SL 1.4 Apply geometric models to compound interest and depreciation | International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

Number and Algebra

Using repeated percentage change to model savings and asset values

When an amount changes by the same percentage each period, it does not usually change by the same number of dollars each time. Instead, each new amount is found by multiplying the previous amount by a constant factor. This repeated multiplication models compound interest and percentage depreciation. The model is useful when its assumptions match the situation: the rate stays constant, the period is clear, and there are no unmodelled deposits, withdrawals, or fees.

What you will learn

Recognize compound interest and percentage depreciation as repeated multiplication by a constant factor.
Choose a geometric model and define its quantities, periods, and units.
Calculate a future value or identify the first complete period at which a target is reached.
Use a numerical table or graphing technology to check a model and interpret its results.

Prior knowledge: percentages and repeated change

To write a percentage as a decimal, divide it by

100

. For example,

4.5\%=0.045

. If an amount increases by a rate

r

per period, its new value is the old value plus

r

times the old value. Factoring out the old value gives the multiplier

1+r

.

For a decrease of

d

per period, written as a decimal, the fraction that remains is

1-d

. A loss of 12%

means

88% remains, so the multiplier is

0.88

. A multiplier above

1

represents growth; a positive multiplier below

1

represents a decrease.

A geometric sequence is a list of values made by repeatedly multiplying by the same number, called the common ratio. The index

n

counts periods, such as years, and the starting amount is the value at period zero.

u_n=u_0q^n

Convert a percentage to a decimal before using it in a multiplier.
Growth multiplier: $1+r$ ; decrease multiplier: $1-d$ .
Make sure the rate period matches the period counted by $n$ .

Models for interest and depreciation

Let

P

be an initial balance,

r

the interest rate per compounding period as a decimal, and

n

the number of periods. If the same rate is added at the end of every period, the balance is multiplied by

1+r

each time. After

n

periods, the initial balance has been multiplied by this factor

n

times.

For depreciation, let

V_0

be the initial value and

d

the fraction lost per period. If the same percentage decrease applies each period, the retained fraction is

1-d

. The value after

n

periods is the starting value multiplied by this retention factor

n

times. When

0<d<1

, the model stays positive and decreases as

n

increases.

The context determines the units and possible values of

n

. If a rate is annual,

n

counts years. If periods are months, use a monthly rate only if the situation provides or supports one. For questions about complete annual periods,

n

is a non-negative integer. The models assume a constant rate and no additional changes unless these are included in the question.

Numerically, a table displays the amount after each period. Graphically, a plot against

n

shows increasing values for compound growth and decreasing values for depreciation. A graphing calculator can help locate a target crossing, but the equation and the context determine how to interpret it.

A_n=P(1+r)^n,\quad V_n=V_0(1-d)^n

Compound interest uses a constant growth factor each period.
Percentage depreciation uses a constant retention factor each period.
State what one period means and whether only whole periods are relevant.

Calculating, checking, and interpreting

For a future value, substitute the initial amount, rate, and number of periods into the appropriate model. Keep intermediate values unrounded where possible, then round the final result as requested. For money, state the currency and use the requested accuracy; if none is stated, the nearest cent is usually appropriate.

If the unknown is the number of periods, set the model equal to the target. A graph or table can show where the model reaches or passes that value. When the question asks for complete periods, compare consecutive whole-period values rather than treating an estimated crossing as a complete period.

For a technology check, enter the model with its starting value and multiplier, then inspect a table or graph over a suitable range of periods. A table is especially helpful for comparing values at neighbouring whole periods. Technology can check arithmetic and estimate a crossing, but it cannot decide which rate period applies or how to interpret the question.

The graph uses the period number on the horizontal axis and the amount on the vertical axis. Since periods in these contexts are counted in whole units when interest or depreciation is applied once per period, interpret a target using the relevant whole-period entries.

A_n=P(1+r)^n

Write the correct model before calculating or checking a target.
Use a table or graph to compare values at consecutive whole periods.
Give the result with appropriate units and explain how it meets the condition.

Growth and depreciation multipliers

Situation	Rate per period	Multiplier	Model
Compound interest	$r$ as a decimal	$1+r$	$A_n=P(1+r)^n$
Percentage depreciation	$d$ as a decimal	$1-d$	$V_n=V_0(1-d)^n$

Worked example

Compound interest over complete years

A savings account starts with CAD 2,400 and earns compound interest at 3.2% per year. Find the balance after 6 years, assuming no deposits or withdrawals. Give the answer to the nearest cent.

Identify the model
The annual rate is $3.2\%=0.032$ , so the balance is multiplied by $1.032$ each year. There are $6$ annual periods.
$P=2400,\quad r=0.032,\quad n=6$
Substitute and evaluate
Use the compound-interest model. Keep the calculator value unrounded until the final step.
$A_6=2400(1.032)^6\approx 2900.2351$
Round and state units
Rounding to the nearest cent gives the balance in Canadian dollars.
$A_6\approx 2900.24$

Answer: The balance after 6 years is approximately CAD 2,900.24.

