Question 3
Mina has $12 350 in a savings account that pays interest at 5.4% p.a., compounding monthly. She will also deposit an equal amount into the account at the end of each month.
Mina wants the account balance to be $64 000 at the end of 5 years.
Determine the monthly deposit Mina must make.
Grow the opening balance, then let the deposits build the rest
The opening balance grows on its own for 60 months, and the gap up to the target is what sixty equal deposits must build.
Test a monthly deposit
Set a monthly deposit. The readout runs $A_{n+1} = 1.0045A_n + d$ from $A_0 = 12\,350$ for 60 months and compares $A_{60}$ with the $64 000 target.
| Term | Balance in dollars |
|---|---|
$A_{60}$ = $
This reaches the $64 000 target, with $ to spare.
This is $ short of the $64 000 target.
$690.00 falls short of $64 000, so the deposit must be larger.
$700.00 goes over $64 000. With $690.00 falling short, the deposit lies between them.
$696.19 leaves the balance 19 cents short of $64 000.
$696.20 is the smallest deposit to the cent that reaches $64 000.
This deposit comes from $d = \dfrac{64\,000}{68.7047} = 931.52$, which leaves out the $12 350 already in the account. The balance overshoots the target.
This deposit comes from $64\,000 - 12\,350 = 51\,650$, which subtracts the $12 350 without growing it, so $d = 751.77$ is too large.
The present value formula gives $47\,831.73 = 52.4796\,d$, so $d = 911.44$. A target balance needs the future value formula.
Rounding $1.0045^{60}$ to 1.31 early makes the opening balance $16 178.50 and the factor 68.8889, so $d = 694.18$ falls short.
Work through the solution one mark at a time
Try each step yourself before you reveal it.
Convert the annual rate to a monthly rate and count the months.
$$i = \frac{5.4}{100 \times 12} = 0.0045$$
$$n = 5 \times 12 = 60$$
The formula-book finance formulas use the rate per compounding period and the number of periods.
This mark may be implied by subsequent working.
Marker: correctly determines the values for $i$ and $n$
Grow the $12 350 for the full 5 years.
$$A = P(1+i)^n$$
$$A = 12\,350(1.0045)^{60} = 16\,168.27$$
The $12 350 already in the account earns compound interest for the full 5 years.
Type 12 350 × 1.0045, then the power key and 60, and press =.
Marker: determines the value of the $12 350 at the end of 5 years
Subtract the grown balance from the $64 000 target.
$$64\,000 - 16\,168.27 = 47\,831.73$$
The deposits only need to build the part of the $64 000 that the opening balance does not.
Follow-through applies from an error in earlier working.
Marker: determines the amount needed from the monthly deposits
Substitute into the future value annuity formula with $A_{FV} = 47\,831.73$.
$$A_{FV} = d\left(\frac{(1+i)^n - 1}{i}\right)$$
$$47\,831.73 = d\left(\frac{1.0045^{60} - 1}{0.0045}\right)$$
$$47\,831.73 = 68.7047\,d$$
Equal deposits at the end of each month form an ordinary annuity, so use the future value annuity formula.
Evaluate the bracket in one line, with brackets around the numerator: (1.0045, power key, 60, − 1) ÷ 0.0045.
Using the recurrence relation $A_{n+1} = 1.0045A_n + d$ with $A_0 = 12\,350$ and trial values of $d$ is an alternative method. Method 2 shows that route.
Marker: substitutes into the future value annuity formula
Divide to find $d$.
$$d = \frac{47\,831.73}{68.7047} = 696.193$$
Round up to the next cent: $696.20.
Round up to the cent so the balance reaches $64 000. A deposit of $696.19 leaves the balance 19 cents short of $64 000.
A deposit of $696.19 also earns this mark.
Marker: determines the monthly deposit
Convert the annual rate to a monthly rate and count the months.
$$i = \frac{5.4}{100 \times 12} = 0.0045$$
$$n = 5 \times 12 = 60$$
The recurrence uses the monthly growth factor $1 + i$ and runs once per month.
This mark may be implied by subsequent working.
Marker: correctly determines the values for $i$ and $n$
Set up the formula-book recurrence for an annuity.
$$A_{n+1} = rA_n + d$$
$$A_{n+1} = 1.0045A_n + d, \quad A_0 = 12\,350$$
The formula-book recurrence for annuities grows the opening balance and adds each deposit in the same step.
Enter 12 350 and press =. Then type Ans × 1.0045 + (trial value of $d$) and press = sixty times. The display after the sixtieth press is the balance at the end of 5 years.
Run it for trial values of $d$.
| trial $d$ | $A_{60}$ |
|---|---|
| $690.00 | $63 574.53 |
| $700.00 | $64 261.57 |
| $696.19 | $63 999.81 |
| $696.20 | $64 000.50 |
Trial values of $d$ trap the deposit that makes $A_{60}$ reach $64 000.
Marks 2, 3 and 4 are implied by the recurrence working, which grows the $12 350 and builds the deposits together.
Follow-through applies from an error in earlier working.
Marker: determines the value of the $12 350 at the end of 5 years
Marker: determines the amount needed from the monthly deposits
Marker: substitutes into the future value annuity formula
From the table, $696.19 gives $63 999.81 and $696.20 gives $64 000.50.
$696.20 is the smallest deposit to the cent that reaches $64 000.
A deposit of $696.19 also earns this mark.
Marker: determines the monthly deposit
If I had to guess where you might lose marks…
- Leaving out the $12 350: $d = \dfrac{64\,000}{68.7047} = 931.52$. This loses Marks 2 and 3.
- Subtracting the $12 350 without growing it: $64\,000 - 12\,350 = 51\,650$, so $d = 751.77$. This loses Mark 2.
- Using the present value formula $A_{PV} = d\left(\dfrac{1-(1+i)^{-n}}{i}\right)$ for a target balance: $47\,831.73 = 52.4796\,d$, so $d = 911.44$. This loses Marks 4 and 5.
- Rounding $1.0045^{60}$ to 1.31 early: the opening balance becomes $16 178.50, the factor 68.8889 and $d = 694.18$. This loses Mark 5.
Only mark this done when you could do it without help. Reading the solution does not count.
A Tangent Tuition mock exam question. It is not a past QCAA question, and Tangent Tuition is not affiliated with the QCAA.
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