The table above shows the population of Greenleaf, Idaho, for the years 2000 and 2010. If the relationship between population...
GMAT Algebra : (Alg) Questions
The table above shows the population of Greenleaf, Idaho, for the years 2000 and 2010. If the relationship between population and year is linear, which of the following functions \(\mathrm{P}\) models the population of Greenleaf \(\mathrm{t}\) years after 2000?
| Population of Greenleaf, Idaho | |
|---|---|
| Year | Population |
| 2000 | 862 |
| 2010 | 846 |
\(\mathrm{P(t) = 862 - 1.6t}\)
\(\mathrm{P(t) = 862 - 16t}\)
\(\mathrm{P(t) = 862 + 16(t - 2,000)}\)
\(\mathrm{P(t) = 862 - 1.6(t - 2,000)}\)
1. TRANSLATE the problem information
- Given information:
- Year 2000: Population = 862
- Year 2010: Population = 846
- Linear relationship
- \(\mathrm{t}\) = years after 2000 (so \(\mathrm{t = 0}\) means year 2000, \(\mathrm{t = 10}\) means year 2010)
2. INFER the function structure
- Since the relationship is linear, we need \(\mathrm{P(t) = mt + b}\)
- We need to find slope \(\mathrm{(m)}\) and y-intercept \(\mathrm{(b)}\)
- The y-intercept occurs when \(\mathrm{t = 0}\), which corresponds to year 2000
3. Find the y-intercept (b)
- When \(\mathrm{t = 0}\) (year 2000), \(\mathrm{P(0) = 862}\)
- Therefore, \(\mathrm{b = 862}\)
4. SIMPLIFY to find the slope
- We have two points: \(\mathrm{(0, 862)}\) and \(\mathrm{(10, 846)}\)
- Slope = \(\mathrm{\frac{846 - 862}{10 - 0} = \frac{-16}{10} = -1.6}\)
5. Write the final function
- \(\mathrm{P(t) = -1.6t + 862 = 862 - 1.6t}\)
Answer: A. \(\mathrm{P(t) = 862 - 1.6t}\)
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak SIMPLIFY execution: Students calculate the slope as \(\mathrm{\frac{846 - 862}{10 - 0} = \frac{-16}{10}}\) but forget to reduce the fraction, leaving the slope as -16 instead of -1.6.
This leads them to write \(\mathrm{P(t) = 862 - 16t}\) and select Choice B (\(\mathrm{P(t) = 862 - 16t}\)).
Second Most Common Error:
Poor TRANSLATE reasoning: Students misinterpret what "t years after 2000" means and think t represents the actual year (like \(\mathrm{t = 2000, t = 2010}\)) rather than the number of years elapsed since 2000.
This confusion leads them to work with expressions involving \(\mathrm{(t - 2000)}\) and select Choice C (\(\mathrm{P(t) = 862 + 16(t - 2,000)}\)) or Choice D (\(\mathrm{P(t) = 862 - 1.6(t - 2,000)}\)).
The Bottom Line:
This problem tests whether students can correctly set up a linear model by understanding the time variable definition and accurately computing slope through fraction reduction.
\(\mathrm{P(t) = 862 - 1.6t}\)
\(\mathrm{P(t) = 862 - 16t}\)
\(\mathrm{P(t) = 862 + 16(t - 2,000)}\)
\(\mathrm{P(t) = 862 - 1.6(t - 2,000)}\)