Check: The balance is greater than the starting amount, as expected for a positive interest rate. The increase is approximately CAD 500.24.

Worked example

Depreciation as a constant percentage decrease

A machine is valued at CAD 18,000 and depreciates by 15% of its value each year. Find its value after 4 years, to the nearest dollar.

Find the retained fraction
A loss of 15% $means$ 85% remains each year. The annual multiplier is therefore $0.85$ .
$1-0.15=0.85$
Apply repeated depreciation
Multiply the initial value by the retention factor once for each of the four years.
$V_4=18000(0.85)^4=9396.1125$
Round and interpret
To the nearest dollar, the modelled value is CAD 9,396. This assumes the same percentage loss applies each year.
$V_4\approx 9396$

Answer: The machine's modelled value after 4 years is approximately CAD 9,396.

Check: The multiplier is positive and less than

1

, so the result should remain positive and be below CAD 18,000.

Worked example

Finding when savings first exceed a target

An investment of CAD 1,500 earns compound interest at 5% per year. After how many complete years will its value first exceed CAD 2,000?

Set up the model
The value after $n$ years is the starting balance multiplied by $1.05$ once for each year. Because the question asks for complete years, check whole-number values of $n$ .
$A_n=1500(1.05)^n$
Check consecutive years
Use a table or graphing-calculator table to evaluate the model at years $5$ and $6$ . These values bracket the target, so no fractional-year interpretation is needed.
$A_5\approx 1914.42,\quad A_6\approx 2010.14$
Interpret the result
At year $5$ the value is below CAD 2,000, while at year $6$ it is above CAD 2,000. Therefore the first complete year that meets the condition is year $6$ .
$1914.42<2000<2010.14$

Answer: The investment first exceeds CAD 2,000 after 6 complete years.

Check: Year 5 does not satisfy the condition, and year 6 does. Therefore, year 6 is the first complete year that satisfies it.

Common mistakes and how to avoid them

Using the percentage rate itself as the multiplier, such as using

0.05

for 5% interest.

Correction: For 5% growth use

1.05

; for a 5% decrease use

0.95

.

Treating compound interest as the same fixed amount added every year.

Correction: Compound interest applies the rate to the current balance, so the amount of interest can change from period to period.

Using an annual rate with a number of monthly periods without matching the rate and period.

Correction: Match the rate period to the period counted by the exponent. Use only rate information supported by the situation.

Rounding a target crossing directly to the nearest whole number when asked when a target is first exceeded.

Correction: Check values at consecutive whole periods and choose the first period that satisfies the condition.

Lesson summary

Repeated percentage change is repeated multiplication, producing a geometric sequence.
Compound interest uses the multiplier $1+r$ ; depreciation at rate $d$ uses the multiplier $1-d$ .
Define the starting value, rate, period, and number of periods before calculating.
Use a table or graphing calculator to check values and locate a target crossing, then interpret the result in context.

Check your understanding

Question 1

An amount of CAD 800 decreases by 10% each year. Which expression gives its value after 3 years?

$800(1.10)^3$
$800(0.90)^3$
$800-0.10(3)$
$800(0.10)^3$

Show answer and explanation

800(0.90)^3

After a 10% decrease, 90% remains each year, so the multiplier is

0.90

and it is applied three times.

Question 2

A balance of CAD 1,000 earns 4% compound interest per year. What is its value after 2 years, to the nearest cent?

CAD 1,080.00
CAD 1,081.60
CAD 1,080.16
CAD 1,040.00

Show answer and explanation

CAD 1,081.60

The model gives

1000(1.04)^2=1081.6

, so the balance is CAD 1,081.60.

Question 3

A value is multiplied by

0.8

each year. Which description matches this multiplier?

It increases by 80% each year.
It decreases by 20% each year.
It decreases by 80% each year.
It increases by 20% each year.

Show answer and explanation

It decreases by 20% each year.

The multiplier

0.8

means 80% remains, so 20% is lost each year.

Key terms

Compound interest: Interest calculated on the current balance, including interest added in earlier periods.
Depreciation: A decrease in an asset's value over time; here, the decrease is a fixed percentage per period.
Multiplier: The factor by which a quantity is multiplied in one period.
Geometric sequence: A sequence in which each term is obtained by multiplying the previous term by the same ratio.

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SL 1.4 · Apply geometric models to compound interest and depreciation

International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

Using repeated percentage change to model savings and asset values

What you will learn

Prior knowledge: percentages and repeated change

Models for interest and depreciation

Calculating, checking, and interpreting

Growth and depreciation multipliers

Compound interest over complete years

Depreciation as a constant percentage decrease

Finding when savings first exceed a target

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through IB AA SL

About this lesson and its review