Financial Mathematics

Start with a numerical example

Question

Invest $1000 at 5% simple interest per year for three years. Find the final balance.

Solution

With simple interest, every year’s interest is calculated from the original $1000, even when the balance has grown. \[\text{Interest each year}=\frac{5}{100}\times1000=\$50.\]

After three years, the total interest is \(3\times\$50=\$150\). The final balance is \(\$1000+\$150=\boxed{\$1150}\).

The same example using PV and FV

Question

Invest $1000 at 5% simple interest per year for three years. Find (a) the total interest and (b) the future value.

Solution

Here \(\mathrm{PV}=1000\), \(r=5\) and \(n=3\). \[\text{(a) Total interest}=\mathrm{PV}\times\frac{r}{100}\times n =1000\times\frac{5}{100}\times3=\boxed{\$150}.\] \[\text{(b) }\mathrm{FV}=\mathrm{PV}\left(1+\frac{nr}{100}\right) =1000\left(1+\frac{3\times5}{100}\right)=\boxed{\$1150}.\] On a Calculator page enter

1000*(1+3*5/100)

This gives the final balance, including the original deposit. The interest alone is \(\mathrm{FV}-\mathrm{PV}=1150-1000=\$150\).

Worked example 1 | Annual compounding

Question

$3000 earns 4.5% per year, compounded annually, for five years. Find (a) its final balance and (b) the interest earned.

Solution

Here PV \(=3000\), \(r=4.5\), \(k=1\) and \(n=5\). \[\text{(a) }\mathrm{FV}=3000\left(1+\frac{4.5}{100}\right)^5=3738.545812\ldots\approx\boxed{\$3738.55}.\] (b) Interest earned \(=\mathrm{FV}-\mathrm{PV}=3738.545812\ldots-3000\approx\boxed{\$738.55}\).

Booklet: interest earned is not given as a separate formula.

Worked example 2 | Monthly compounding

Question

$5000 earns a nominal annual rate of 4.8%, compounded monthly, for three years. Find (a) the final balance and (b) the interest earned.

Solution

Monthly means \(k=12\). The multiplier is \(1+4.8/1200=1.004\) and there are \(kn=12\times3=36\) periods. \[\text{(a) }\mathrm{FV}=5000\left(1+\frac{4.8}{100\times12}\right)^{36}=5772.762169\ldots\approx\boxed{\$5772.76}.\] (b) Interest earned \(=5772.762169\ldots-5000\approx\boxed{\$772.76}\).

On a Calculator page enter

5000*(1+4.8/1200)^36

Press Enter; use Ctrl+Enter if a decimal approximation is needed.

Open the Finance Solver

1. Open a Calculator page.

2. Press Menu.

3. Select 8: Finance, then 1: Finance Solver.

Menu numbering can vary with software; follow the menu names.

Worked example 3 | How much should be invested now?

Question

An investment must grow to $8000 in five years. The nominal annual rate is 3.6%, compounded quarterly. Find the initial deposit to the nearest cent.

Solution

Let \(x\) be the initial deposit, in dollars. Use \(\mathrm{FV}=\mathrm{PV}(1+r/(100k))^{kn}\) with \(r=3.6\), \(k=4\) and \(n=5\): \[8000=x\left(1+\frac{3.6}{100\times4}\right)^{4\times5}=x(1.009)^{20}.\]

Booklet: the compound-interest formula is given in 1.4. Here \(x\) is the positive amount invested.

TI-Nspire CX | Finance Solver

There are \(4\times5=\textbf{20 quarters}\). Set PpY = CpY = 4 and Pmt = 0. We know FV and need PV. Enter the known values, then solve the PV field.

1. Enter the known values

2. Check the lower settings

3. Select PV and press Enter

4. Interpret the result

The solver gives \[\mathrm{PV}=-6687.544151\ldots\] The minus sign records money leaving us when we make the deposit.

The amount invested is therefore $6687.54 to the nearest cent.

Read the wording: if asked for the minimum deposit in cents to reach at least $8000, round the unrounded amount up to $6687.55.

