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A car is purchased for $25,000.At the end of each 6-month period, the car's value is 80% of its value...

GMAT Advanced Math : (Adv_Math) Questions

Source: Prism
Advanced Math
Nonlinear functions
MEDIUM
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Notes
Post a Query
  1. A car is purchased for \(\$25,000\).
  2. At the end of each 6-month period, the car's value is \(80\%\) of its value at the end of the preceding 6-month period.
  3. Which of the following functions best models \(\mathrm{V}\), the value of the car in dollars, \(\mathrm{t}\) years after purchase?
  4. Assume the depreciation rate remains constant over time.
A
\(\mathrm{V = 25,000(0.20)^t}\)
B
\(\mathrm{V = 25,000(0.80)^t}\)
C
\(\mathrm{V = 25,000(0.80)^{2t}}\)
D
\(\mathrm{V = 25,000(1.20)^{2t}}\)
E
\(\mathrm{V = 25,000(1.80)^t}\)
Solution

1. TRANSLATE the problem information

  • Given information:
    • Initial value: \(\$25,000\)
    • Every 6 months: value becomes 80% of previous value
    • Need: function V(t) where t = years after purchase
  • What this tells us: We have exponential decay happening every 6 months, but need to express it in terms of years.

2. INFER the compounding pattern

  • Key insight: The 80% reduction happens every 6 months, not every year
  • In 1 year, there are 2 six-month periods
  • In t years, there are 2t six-month periods
  • Each period multiplies the value by 0.80

3. INFER the mathematical relationship

  • After 1 six-month period: \(\$25,000 \times 0.80\)
  • After 2 six-month periods (1 year): \(\$25,000 \times (0.80)^2\)
  • After 2t six-month periods (t years): \(\$25,000 \times (0.80)^{2t}\)

4. Verify against answer choices

Looking at the choices, \(\mathrm{V} = 25,000(0.80)^{2t}\) matches choice (C) exactly.

Answer: C




Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students recognize the 80% decay but fail to account for the frequency difference between the given 6-month periods and the requested yearly variable.

They think: "The value is 80% each period, and t represents time, so it should be \((0.80)^t\)."

This leads them to select Choice B [\(\mathrm{V} = 25,000(0.80)^t\)] because they treat the decay as if it happens once per year instead of twice per year.

Second Most Common Error:

Poor TRANSLATE reasoning: Students misinterpret "80% of previous value" as meaning the car loses 80% of its value (keeping only 20%).

They incorrectly use \(0.20\) as the multiplication factor instead of \(0.80\).

This may lead them to select Choice A [\(\mathrm{V} = 25,000(0.20)^t\)].

The Bottom Line:

This problem tests whether students can correctly handle exponential functions when the compounding period differs from the time variable units. The key is recognizing that 6-month periods must be converted to match the yearly time variable.

Answer Choices Explained
A
\(\mathrm{V = 25,000(0.20)^t}\)
B
\(\mathrm{V = 25,000(0.80)^t}\)
C
\(\mathrm{V = 25,000(0.80)^{2t}}\)
D
\(\mathrm{V = 25,000(1.20)^{2t}}\)
E
\(\mathrm{V = 25,000(1.80)^t}\)
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