and that P/Y and C/Y are set correctly before solving.

Common mistake

Entering PV and FV with the same sign, which gives an error, or entering N in years when compounding is monthly.

8 Loans and annuities

Regular payments.

A loan is repaid with regular payments that cover both interest and principal: this is amortization. The TVM solver finds the regular payment (PMT). For a loan, PV is positive (you receive the money), PMT is negative (you pay) and FV = 0 (the loan is fully repaid).

An annuity is a series of regular payments into or out of an account, for example regular savings or a pension. For regular savings, PMT is negative and FV is the accumulated value.

Total amount paid = PMT × N; total interest = total paid − amount borrowed.

WORKED EXAMPLE 12 Loan repayments (standard)

A $20 000 loan is repaid monthly over 5 years at 6% per annum compounded monthly. Find the monthly repayment and the total interest.

Solution

N = 60, I% = 6, PV = 20 000, FV = 0, P/Y = C/Y = 12M1

PMT = −386.66: $386.66 per monthA1

Total paid 60 × 386.66 = $23 199.60; interest $3199.60A1

WORKED EXAMPLE 13 Regular savings (standard)

Ana deposits $200 at the end of every month into an account paying 4.8% per annum compounded monthly. Find the value after 10 years.

Solution

N = 120, I% = 4.8, PV = 0, PMT = −200, P/Y = C/Y = 12M1

FV = $30 726.39A1

Total deposited $24 000; interest earned $6726.39A1

WORKED EXAMPLE 14 Loan term (exam level)

A $250 000 mortgage at 5.2% per annum compounded monthly is repaid at $1800 per month. How many years will it take?

Solution

I% = 5.2, PV = 250 000, PMT = −1800, FV = 0, P/Y = C/Y = 12M1

N = 213.0 monthsA1

17.7 years (213 payments)A1

Practice exercise 7

1. Find the monthly repayment on a $15 000 loan over 4 years at 7.5% compounded monthly.

2. How much interest is paid in total on that loan?

3. Find the value after 20 years of saving $150 per month at 5% compounded monthly.

4. What monthly payment is needed to repay $8000 in 2 years at 9% compounded monthly?

9 Compound interest as a geometric sequence

Linking finance to sequences.

With annual compounding at r%, the balances at the end of each year form a geometric sequence with first term PV(1 + r/100) and ratio 1 + r/100. This explains the formula FV = PV(1 + r/100)n, and allows questions about 'which year' to be answered with sequence techniques or tables.

WORKED EXAMPLE 15 Year of reaching a target (standard)

$4000 is invested at 3.8% compounded annually. In which year does the balance first exceed $6000?

Solution

4000(1.038)n > 6000M1

n > ln 1.5/ln 1.038 = 10.87A1

During year 11A1

WORKED EXAMPLE 16 Comparing simple and compound (standard)

Compare $5000 at 6% simple interest with $5000 at 5% compound interest after 10 and 20 years.

Solution

Simple: 8000 and 11 000A1

Compound: 8144.47 and 13 266.49A1

Compound overtakes simple before year 10R1

Practice exercise 8

1. Find the value of $2500 after 12 years at 4.5% compounded annually.

2. How many years does it take for money to triple at 7% compounded annually?

3. Find the rate needed for $1000 to grow to $1500 in 8 years with annual compounding.

10 Compounding frequency and nominal rates

Monthly, quarterly, daily.

A nominal annual rate of r% compounded k times per year gives r/k % per period and kn periods in n years. The more frequent the compounding, the larger the final value, but the gains diminish quickly.

Compounding	k	Value of $1000 after 1 year at 6%
annually	1	$1060.00
quarterly	4	$1061.36
monthly	12	$1061.68
daily	365	$1061.83

WORKED EXAMPLE 17 Quarterly compounding (standard)

Find the value of $7500 after 6 years at 4.8% compounded quarterly.

Solution

7500(1 + 0.048/4)24M1

= $9986.05A1

Practice exercise 9

1. Find the value of $3000 after 5 years at 3.6% compounded monthly.

2. Find the interest earned on $10 000 at 2.4% compounded quarterly for 3 years.

11 Effective annual rate

Comparing offers fairly.

The effective annual rate is the annual compound rate that gives the same growth as the nominal rate with its compounding. Banks quote it (often as the AER or APY) so that savers can compare products.

WORKED EXAMPLE 18 Comparing two accounts (exam level)

Account A pays 4.6% compounded annually; account B pays 4.5% compounded monthly. Which is better?