Why is the equation’s \(x\) positive but the solver’s PV negative? The equation describes an account balance; the solver records the direction of the cash flow.

Worked example 4 | The unknown is a percentage rate

Question

$6000 grows to $7000 in three years with monthly compounding. Find the nominal annual interest rate to three significant figures.

Solution

Let \(r\%\) be the nominal annual rate. Starting from the compound-interest formula, \[7000=6000\left(1+\frac{r}{100\times12}\right)^{12\times3} =6000\left(1+\frac{r}{1200}\right)^{36}.\]

Booklet: compound interest is given in 1.4; the Finance Solver finds the unknown rate.

TI-Nspire CX | Finance Solver

There are \(12\times3=\textbf{36 months}\). Enter PV = \(-6000\), FV = 7000 and Pmt = 0. Set PpY = CpY = 12, then solve I(%).

1. Enter the known values

2. Check the lower settings

3. Select I(%) and press Enter

4. Interpret the result

The solver gives \[\mathrm{I}(\%)=5.149372839\ldots\] The nominal annual interest rate is therefore 5.15% to three significant figures.

This output is already an annual percentage. Do not multiply it by 100 or by 12.

Monthly compounding is handled by PpY = CpY = 12.

Same equation, a simpler calculator routine: the Finance Solver finds the rate directly from the financial inputs. Optional graphical and nSolve methods appear in Appendices C–D.

Worked example 5 | The first complete month

Question

$6000 earns a nominal annual rate of 4.5%, compounded monthly. Interest is credited at month-end. Find the minimum number of complete months needed to reach at least $8000.

Solution

With \(m\) months, the monthly multiplier is \(1+4.5/1200=1.00375\): \[V(m)=6000(1.00375)^m,\qquad 6000(1.00375)^m\ge8000.\] First solve the boundary equality \(6000(1.00375)^m=8000\).

Booklet: the model uses 1.4; interpreting the inequality and complete months is our step.

TI-Nspire CX | Finance Solver

Set PpY = CpY = 12, so N counts months. Enter the known values below, then solve N.

1. Enter the known values

2. Check the lower settings

3. Select N and press Enter

4. Interpret the result

The solver gives \[\mathrm{N}=76.858970624\ldots\] This is a boundary time in months, not years.

Interest is credited only at month-end. The first possible complete month is therefore 77.

We must check months 76 and 77 against the target before concluding.

Check the neighbouring whole months

On a Calculator page, evaluate

6000*1.00375^76

and

6000*1.00375^77

: \[V(76)\approx\$7974.32<\$8000,\qquad V(77)\approx\$8004.22>\$8000.\] The first qualifying month-end is after 77 months, or 6 years and 5 months. For an increasing balance, round a non-integer boundary up to reach the target.

Worked example 6 | A car's value and total loss

Question

A car costs $24000 and depreciates by 15% of its current value each year. Find (a) its value after four years and (b) the total value lost.

Solution

(a) Use \(V_n=V_0(1-p/100)^n\) with \(V_0=24000\), \(p=15\) and \(n=4\): \[V_4=24000\left(1-\frac{15}{100}\right)^4=24000(0.85)^4=\boxed{\$12528.15}.\] On a Calculator page, enter

24000*(1-15/100)^4

(b) Total value lost \(=V_0-V_4\): \[24000-12528.15=\boxed{\$11471.85}.\]

Booklet: total value lost is not given as a separate formula.

Each entry is the previous entry multiplied by \(0.85\). The monetary loss gets smaller each year because it is 15% of a smaller value.

A fixed percentage is not a fixed amount

Do not subtract \(4\times15\%=60\%\) of the original price after four years. Each year’s percentage applies to the current value. A fixed monetary loss would instead give an arithmetic model.

Worked example 7 | Set up a decay equation

Question

A machine falls from $20000 to $14580 in three years. Find its constant annual percentage depreciation.

Solution

Let \(p\%\) be the annual depreciation rate. Use \(V_n=V_0(1-p/100)^n\): \[\boxed{14580=20000\left(1-\frac{p}{100}\right)^3},\qquad 0\le p\le100.\]

Booklet: this depreciation model is adapted from 1.3; it is not a separate booklet formula.

