A linear model estimates the population of a city from 1991 to 2015. The model estimates the population was 57...
GMAT Algebra : (Alg) Questions
A linear model estimates the population of a city from \(1991\) to \(2015\). The model estimates the population was \(57\) thousand in \(1991\), \(224\) thousand in \(2011\), and \(x\) thousand in \(2015\). To the nearest whole number, what is the value of \(x\)?
1. TRANSLATE the problem information
- Given information:
- Linear model estimates population from 1991 to 2015
- 1991 population: 57 thousand
- 2011 population: 224 thousand
- 2015 population: x thousand (unknown)
- What this tells us: Since it's linear, the population changes at a constant rate over time
2. INFER the solution approach
- Key insight: Linear models have constant rate of change (slope)
- Strategy: Find the rate of change using the two known data points, then apply it to find the 2015 value
- We need to work with the time periods and population changes
3. SIMPLIFY to find the rate of change
- Time period from 1991 to 2011: \(2011 - 1991 = 20\) years
- Population change: \(224 - 57 = 167\) thousand
- Rate of change = \(167 \div 20 = 8.35\) thousand per year (use calculator)
4. SIMPLIFY to find the 2015 population
- Time from 2011 to 2015: \(2015 - 2011 = 4\) years
- Population increase over 4 years: \(8.35 \times 4 = 33.4\) thousand (use calculator)
- 2015 population: \(224 + 33.4 = 257.4\) thousand
5. APPLY CONSTRAINTS for the final answer
- Round to nearest whole number as requested: \(257.4 \to 257\)
Answer: 257
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak INFER skill: Students don't recognize they need to find the rate of change first. Instead, they might try to work directly with the three data points or attempt to guess a pattern without using the linear model concept. This leads to confusion and guessing rather than systematic solution.
Second Most Common Error:
Poor SIMPLIFY execution: Students make arithmetic errors in the multi-step calculations, particularly in computing \(167 \div 20 = 8.35\) or in the final multiplication and addition steps. Small errors early in the calculation cascade to significantly wrong final answers.
The Bottom Line:
This problem tests whether students understand that linear models have constant rates of change and can systematically apply rate calculations across different time periods. Success requires both strategic thinking about linear relationships and careful arithmetic execution.