Solution

A: 4.6%; B: (1 + 0.045/12)12 − 1 = 4.594%M1 A1

A is very slightly betterR1

Practice exercise 10

1. Find the effective rate of 7% compounded quarterly.

2. Find the nominal rate compounded monthly that gives an effective rate of 5%.

12 Depreciation

Loss of value as a percentage.

Assets such as cars and machinery lose value over time. Reducing-balance depreciation at r% per year gives V = V0(1 − r/100)n, a geometric decrease. Straight-line depreciation (a fixed amount each year) is arithmetic.

WORKED EXAMPLE 19 Depreciation (standard)

A machine bought for $64 000 depreciates by 12% per year. Find its value after 6 years and when it first falls below $20 000.

Solution

64 000(0.88)6 = $29 721.86A1

0.88n < 0.3125: n > 9.10M1

After 10 yearsA1

WORKED EXAMPLE 20 Finding the rate (standard)

A car bought for $28 000 is worth $15 400 after 4 years. Find the annual depreciation rate.

Solution

(1 − r/100)4 = 0.55M1

r = 13.9%A1

Practice exercise 11

1. A laptop costs $1600 and depreciates by 25% per year. Find its value after 3 years.

2. Compare 20% reducing-balance depreciation with straight-line depreciation of $3000 per year for a $15 000 van after 4 years.

13 Inflation and real value

What will money buy?

Inflation reduces purchasing power. To find the real value of a future amount in today's money, divide by (1 + i/100)n, where i% is the inflation rate. Approximately, the real interest rate ≈ nominal rate − inflation rate; the exact real rate is (1 + r)/(1 + i) − 1.

WORKED EXAMPLE 21 Real value (exam level)

$12 000 is invested for 8 years at 5% compounded annually. Inflation averages 2.5% per year. Find the real value in today's money.

Solution

FV = 12 000(1.05)8 = 17 729.47A1

Real value = 17 729.47/(1.025)8 = $14 551.40M1 A1

Practice exercise 12

1. A price of $80 rises with 3% inflation. Find the price in 10 years.

2. Find the exact real rate when the nominal rate is 6% and inflation 4%.

14 Loans and repayments

Paying back with interest.

A loan of PV is repaid by equal payments PMT at the end of each period. The TVM solver finds any one of N, I%, PV, PMT and FV given the others. For a loan, PV is positive (money received), PMT negative and FV = 0. The total interest is N × |PMT| − PV.

WORKED EXAMPLE 22 Car loan (standard)

A $26 000 car loan at 7.2% compounded monthly is repaid monthly over 5 years. Find the payment and total interest.

Solution

N = 60, I% = 7.2, PV = 26 000, FV = 0, P/Y = C/Y = 12M1

PMT = −$517.29A1

Interest = 60 × 517.29 − 26 000 = $5037.40A1

Practice exercise 13

1. Find the monthly payment on a $12 000 loan at 9% compounded monthly over 4 years.

2. Find the total interest on that loan.

15 Mortgages and the effect of the term

Longer terms, more interest.

Increasing the term of a loan reduces each payment but increases the total interest. Comparing terms is a common examination task.

(Values in the table are for a $200 000 mortgage at 6.5% compounded monthly.)

Term (years)	Monthly payment	Total interest
15	$1742.21	$113 599
20	$1491.15	$157 875
25	$1350.41	$205 124
30	$1264.14	$255 089

WORKED EXAMPLE 23 Choosing a term (exam level)

For the $200 000 mortgage, find the extra interest paid by choosing 30 years rather than 20 years.

Solution

255 089 − 157 875M1

= $97 214A1

Practice exercise 14

1. For the same mortgage over 20 years, find the balance after 5 years.

16 Savings plans and annuities

Regular deposits.

Regular deposits PMT at the end of each period grow to a future value FV. In the TVM solver, PV = 0, PMT is negative (money paid in) and FV is positive. An annuity paid to you from a lump sum has PV negative (you invest it) and PMT positive (you receive payments).

WORKED EXAMPLE 24 Retirement saving (standard)

$400 is saved at the end of each month for 25 years at 5% compounded monthly. Find the final value.

Solution

N = 300, I% = 5, PV = 0, PMT = −400M1

FV = $238 204A1

WORKED EXAMPLE 25 Annuity payout (exam level)

A lump sum of $250 000 is invested at 4% compounded monthly to provide monthly payments for 20 years. Find the payment.