Method 1: nSolve. Let \(x\) represent \(p\) and enter

The output is \(x=10\). The annual depreciation rate is 10%.

Check: \(20000(1-10/100)^3=20000(0.9)^3=14580\).

Method 2 | Graph the decay equation

Graph \(f_1(x)=20000(1-x/100)^3\) and \(f_2(x)=14580\).

Use \(x_{\min}=0\), \(x_{\max}=30\), \(y_{\min}=5000\), \(y_{\max}=21000\). Find the intersection, using bounds 5 and 15.

The intersection is \((10,14580)\). Here the \(x\)-coordinate represents the annual depreciation percentage, not the number of years.

On the meaningful interval \(0\le x\le100\), increasing the depreciation rate decreases the final value, so there is one solution.

A depreciation percentage and a signed growth rate differ

In our formula \(1-p/100\), a loss of 10% means \(p=10\). In the Finance Solver, I(%) would be \(-10\), with annual settings. Both describe the same decrease; read the meaning of the variable before interpreting its sign.

Worked example 8 | Balance growth versus buying power

Question

$10000 earns 4% per year, compounded annually, for five years. Inflation is 2.5% per year. Give money to the nearest cent. Find:

1. the nominal final balance;

2. its real value in today’s money;

3. the gain in buying power;

4. the real annual rate (the annual percentage growth in buying power), to three significant figures. Use the unrounded rate to check part (b).

Solution

(a) Grow the investment using compound interest: \[\mathrm{FV}=10000\left(1+\frac{4}{100}\right)^5=12166.529024\approx\boxed{\$12166.53}.\] (b) Divide by the price multiplier: \[\text{real value}=\frac{10000(1.04)^5}{(1.025)^5}=10753.438843\ldots\approx\boxed{\$10753.44}.\] (c) Gain in buying power \(=10753.438843\ldots-10000\approx\boxed{\$753.44}\).

Inflation divides; depreciation multiplies

Undo 2.5% annual inflation by dividing by \(1.025^n\), not multiplying by \(0.975^n\). With monthly interest over five years, compound interest uses 60 months, but annual inflation still uses an exponent of 5 years.

Worked example 8 | Continued: a real annual rate

Solution

(d) How much does buying power grow in one year?

Step 1: start with $100 to make the percentage clear. After one year, 4% interest gives \(100\times1.04=\$104\). Prices have multiplied by \(1.025\), so its value in today’s money is \[104\div1.025=101.463414634\ldots.\] Step 2: express the gain as a percentage of the original $100. \[\frac{101.463414634\ldots-100}{100}\times100\% =1.463414634\ldots\%\approx\boxed{1.46\%}.\] This is the real annual rate: buying power grows by about 1.46% each year. In Scratchpad, enter

((104/1.025-100)/100)*100

Keep the unrounded result for the next step.

Booklet: this real-rate calculation is not given separately. Use the percentage-change calculation; no additional general formula is needed here.

Step 3: use the annual growth to check part (b). A gain of \(1.463414634\ldots\%\) means multiplying each year by \[1+\frac{1.463414634\ldots}{100} =1.014634146\ldots=\frac{1.04}{1.025}.\] Apply this multiplier five times to the original $10000: \[\text{real value}=10000\left(\frac{1.04}{1.025}\right)^5 =10753.438843\ldots\approx\boxed{\$10753.44}.\] In Scratchpad, enter

10000*(1.04/1.025)^5

This gives the same answer as part (b): dividing by the inflation multiplier each year is equivalent to dividing by the five-year inflation multiplier at the end.

Worked example 9 | Understand the effective rate first

Question

$5000 is invested for three years. A pays 4.85% per year, compounded annually. B pays a nominal annual rate of 4.8%, compounded monthly.

