A company uses a model to estimate the value of its equipment. The model assumes that at the end of...
GMAT Advanced Math : (Adv_Math) Questions
A company uses a model to estimate the value of its equipment. The model assumes that at the end of each year, the value of a piece of equipment is \(20\%\) less than its value at the end of the previous year. According to the model, a specific piece of equipment had a value of \(\$6{,}400\) at the end of 2018. Which of the following equations represents this model, where \(\mathrm{V}\) is the estimated value, in dollars, of the equipment \(\mathrm{t}\) years after the end of 2017 and \(\mathrm{t} \geq 0\)?
1. TRANSLATE the decay information
- Given information:
- Value decreases by 20% each year
- Value was $6,400 at end of 2018
- t represents years after end of 2017
- What this tells us: If value decreases by 20%, it retains \(80\% = 0.8\) of its value each year
2. INFER the model structure
- This is exponential decay with the form \(\mathrm{V = P(0.8)^t}\)
- We need to find P (the initial value when \(\mathrm{t = 0}\))
- The decay factor is 0.8, not 0.2
3. TRANSLATE the time reference
- "t years after end of 2017" means:
- \(\mathrm{t = 0}\) corresponds to end of 2017
- \(\mathrm{t = 1}\) corresponds to end of 2018
- Since value was $6,400 at end of 2018, we have \(\mathrm{V = 6{,}400}\) when \(\mathrm{t = 1}\)
4. INFER how to find the initial value
- Substitute our known values into \(\mathrm{V = P(0.8)^t}\):
\(\mathrm{6{,}400 = P(0.8)^1}\)
\(\mathrm{6{,}400 = 0.8P}\)
5. SIMPLIFY to find P
- Solve for P: \(\mathrm{P = 6{,}400 \div 0.8 = 8{,}000}\) (use calculator)
- Therefore: \(\mathrm{V = 8000(0.8)^t}\)
Answer: D
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak TRANSLATE skill: Students confuse the decay factor with the decay rate. They see "20% less" and incorrectly use 0.2 as the base instead of 0.8.
They think: "20% decrease means multiply by 0.2 each year"
This leads them to choose equations with \(\mathrm{(0.2)^t}\), selecting Choice A (\(\mathrm{V = 6400(0.2)^t}\)) or Choice C (\(\mathrm{V = 8000(0.2)^t}\)).
Second Most Common Error:
Poor INFER reasoning: Students assume the given value of $6,400 is the initial value (when \(\mathrm{t = 0}\)) rather than recognizing they need to work backward from the \(\mathrm{t = 1}\) data point.
They think: "The equipment is worth $6,400, so that must be P"
This may lead them to select Choice B (\(\mathrm{V = 6400(0.8)^t}\)), missing that they need to find what the value was one year earlier.
The Bottom Line:
This problem requires careful attention to what the percentages mean (decay factor vs. decay rate) and understanding that you might need to work backward from given data points to find initial conditions in exponential models.