Solution

N = 240, PV = −250 000, FV = 0M1

PMT = $1514.95A1

Practice exercise 15

1. How much must be saved each month for 10 years at 3% compounded monthly to reach $30 000?

2. How long will $100 000 at 5% compounded monthly last with withdrawals of $1000 per month?

17 Finding the number of payments or the rate

Solving for N or I%.

The TVM solver can also find N or I%. A non-integer N means the final payment is smaller than the others. Finding I% shows the true cost of a loan with fees or a payment plan.

WORKED EXAMPLE 26 Finding the rate (exam level)

A shop offers a $2400 television for 24 monthly payments of $112. Find the annual interest rate (compounded monthly).

Solution

N = 24, PV = 2400, PMT = −112, FV = 0M1

I% = 11.1%A1

Practice exercise 16

1. A $15 000 loan at 6% compounded monthly is repaid at $350 per month. Find the number of payments.

2. Find the rate for a $5000 loan repaid by 36 monthly payments of $165.

18 Investment decisions

Comparing options.

Real decisions compare options with different rates, terms and fees. Convert each option to a common basis (effective rate, total cost or future value at the same date) before comparing, and consider risk and inflation.

WORKED EXAMPLE 27 Lump sum or payments (exam level)

A prize is offered as $50 000 now or $600 per month for 8 years. With an interest rate of 5% compounded monthly, which is worth more today?

Solution

PV of payments: N = 96, PMT = 600, I% = 5: PV = $47 394M1 A1

$50 000 now is worth moreR1

Practice exercise 17

1. Option A: $20 000 now. Option B: $24 000 in 4 years. Which is better if money earns 4% compounded annually?

19 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 28 Mixed problem 1 (exam level)

Sara borrows $18 000 at 8% compounded monthly and repays $450 per month. (a) How many payments are needed? (b) What is the balance after 2 years?

Solution

(a) N = 46.7: 47 payments (the last one smaller)M1 A1

(b) N = 24: balance $9442 (approximately)M1 A1

WORKED EXAMPLE 29 Mixed problem 2 (exam level)

An investment of $8000 doubles in 9 years with monthly compounding. Find the nominal annual rate.

Solution

(1 + r/1200)108 = 2M1

r = 7.73%A1

Practice exercise 18

1. Find the effective rate for 12% compounded monthly.

2. A $300 000 mortgage at 5% compounded monthly over 25 years: find the payment.

20 Additional exam-style practice

Questions in the style of Paper 2.

WORKED EXAMPLE 30 Paper 2 question (exam level)

Tom invests $5000 at 3.5% compounded quarterly. Jan invests $5000 at 3.6% compounded annually. (a) Find each value after 10 years. (b) Find the first full year in which Jan's value exceeds Tom's.

Solution

(a) Tom $7084.54; Jan $7121.44A1 A1

(b) Compare yearly: Jan's effective rate 3.6% > Tom's 3.546%, so Jan is ahead from year 1R1 A1

Practice exercise 19

1. A car costs $32 000. Option 1: pay cash with a 5% discount. Option 2: a loan at 3.9% compounded monthly over 4 years. Find the total cost of option 2 and compare.

21 Exam-style questions with markscheme

Attempt each question under timed conditions before reading the markscheme.

Question 1 (Paper 2 style) (about 15 minutes)

Option A: $12 000 invested at 5% per annum compounded annually. Option B: $12 000 invested at 4.8% per annum compounded monthly.

(a) Find the value of option A after 10 years. [2]

(b) Find the value of option B after 10 years. [2]

(d) Inflation is 2% per year. Estimate the real value of option A after 10 years. [2]

Markscheme

Part	Answer	Marks
(a)	12 000 × 1.0510	M1
	$19 546.74	A1
(b)	12 000 × (1 + 4.8/1200)120	M1
	$19 374.33	A1
(c)	12 000 × 1.05n > 25 000	M1
	n > 15.04	A1
	16 years	A1
(d)	12 000 × 1.0310	M1
	$16 127.00	A1

Examiner's comments. In (c), the integer answer must be given; 15.04 years would lose the final mark.

Question 2 (Paper 2 style) (about 12 minutes)

Mei borrows $18 000 to buy a car. The loan is repaid with equal monthly payments over 4 years at 6.6% per annum compounded monthly.