1. Find each balance after one year and its percentage gain. Which grows faster?

2. Find both three-year balances and their difference.

Solution

(a) Use the compound-interest formula for one year: \[\mathrm{FV}_A=5000(1.0485)=5242.50,\quad \mathrm{FV}_B=5000\left(1+\frac{4.8}{1200}\right)^{12}=5245.351037\ldots\] Percentage gain \(=(\text{final}-\text{initial})/\text{initial}\times100\%\): \[A:\ \frac{5242.50-5000}{5000}\times100\%=4.85\%,\qquad B:\ \frac{5245.351037\ldots-5000}{5000}\times100\%\approx4.90702\%.\] These actual one-year percentage gains are the effective annual rates. B grows faster despite its lower quoted rate.

(b) After three years, \(5000(1.0485)^3\approx\$5763.35\) and \(5000(1.004)^{36}\approx\$5772.76\). Using unrounded balances, B gives $9.41 more.

From the example to the formula

B’s one-year multiplier is \(1.004^{12}\). Subtract 1 to isolate the gain, then multiply by 100 to express it as a percentage. In general, \[\boxed{\text{effective annual rate}=100\left[\left(1+\frac{r}{100k}\right)^k-1\right]\%}.\]

Booklet: not given separately; it follows from compound interest over one year.

Apply to B: \(100[(1+4.8/1200)^{12}-1]\%=4.907020753\ldots\%\), agreeing with the example. This shortcut performs the same steps in one calculation.

Compare before rounding. Close rates can round to the same value. Compare annual inflation with effective annual growth to judge whether buying power increases.

Set up the mathematics, then use the appropriate calculator tool

1. Define the unknown and its units. Is it money, an annual percentage, years or months?

2. Write the formula and substitute. Keep the unknown visible in the equation.

3. Calculate or solve.

   All quantities known: enter the expression directly.

   Standard compound interest: use the Finance Solver. Set Pmt = 0 and PpY = CpY = \(k\); solve PV, FV, I(%) or N. N counts compounding periods, and I(%) is the nominal annual rate.

   For other equations, use the numerical or graphical methods shown in the lesson. Optional investment alternatives are in Appendices A–E.

4. Interpret and check. Include units, use the requested accuracy, and verify whole-period conditions using neighbouring values.

Related video | Compound interest

Annual and monthly compounding:

https://youtu.be/YsgT-9gYJ90

Worked example 3 revisited | An optional nSolve solution

Question

An investment must grow to $8000 in five years. The nominal annual rate is 3.6%, compounded quarterly. Find the initial deposit to the nearest cent.

Solution

Let \(x\) be the initial deposit, in dollars. Use \(\mathrm{FV}=\mathrm{PV}(1+r/(100k))^{kn}\) with \(r=3.6\), \(k=4\) and \(n=5\): \[\boxed{x\left(1+\frac{3.6}{100\times4}\right)^{4\times5}=8000}, \qquad\text{or}\qquad x(1.009)^{20}=8000.\]

Booklet: the compound-interest formula is given in 1.4; this is its substitution for this question.

TI-Nspire CX | Numerical Solve

Open a Calculator page or the calculator in Scratchpad. Use a variable such as \(x\) that has no stored value.

1. Menu \(\to\) Algebra

2. Choose Numerical Solve

Enter the equation, a comma, then the unknown \(x\):

3. Check the complete entry

4. Press Enter for the result

The calculator displays \(6687.54\). The unrounded solution is \(x=6687.544151\ldots\), so the required deposit is $6687.54 to the nearest cent. Here \(x\) is a positive deposit, not the Finance Solver’s signed PV.

nSolve returns one solution at a time

Here increasing \(x\) increases the final balance, so there is exactly one positive deposit. When an equation may have several solutions, a single nSolve result is not a complete answer: inspect the graphs and find the relevant intersections.

To restrict the search, use

nSolve(equation,x,lower,upper)

; adding

,0,8000

after \(x\) searches from 0 to 8000.

Interpret rounding: for the minimum deposit in cents to reach at least $8000, use the unrounded result and round up to $6687.55.

Worked example 3 revisited | A graphical solution

Question

An investment must grow to $8000 in five years at a nominal annual rate of 3.6%, compounded quarterly. Find the initial deposit to the nearest cent.

