AI HL · 12.2
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Finance Solver: compound interest and depreciation
12.2.1
TI-Nspire Finance Solver. Money you hand over is negative. Money you get back is positive. N = number of years × periods per year.
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The same money at the same annual rate, compounded yearly, quarterly, monthly and daily. Put one column into your Finance Solver and find FV, then press Show FV to check.
| Box | Yearly | Quarterly | Monthly | Daily |
|---|
Move a slider to compare the four.
A motorbike costs £14 500 and loses 18% of its value each year. Find its value after 4 years.
Have a go first. Then open the working one line at a time.
N = 4
Once a year for 4 years.
I(%) = −18
Losing 18% a year is interest of −18%.
PV = −14 500, Pmt = 0
The money at the start, handed over, so negative. Nothing is paid in or out after that.
PpY = CpY = 1 → FV = 6555.77
The motorbike is worth £6555.77.
A car is bought for £21 000. It loses 11% of its value each year. After how many complete years is it first worth less than half of its price?
Have a go first. Then open the working one line at a time.
I(%) = −11, PV = −21 000, Pmt = 0, FV = 10 500, PpY = CpY = 1
Depreciation is a negative I(%). Half of £21 000 is £10 500.
N = 5.95…
The value passes £10 500 part-way through the 6th year.
N = 5 → FV = 11 726.52; N = 6 → FV = 10 436.61
Complete years: 6.
$6400 is invested at 2.7% per year, compounded monthly, for 4 years. What is N?
Have a go first. Then open the working one line at a time.
N = number of years × periods per year
Monthly means 12 periods per year.
N = 4 × 12 = 48
48 months.
$3000 is invested at 3.5% per year, compounded monthly, and left for 18 months. What is N?
Have a go first. Then open the working one line at a time.
N = 18
If the question already counts in months or quarters, that count is N.
£1000 is invested at 6% per year, compounded monthly, for 2 years. Find its value.
Have a go first. Then open the working one line at a time.
N = 2 × 12 = 24
Monthly for 2 years.
I(%) = 6
Type the annual rate. The solver shares it between the months.
PV = −1000, Pmt = 0, PpY = CpY = 12
£1000 handed over. No regular payment.
FV = 1127.16
The investment is worth £1127.16.
£1000 is invested at 4% per year, compounded quarterly. How long until it doubles?
Have a go first. Then open the working one line at a time.
I(%) = 4, PV = −1000, Pmt = 0, FV = 2000, PpY = CpY = 4
Doubling means FV is 2000. Money in is negative, money back is positive.
N = 69.66…
Interest is added at the end of a quarter, so it takes 70 quarters.
70 ÷ 4 = 17.5 years
N comes back in periods, not years.
How much must be invested now at 2.4% per year, compounded quarterly, to have €5000 in 5 years?
Have a go first. Then open the working one line at a time.
N = 5 × 4 = 20
Quarterly for 5 years.
I(%) = 2.4, Pmt = 0, FV = 5000, PpY = CpY = 4
€5000 is what you get back, so it is positive.
PV = −4436.19
The sign flips: money in is negative. Invest €4436.19.
€2500 is invested at 4.8% per year. (a) Determine whether annual or monthly compounding gives the better return. (b) Find the difference between the two values after 8 years.
Have a go first. Then open the working one line at a time.
N = 96, I(%) = 4.8, PV = −2500, Pmt = 0, PpY = CpY = 12 → FV = 3667.55
8 years of months is 96 periods.
N = 8, I(%) = 4.8, PV = −2500, Pmt = 0, PpY = CpY = 1 → FV = 3637.73
Only N, PpY and CpY change.
€3667.55 against €3637.73
Monthly compounding gives the better return.
€3667.55 − €3637.73 = €29.82
The difference after 8 years.
A delivery van is bought for €42 000. Its value depreciates by 14% per year. (a) Find the value of the van after 6 years. (b) Find the number of complete years after which the van is first worth less than €15 000. (c) Inflation over the 6 years is 2.2% per year. Using r = c − i, find the value of the van after 6 years in real terms.
Have a go first. Then open the working one line at a time.
