Question 15
The resale value of a boat shows geometric decay.
| Years after purchase | 0 | 1 | 2 |
|---|---|---|---|
| Resale value ($) | 50 000 | 40 000 | 32 000 |
Determine the resale value four years after purchase.
Not quite. One way to get this is to keep subtracting the last drop, $8000, for two more years. That is arithmetic decay. In geometric decay each year takes 20% of a smaller value, so the drops keep shrinking. In 2023, 11.7% of Queensland students chose (A).
That’s it. Step through the hints below to see why. In 2023, 41.4% of Queensland students chose (B).
Not quite. One way to get this is to see drops of 10 000 and 8000 and keep shrinking them by 2000, to 6000 and then 4000. The drops do shrink, but by the factor 0.8 each time, not by a fixed amount: 10 000, 8000, 6400, 5120. In 2023, 26.9% of Queensland students chose (C).
Not quite. $25 600 is the value three years after purchase. Year 0 is the first term, so year 4 is $t_5=50\,000\times0.8^4$. Using $n=4$ leaves you one factor of 0.8 short. In 2023, 18.9% of Queensland students chose (D).
Each year the boat keeps 80% of the value it had the year before. Year 3 is worth $25 600, which is option (D). Year 4 is worth $20 480, option (B).
Taking the last drop, $8000, off every year is arithmetic decay. It reaches $16 000 at year 4, option (A). The table’s drops are already shrinking, so this can’t be the pattern.
Shrinking each drop by $2000 fits the first two drops and lands on $22 000, option (C). A geometric sequence shrinks its drops by the same factor, 0.8, never by the same amount.
Geometric decay means multiplying by the same ratio every year. Divide each value by the one before it.
$$r=\frac{40\,000}{50\,000}=0.8,\qquad\frac{32\,000}{40\,000}=0.8$$
The boat keeps 80% of its value each year. The drops in dollars shrink, because 20% of a smaller value is smaller.
The formula book gives $t_n=t_1r^{n-1}$. The first term is the value at year 0, so the term number is always one more than the year.
$$t_1=50\,000\ (\text{year }0),\qquad t_5\ (\text{year }4)$$
Using $n=4$ for year 4 gives $50\,000\times0.8^3=25\,600$, which is option (D) and the value at year 3.
$$\begin{aligned}t_5&=50\,000\times0.8^{5-1}\\&=50\,000\times0.8^4\\&=20\,480\end{aligned}$$
Check by carrying the table on, one factor of 0.8 per year:
$$32\,000\xrightarrow{\ \times0.8\ }25\,600\xrightarrow{\ \times0.8\ }20\,480$$
$$\begin{aligned}r&=\frac{40\,000}{50\,000}=0.8\\t_5&=50\,000\times0.8^4\\&=\text{\textdollar}20\,480\end{aligned}$$
Year 0 is the first term, so four years after purchase is the fifth term. That means four factors of 0.8.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2023 QCAA General Mathematics external assessment, © State of Queensland (QCAA) 2023, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
Spot a mistake on this page? Report it