VertexMath · MCR3U · Free worksheet

Financial Applications Review

Try each problem first. You don’t need to finish everything.

Core path ~55 min Full sheet ~1 h 50 min

Recommended: K1–K5, then one A and one T problem

K = Knowledge & Understanding · A = Application · T = Thinking · Boss = Challenge

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Core Skills (K)

Simple Interest · Compound Interest · Present Value · Annuities · Loans

K1 Simple Interest

Each investment or loan earns simple interest.

  1. a)
    Determine the interest earned on $2400 invested at 3.5%/a for 8 months.
  2. b)
    Determine the total amount owed on a $1500 loan at 4.2%/a after 30 months.
  3. c)
    The graph shows the amount AA, in dollars, of an investment after tt years. Each grid square is 1 year by $300. Determine the principal and the annual interest rate.
    A straight line starting at the point (0, 1200) and rising through the point (5, 1500). The horizontal axis is t in years, the vertical axis is A in dollars.tA12345678300600900120015001800
Stuck? Peek at a hint

Write the rate as a decimal and the time in years before substituting. For the graph, read AA at t=0t=0 first.

K2 Compound Interest

Determine the amount of each investment.

  1. a)
    $2500 at 3.5%/a compounded annually for 6 years.
  2. b)
    $4000 at 5.2%/a compounded semi-annually for 7 years.
  3. c)
    $1800 at 7.2%/a compounded monthly for 30 months. Also find the interest earned.
Stuck? Peek at a hint

Find the rate per period i=rmi=\frac{r}{m} and the number of periods n=mtn=mt before using A=P(1+i)nA=P(1+i)^n.

K3 Present Value

  1. a)
    Determine the present value of $6500 due in 4 years at 5%/a compounded annually.
  2. b)
    Determine the present value of $15 000\$15\,000 due in 8 years at 4.4%/a compounded quarterly.
  3. c)
    $2500 grows to $3600 in 4 years, compounded annually. Determine the annual interest rate, to two decimal places.
Stuck? Peek at a hint

Present value is the amount today: PV=A(1+i)−nPV=A(1+i)^{-n}. In (c), isolate (1+i)4(1+i)^4 first.

K4 Annuities: Future Value

Each deposit is made at the end of its period.

  1. a)
    Determine the future value of $600 deposited every year for 5 years at 4%/a compounded annually.
  2. b)
    Determine the future value of $250 deposited every 3 months for 6 years at 6%/a compounded quarterly.
  3. c)
    Determine the monthly deposit needed to have $15 000\$15\,000 in 5 years at 4.8%/a compounded monthly.
Stuck? Peek at a hint

Identify RR, ii and nn first. In (c), the unknown is RR: substitute into the future value formula, then isolate RR.

K5 Annuities: Present Value & Loans

Each payment is made at the end of its period.

  1. a)
    Determine the present value of $400 paid every 3 months for 5 years at 8%/a compounded quarterly.
  2. b)
    A car loan of $18 000\$18\,000 at 7.2%/a compounded monthly is repaid over 4 years. Determine the monthly payment.
  3. c)
    Determine the total interest paid on the loan in (b).
Stuck? Peek at a hint

A loan amount is a present value. In (b), substitute it for PVPV and solve for RR.

Real World (A)

Doubling Time · Loan Payments · Savings Plans

A1 Doubling Time

Amira deposits $4000 in a savings account at a Sudbury credit union. It pays 4.8%/a compounded monthly, and interest is credited at the end of each month.

  1. a)
    Write an equation for the balance AA after nn months. Then find the balance after 5 years.
  2. b)
    Use a table of values or graphing technology to solve 4000(1.004)n=80004000(1.004)^n=8000, to one decimal place.
  3. c)
    Is the balance ever exactly $8000? At which month-end does it first reach $8000, how many years is that, and what is the balance then?
Stuck? Peek at a hint

Find the interest rate per month first; each month-end multiplies the balance by 1+i1+i.

A2 Credit Card Balance

Jordan owes $3000 on a store credit card that charges 19.2%/a compounded monthly.

  1. a)
    Find the monthly payment that pays off the balance in 2 years.
  2. b)
    Find the total interest Jordan pays with this plan.
  3. c)
    Jordan pays $45 a month instead. Find the interest charged in the first month. Will the balance ever reach $0? Explain.
Stuck? Peek at a hint

The $3000 owed today is a present value.