Solution

Turn the equation into an intersection problem. Let \(x\) be the initial deposit. We need \(x(1.009)^{20}=8000\). Graph the left and right sides as two functions: \[f_1(x)=x(1.009)^{20},\qquad f_2(x)=8000.\] At their intersection, the two outputs are equal. The \(x\)-coordinate is the required initial deposit; the \(y\)-coordinate is the final balance of $8000.

TI-Nspire CX | Graph intersection, step by step

1. Add a Graphs page. After the \(f_1(x)=\) prompt, enter

   x*(1+3.6/(100*4))^(4*5)

   and press Enter. Then enter

   8000

   after \(f_2(x)=\). These are the two functions above.

   

   

2. Choose Menu \(\to\) Window/Zoom \(\to\) Window Settings. Match the capture below: \(x_{\min}=-1\), \(x_{\max}=9000\), \(y_{\min}=-1\), \(y_{\max}=11000\); leave both scales on Auto.

3. Choose Menu \(\to\) Analyze Graph \(\to\) Intersection. If prompted, select the two graphs. Select a lower bound to the left of the crossing, then an upper bound to its right; for example \(x=6000\) and \(x=7500\).

Set a suitable viewing window

Find the intersection

Use enough digits before rounding

Read the \(x\)-coordinate: the capture displays \((6688,8000)\), with rounded coordinates. For a nearest-cent answer, increase the displayed precision or use nSolve, as on the previous page: \(x=6687.544151\ldots\). The required deposit is $6687.54. Both methods solve the same equation.

The standard \(-10\) to \(10\) window cannot show a final balance of $8000. Use a suitable window, then Intersection; do not estimate the crossing from pixels.

Another valid approach: graph \(y=x(1.009)^{20}-8000\) and find its zero with Analyze Graph \(\to\) Zero. Its \(x\)-intercept gives the same deposit.

Worked example 4 revisited | A graphical solution

Question

$6000 grows to $7000 in three years with monthly compounding. Find the nominal annual interest rate to three significant figures.

Solution

Let \(r\%\) be the nominal annual rate. Starting from the booklet’s compound-interest formula, \[7000=6000\left(1+\frac{r}{100\times12}\right)^{12\times3} =6000\left(1+\frac{r}{1200}\right)^{36}.\]

Booklet: compound interest is given in 1.4; we solve the substituted equation graphically.

TI-Nspire CX | Find the intersection

Use \(x\) on the calculator to represent the percentage \(r\). Graph the two sides of the equation: \[f_1(x)=6000(1+x/1200)^{36},\qquad f_2(x)=7000.\]

1. Add a Graphs page. After \(f_1(x)=\), enter

   6000*(1+x/(100*12))^(12*3)

   and press Enter. Then enter

   7000

   after \(f_2(x)=\).

2. Choose Menu \(\to\) Window/Zoom \(\to\) Window Settings. Use a window showing both graphs and their crossing; for example \(0\le x\le15\) and \(0\le y\le12000\).

3. Choose Menu \(\to\) Analyze Graph \(\to\) Intersection. If prompted, select the two graphs. Choose a lower bound to the left of the crossing and an upper bound to its right, such as \(x=4\) and \(x=6\).

The calculator displays the intersection as \((5.149,7000)\). The \(x\)-coordinate gives the annual percentage rate; the \(y\)-coordinate is the final balance of $7000. Thus \[r\approx5.149\quad\Longrightarrow\quad\boxed{\text{nominal annual interest rate}=5.15\%\text{ (3 s.f.)}}.\]

Know what your variable represents

We wrote \(1+r/1200\), so \(r\) is already an annual percentage. Do not multiply the answer by 100 or 12. If you had chosen a decimal monthly rate instead, its interpretation would be different.

Finance Solver check: N = 36, PV = \(-6000\), Pmt = 0, FV = 7000, PpY = CpY = 12, END. Solve I(%) to get the same rate.

Worked example 4 revisited | Give nSolve a starting point

Question

$6000 grows to $7000 in three years with monthly compounding. Find the nominal annual interest rate to three significant figures.