(a) Find the monthly payment. [3]

(b) Find the total interest paid. [2]

Markscheme

Part	Answer	Marks
(a)	N = 48, I% = 6.6, PV = 18 000, FV = 0, P/Y = C/Y = 12	M1
	PMT = −427.70	A1
	$427.70	A1
(b)	48 × 427.70 − 18 000	M1
	$2529.60	A1
(c)	N = 24 with same PV and PMT; solve FV	M1
	$9591.51	A1

Examiner's comments. Part (c) requires recognizing that the outstanding balance is the future value after 24 payments.

22 Common misconceptions and links

Students often think…	Actually…
more frequent compounding doubles interest	the gain is small
depreciation uses 1 + r/100	use 1 − r/100
N is the number of years	N is the number of periods
PV and PMT have the same sign for a loan	borrowed is positive, repayments negative

How 1.4 connects to the rest of the course

Topic	Connection
1.1 Sequences and series	geometric sequences
1.5 Exponents and logarithms	solving for n
2.8 Exponential functions	continuous growth
5.18 Differential equations (HL)	continuous models

23 Check your understanding

Section A: multiple choice

1. $1000 at 10% compound interest for 2 years becomes

A. $1210 B. $1200 C. $1100 D. $1221

2. The multiplier for a 7% depreciation is

A. 1.07 B. 0.07 C. 0.93 D. 0.7

3. For quarterly compounding over 3 years, N =

A. 3 B. 12 C. 4 D. 36

4. A deposit in the TVM solver is entered as

A. a positive PV B. PMT C. FV D. a negative PV

5. Real rate with 5% interest and 3% inflation ≈

A. 8% B. 2% C. 15% D. 1.67%

6. Compound interest balances form

A. a geometric sequence B. an arithmetic sequence C. a constant sequence D. a quadratic sequence

Section B: structured

7. Compare simple and compound interest on $2000 at 6% for 15 years. [3]

8. Find the annual rate needed to double money in 12 years. [2]

9. Explain why the TVM sign convention matters. [2]

Answers

1. A 2. C 3. B 4. D 5. B 6. A

7. Simple: $3800; compound: $4793.12; compound is larger by $993.12.

8. 21/12 = 1.0595: 5.95%.

9. Money in and out must have opposite signs; otherwise the solver cannot find a solution.

24 Summary and topic map

Idea	Remember
Simple vs compound	arithmetic vs geometric growth
Compound interest	FV = PV(1 + r/100k)kn
Effective rate	(1 + r/100k)k − 1
Depreciation	V = V0(1 − r/100)n
Inflation	real rate ≈ nominal − inflation
TVM	N, I%, PV, PMT, FV, P/Y, C/Y; sign convention
Loans	PV positive, PMT negative, FV 0

Before the exam, make sure you…

•use kn periods and r/k per period
•give money to 2 d.p.
•round years up when a target must be reached
•list TVM inputs in your answer

Financial Applications on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 2. Topic map for Financial Applications.

25 Answers to practice exercises

Check your working against these answers. Full methods are expected in the exam.

Practice exercise 1

1. $504

2. $2950.21

3. arithmetic: the same amount is added each year

Practice exercise 2

1. $798.60

2. $20 771.76

3. $2791.97

Practice exercise 3

1. $3379.48

2. 8.24%

3. $2321.08

4. 5.85% monthly gives 6.01% effective: slightly better

Practice exercise 4

1. 11.9

2. 9.01, so 10 years

3. 5.36%

4. 33 quarters

Practice exercise 5

1. $411.60

2. 9 years (n > 8.31)

3. 14.8%

Practice exercise 6

1. 2.5%

2. $5523.11

3. $6.81

Practice exercise 7

1. $362.69

2. 48 × 362.68 − 15 000 = $2408.64

3. $61 655.05

4. $365.48

Practice exercise 8

1. $4239.70

2. 16.2, so 17 years

3. 5.20%

Practice exercise 9

1. $3590.68

2. $744.24

Practice exercise 10

1. 7.19%

2. 4.89%

Practice exercise 11

1. $675

2. $6144 vs $3000

Practice exercise 12

1. $107.51

2. 1.92%

Practice exercise 13

1. $298.62

2. $2333.76

Practice exercise 14

1. N = 60: balance $171 178 (approximately)

Practice exercise 15

1. $214.69

2. N = 129.6, so about 130 months (10.8 years)

Practice exercise 16

1. N = 48.4 (49 payments, the last smaller)

2. I% = 11.6%

Practice exercise 17

1. PV of B = 24 000/1.044 = $20 515: B is better

Practice exercise 18

1. 12.68%

2. $1753.77

Practice exercise 19

1. PMT = $721.11; total $34 613 vs $30 400: cash is cheaper

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