The value of a piece of industrial equipment, V, in dollars, is modeled as a function of the number of...
GMAT Advanced Math : (Adv_Math) Questions
The value of a piece of industrial equipment, \(\mathrm{V}\), in dollars, is modeled as a function of the number of years, \(\mathrm{t}\), since it was purchased. For each year that passes, the value of the equipment decreases by \(20\%\). The initial purchase price of the equipment was \(\$50,000\). Which of the following equations best models the value of the equipment \(\mathrm{t}\) years after it was purchased?
1. TRANSLATE the problem information
- Given information:
- Initial equipment value: \(\$50,000\)
- Equipment "decreases by 20%" each year
- Need to find V(t) after t years
- What "decreases by 20%" means: If something loses 20% of its value, it keeps \(100\% - 20\% = 80\%\) of its original value each year.
2. INFER the mathematical model needed
- This is exponential decay because:
- Value changes by a constant percentage each year
- We multiply by the same factor repeatedly
- Standard exponential model: \(\mathrm{V(t)} = \mathrm{a}(\mathrm{b})^\mathrm{t}\)
- a = initial value
- b = decay factor (what we multiply by each year)
- t = time in years
3. TRANSLATE percentage to decimal factor
- "Keeps 80% each year" means multiply by 0.8 each year
- So our decay factor \(\mathrm{b} = 0.8\)
4. Construct the final equation
- \(\mathrm{V(t)} = 50,000(0.8)^\mathrm{t}\)
Answer: B
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak TRANSLATE skill: Students confuse the decrease percentage with the decay factor.
They see "decreases by 20%" and think the base should be 0.2, leading them to write \(\mathrm{V(t)} = 50,000(0.2)^\mathrm{t}\). This misses the key insight that if something decreases by 20%, you multiply by the remaining 80% = 0.8, not by the lost 20% = 0.2.
This leads them to select Choice A (\(50,000(0.2)^\mathrm{t}\)).
Second Most Common Error:
Conceptual confusion about exponential growth vs decay: Students might think "20%" automatically means growth.
They incorrectly reason that since there's a 20% change, they should add it to 100%, getting \(120\% = 1.2\) as the base. This completely ignores the word "decreases" in the problem.
This may lead them to select Choice C (\(50,000(1.2)^\mathrm{t}\)).
The Bottom Line:
The key challenge is correctly translating percentage language into mathematical operations. "Decreases by X%" means "multiply by \((100-\mathrm{X})\%\)", not "multiply by \(\mathrm{X}\%\)".