A3 Education Fund

From the month Maya is born, her parents deposit $150 at the end of every month into an education savings plan earning 4.8%/a compounded monthly. A government grant adds 20% of each deposit to the plan.

  1. a)
    Find the total amount added to the plan each month, and the value of the plan after 17 years.
  2. b)
    How much of that value is interest?
  3. c)
    Maya's grandfather puts $150 a month into a jar at home: no grant, no interest. What does the future value formula give with i=0i=0? How much is in the jar after 17 years?
Stuck? Peek at a hint

Find the total deposit per month first, then count how many monthly deposits fit in 17 years.

Think It Through (T)

Error Analysis · Counterexamples

T1 Spot the Flaw

A student finds the value after 4 years of $800 deposited at the end of each year at 5%/a compounded annually. The student's table is below.
"Adding the last row: 800(1.05)4+800(1.05)3+800(1.05)2+800(1.05)≈$3620.51800(1.05)^4+800(1.05)^3+800(1.05)^2+800(1.05)\approx\$3620.51."

Deposit\text{Deposit} 11223344
Made at end of year\text{Made at end of year} 11223344
Years of growth\text{Years of growth} 44332211
Value at year 4\text{Value at year }4 800(1.05)4800(1.05)^4800(1.05)3800(1.05)^3800(1.05)2800(1.05)^2800(1.05)800(1.05)
  1. a)
    Identify the error in the table and correct the "Years of growth" row.
  2. b)
    Find the correct value after 4 years, and check it with the future value formula.
  3. c)
    Explain why the student's total is exactly 1.051.05 times the correct value. When would the student's total be correct?
Stuck? Peek at a hint

For (a), place each deposit on a timeline from year 0 to year 4.

T2 True or False?

Decide whether each claim is true. If it is false, give a counterexample with numbers.

  1. a)
    "Leaving an investment twice as long earns twice as much interest."
  2. b)
    "If you invest twice as much, it takes twice as long to double."
  3. c)
    "Paying off a loan in half the time doubles the monthly payment."
Stuck? Peek at a hint

Test each claim with round numbers first, such as $1000 at 10%/a compounded annually.

Boss Level

Optional challenge

Pay Now or Pay Later

A laptop costs $1000 cash. The store also offers a pay-later plan: $550 one year from now and $605 two years from now.

  1. a)
    Find the annual interest rate, compounded annually, at which the two plans are worth the same today.
  2. b)
    At that rate, the store switches to two equal payments, one and two years from now. Find the payment.
  3. c)
    Your savings earn 4%/a compounded annually. Which plan is better for you, and by how much in today's dollars?
Stuck? Peek at a hint

Bring each payment back to today separately: the cash price equals the sum of their present values. Let x=1+ix=1+i.

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Answer key

Final answers only. Try first, then check.

Core Skills

K1
  • a) $56.00
  • b) $1657.50
  • c) P=$1200P=\$1200; r=5%r=5\%/a
K2
  • a) $3073.14
  • b) $5729.58
  • c) $2153.83; interest $353.83
K3
  • a) $5347.57
  • b) $10 569.50\$10\,569.50
  • c) 9.54%9.54\%/a
K4
  • a) $3249.79
  • b) $7158.38
  • c) $221.70 per month
K5
  • a) $6540.57
  • b) $432.70 per month
  • c) $2769.60

Real World

A1
  • a) A=4000(1.004)nA=4000(1.004)^n; $5082.56
  • b) n≈173.6n\approx173.6
  • c) No: the balance only changes at month-ends, and 173.6173.6 is not one. It is $7979.81 after month 173 and first passes $8000 at the end of month 174 (14.5 years), at $8011.72.
A2
  • a) $151.52 per month
  • b) $636.48
  • c) $48.00. No: each payment is smaller than the interest charged, so the balance grows ($3003.00 after one month) and is never paid off. The formula would need 1−1.016−n=4845>11-1.016^{-n}=\frac{48}{45}>1, which is impossible.
A3
  • a) $180 per month; $56 599.12\$56\,599.12
  • b) $19 879.12\$19\,879.12 (deposits and grants total $36 720\$36\,720)
  • c) 00\frac{0}{0}: the formula is undefined at i=0i=0. The jar holds 150×204=$30 600150\times204=\$30\,600.

