A new car is purchased for $25,000. The value of the car depreciates by 15% each year after it is...
GMAT Advanced Math : (Adv_Math) Questions
A new car is purchased for \(\$25,000\). The value of the car depreciates by \(15\%\) each year after it is purchased. Which of the following functions gives the value of the car, \(\mathrm{V}\), in dollars, \(\mathrm{t}\) years after the purchase?
- \(\mathrm{V} = 25,000(0.15)^{\mathrm{t}}\)
- \(\mathrm{V} = 25,000(0.85)^{\mathrm{t}}\)
- \(\mathrm{V} = 25,000(1.15)^{\mathrm{t}}\)
- \(\mathrm{V} = 25,000(1 - 0.15\mathrm{t})\)
1. TRANSLATE the problem information
- Given information:
- Initial car value: \(\$25,000\)
- Car depreciates by \(15\%\) each year
- Need to find function \(\mathrm{V(t)}\) for value after \(\mathrm{t}\) years
2. INFER the mathematical model needed
- Key insight: "Depreciates by \(15\%\) each year" describes exponential decay
- If the car loses \(15\%\) of its value each year, then it retains \(85\%\) of its value
- This means each year the value is multiplied by \(0.85\)
- This fits the exponential decay model: \(\mathrm{V(t) = a(b)^t}\)
3. INFER the specific values for the exponential function
- Initial value \(\mathrm{(a) = 25,000}\)
- Decay factor \(\mathrm{(b) = 0.85}\) (since \(\mathrm{100\% - 15\% = 85\% = 0.85}\))
- Function: \(\mathrm{V(t) = 25,000(0.85)^t}\)
4. TRANSLATE to match the answer choices
- Looking at the options, this matches choice (B): \(\mathrm{V = 25,000(0.85)^t}\)
Answer: B
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak INFER skill: Students confuse the decay rate with the decay factor
Many students see "depreciates by \(15\%\)" and think the base should be \(0.15\), reasoning that since \(15\%\) is lost each year, they should use \(15\%\) in the function. However, exponential functions use the factor by which the quantity is multiplied each time period, not the amount that disappears.
This leads them to select Choice A (\(\mathrm{V = 25,000(0.15)^t}\))
Second Most Common Error:
Poor TRANSLATE reasoning: Students misinterpret depreciation as growth
Some students see the percentage and assume any percentage means growth, or they incorrectly think that \(15\%\) depreciation means the value becomes \(115\%\) of what it was (adding instead of subtracting the depreciation).
This may lead them to select Choice C (\(\mathrm{V = 25,000(1.15)^t}\))
The Bottom Line:
The key insight that separates successful students from struggling ones is understanding that in exponential decay, you use the fraction that remains each period (\(0.85\)), not the fraction that disappears (\(0.15\)). Students who master this concept can quickly eliminate wrong answers and confidently select the correct exponential model.