Solution

As on the previous page, let \(x\) represent the annual percentage rate. Substituting into the compound-interest formula gives \[6000(1+x/1200)^{36}=7000.\]

1. A correct equation can still give an unsuccessful search

With only the equation and \(x\), this attempt returns “No solution found”.

nSolve searches numerically. This message means that its search did not find a solution; it does not prove that no solution exists. Our graph already shows the intersection.

2. Add an initial guess and lower and upper bounds

The successful entry in the screenshots is:

• x=3

  : start the search near 3%. This is a guess, not the answer.

• 0

  : use 0% as the lower bound for the rate.

• 10

  : use 10% as the upper bound for the rate.

Successful calculation: left view

Scroll right to see the bounds

These are two views of the same successful calculation. The displayed result is \(x=5.14937\), giving an annual rate of 5.15% to three significant figures, as found graphically.

Why are 0 and 10 suitable bounds here?

At 0%, the final balance is $6000; at 10%, it is \(6000(1+10/1200)^{36}\approx\$8089.09\). As the rate increases, the final balance increases. The rate producing $7000 therefore lies between 0% and 10%.

When to choose the graphical method

Use the Finance Solver first for standard compound-interest questions. When choosing between these optional equation-solving methods, prefer graphs if you are not comfortable choosing an initial guess and bounds. Plot the two sides and use Intersection, as on the previous page. An unsuccessful nSolve search does not prove that no solution exists.

Worked example 5 revisited | nSolve and graphical methods

Question

$6000 earns a nominal annual rate of 4.5%, compounded monthly. Interest is credited at month-end. Find the minimum number of complete months needed to reach at least $8000.

Solution

If \(m\) counts months, then \(n=m/12\) years and \(kn=12(m/12)=m\). The monthly multiplier is \(1+4.5/1200=1.00375\). Hence \[V(m)=6000(1.00375)^m,\qquad 6000(1.00375)^m\ge8000.\] First solve the boundary equality \(6000(1.00375)^m=8000\).

Booklet: the model uses 1.4; changing the time unit and interpreting the inequality are our steps.

nSolve: let \(x\) represent months and enter

This gives \(x=76.858970624\ldots\) months.

Graphical alternative | Find the boundary crossing

Graph \(f_1(x)=6000(1.00375)^x\) and \(f_2(x)=8000\). Use \(0\le x\le100\) and \(5000\le y\le9500\). Find the intersection with bounds 70 and 85.

The \(x\)-coordinate is again \(76.858970624\ldots\). The smooth curve extends the model between month-ends; the question asks for whole month-ends.

Check the neighbouring whole months

On a Calculator page, evaluate the model at \(m=76\) and \(m=77\): \[V(76)\approx\$7974.32<\$8000,\qquad V(77)\approx\$8004.22>\$8000.\] The first qualifying month-end is after 77 months, or 6 years and 5 months. For an increasing balance, round a non-integer boundary up to reach the target.

Do not round the time before interpreting it. A calculator approximation close to a whole number should be checked against the original condition.

Worked example 2 revisited | The Finance Solver

Question

$5000 is invested for three years at a nominal annual rate of 4.8%, compounded monthly. Find (a) the final balance and (b) the total interest earned, to the nearest cent.

Solution

We need FV. Three years gives \(\mathrm{N}=12\times3=36\) months.

1. Enter the known values

N = 36; I(%) = 4.8; PV = \(-5000\); Pmt = 0; PpY = 12. The displayed FV = 0 has not yet been solved.

2. Check the lower settings

Set CpY = 12 and check PpY = 12. Leave PmtAt at END. N now counts months.

3. Select FV and press Enter

Place the cursor inside the FV cell. Its selected 0 is a starting value, not the answer.

4. Interpret the answer

\(\mathrm{FV}=5772.762169\ldots\). Its positive sign represents savings returned to us.

Finish with the balance and the interest

(a) The balance is $5772.76.

(b) Interest earned \(=5772.762169\ldots-5000=\boxed{\$772.76}\).

Keep precision: use the unrounded output for the next calculation. On a Calculator page in the same problem, the saved value is tvm.fv; enter tvm.fv-5000. You can also re-enter the full calculation rather than a rounded result.