Think It Through

T1
  • a) Deposit 4 is made at the end of year 4, so it earns no interest. The row should read 3,2,1,03,2,1,0.
  • b) $3448.10
  • c) Every term has one extra factor of 1.051.05: 3448.10×1.05≈3620.513448.10\times1.05\approx3620.51. It is correct if each deposit is made at the start of each year.
T2
  • a) False for compound interest: e.g. $1000 at 10%/a compounded annually earns $100 in 1 year but $210 in 2 years. (True for simple interest.)
  • b) False: P(1+i)n=2PP(1+i)^n=2P gives (1+i)n=2(1+i)^n=2, which does not involve PP. E.g. at 8%/a compounded annually, $1000 and $2000 both first double after 10 years.
  • c) False: e.g. $10 000\$10\,000 at 6%/a compounded monthly costs $193.33 a month over 5 years but $359.79 a month over 2.5 years, less than 2×193.33=$386.662\times193.33=\$386.66.

Boss Level

Boss
  • a) 10%10\%/a
  • b) $576.19
  • c) Pay cash: the pay-later plan is worth $1088.20 today, so cash saves about $88.20.

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Most testedFind the monthly payment on a $15 000 loan over 55 years at 6%6\% compounded monthly. (card 5)

1Simple interest

I=PrtA=P(1+rt)I=Prt\qquad A=P(1+rt)

Userr as a decimal, tt in years. Solving for PP, rr or tt: divide by the whole product, in brackets.

Example$1500 at 4%4\% for 99 months: I=1500(0.04)(0.75)=I=1500(0.04)(0.75)= $45.

TrapTime is in years: 88 months is t=8/12t=8/12, not t=8t=8.

2Compound interest

A=P(1+i)ni=rmn=mtA=P(1+i)^n\qquad i=\dfrac rm\qquad n=mt

Usemm = compounding periods per year. nn counts periods, not years: 33 years monthly is n=36n=36.

Example$3000 at 6%6\% quarterly for 44 years: 3000(1.015)16≈3000(1.015)^{16}\approx $3806.96.

TrapAdjust both: 8%8\% quarterly for 22 years is (1.02)8(1.02)^8, not (1.02)2(1.02)^2.

Trap1.02n=21.02^n=2 gives n≈35n\approx35 quarters, about 8.758.75 years, when i=8%/4i=8\%/4.

3Present value

PV=A(1+i)−n=A(1+i)nPV=A(1+i)^{-n}=\dfrac{A}{(1+i)^n}

UsePVPV is always less than AA. Compare options at the same moment: all present values, or all future values.

Example$8000 in 66 years at 4.5%4.5\% annually: 8000(1.045)−6≈8000(1.045)^{-6}\approx $6143.17.

Trap4500=3000(1+i)54500=3000(1+i)^5: 1+i=1.51/51+i=1.5^{1/5}, so i≈8.4%i\approx8.4\%. Never divide 1.51.5 by 55.

4Annuities: future value

FV=R[(1+i)n−1]iFV=\dfrac{R\left[(1+i)^n-1\right]}{i}

R=FV⋅i(1+i)n−1R=\dfrac{FV\cdot i}{(1+i)^n-1}

UsePayments at the end of each period; the last one earns no interest. Payment and compounding periods must match.

Example$200 a month for 33 years at 6%6\% monthly: 200(1.00536−1)/0.005≈200(1.005^{36}-1)/0.005\approx $7867.22.

TrapOn a calculator, bracket (1.00536−1)(1.005^{36}-1) before dividing by ii.

5Annuities: present value and loans

PV=R[1−(1+i)−n]iPV=\dfrac{R\left[1-(1+i)^{-n}\right]}{i}

R=PV⋅i1−(1+i)−nR=\dfrac{PV\cdot i}{1-(1+i)^{-n}}

UseA loan amount is a present value. Total interest = (number of payments)×R−PV\times R-PV.

Example$15 000, 55 years, 6%6\% monthly: R=75/(1−1.005−60)≈R=75/(1-1.005^{-60})\approx $289.99; interest ≈\approx $2399.40.

TrapTotal paid uses the number of payments: 6060 monthly payments, not 55.

6Financial decisions

UseBring every cash flow to the same moment (usually now) at the same rate: down payments, payments and resale values. Then compare.

TrapDiscount the resale value too: $8000 received in 44 years is worth less than $8000 today.

Which formula?

One deposit, value later A=P(1+i)nA=P(1+i)^n
One amount later, value now PV=A(1+i)−nPV=A(1+i)^{-n}
Equal regular payments Annuity FV or PV

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