N = 6, I(%) = −14, PV = −42 000, Pmt = 0, PpY = CpY = 1 → FV = 16 991.82
€16 991.82, which is €17 000 to 3 s.f.
I(%) = −14, PV = −42 000, FV = 15 000 → N = 6.83…
After 6 years it is still worth more than €15 000, so the answer is after 7 years.
r = c − i = −14 − 2.2 = −16.2
The change is −14% and inflation is 2.2%.
I(%) = −16.2, N = 6 → FV = 14 544.99
€14 544.99 in real terms, which is €14 500 to 3 s.f.
£8000 is invested for 6 years. Bank A pays 3.9% per year, compounded monthly. Bank B pays 3.95% per year, compounded annually. (a) Which bank gives the greater value after 6 years? (b) Find the difference between the two values.
Have a go first. Then open the working one line at a time.
N = 72, I(%) = 3.9, PV = −8000, Pmt = 0, PpY = CpY = 12 → FV = 10 105.32
6 years of months is 72 periods.
N = 6, I(%) = 3.95, PV = −8000, Pmt = 0, PpY = CpY = 1 → FV = 10 093.39
Only N, I(%), PpY and CpY change.
£10 105.32 against £10 093.39
Bank A is better, although its rate is lower: interest is added more often.
£10 105.32 − £10 093.39 = £11.93
The difference after 6 years.
$7200 is invested at 3.9% per year, compounded monthly, for 5 years. Find the interest earned.
Have a go first. Then open the working one line at a time.
N = 5 × 12 = 60
Monthly for 5 years.
I(%) = 3.9, PV = −7200, Pmt = 0, PpY = CpY = 12
$7200 is handed over, so PV is negative. No regular payment.
FV = 8747.47
The investment is worth $8747.47.
8747.47 − 7200 = 1547.47
What comes back, less what went in. $1547.47.
€9000 is invested in an account where interest is compounded half-yearly. After 6 years it is worth €11 400. Find the annual interest rate, to 3 significant figures.
Have a go first. Then open the working one line at a time.
N = 6 × 2 = 12
Half-yearly means 2 periods per year.
PV = −9000, FV = 11 400
Money in is negative, money back is positive. If both have the same sign the solver shows an error.
Pmt = 0, PpY = CpY = 2
No regular payment.
I(%) = 3.9789…
The annual rate is 3.98%. The solver always gives the rate per year.
A tractor is bought for $68 000. Its value falls by the same percentage each year. After 5 years it is worth $31 500. Find the annual rate of depreciation, to 3 significant figures.
Have a go first. Then open the working one line at a time.
N = 5
Once a year for 5 years.
PV = −68 000, FV = 31 500
The price is paid out, so PV is negative. The value at the end is positive.
Pmt = 0, PpY = CpY = 1
Nothing is paid in or out in between.
I(%) = −14.2646…
A negative rate is depreciation. The tractor loses 14.3% of its value each year.
£12 000 is invested at 5.2% per year, compounded annually, for 8 years. Inflation over the 8 years is 2.4% per year. (a) Using r = c − i, find the value of the investment after 8 years in real terms. (b) Find how much would have to be invested for the investment to be worth £20 000 in real terms after 8 years.
Have a go first. Then open the working one line at a time.
r = c − i = 5.2 − 2.4 = 2.8
The interest rate less the inflation rate. Type 2.8 as I(%).
N = 8, I(%) = 2.8, PV = −12 000, Pmt = 0, PpY = CpY = 1 → FV = 14 966.70
£14 966.70 in real terms. The account will show £18 001.44, but that money buys less than it does today.
N = 8, I(%) = 2.8, Pmt = 0, FV = 20 000, PpY = CpY = 1 → PV = −16 035.59
Now the real value is known and the amount invested is the unknown. It is negative because it is paid in. Invest £16 035.59.
A laptop costs €900 today. Prices rise by 3.4% per year. Find the price of the same laptop in 5 years' time.
Have a go first. Then open the working one line at a time.
N = 5, I(%) = 3.4, PV = −900, Pmt = 0, FV = ?, PpY = CpY = 1
Inflation works like interest: the price grows by 3.4% each year.
FV = 1063.76
The laptop costs €1063.76. The price has risen by €163.76.