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The population of a town is currently 50{,000}, and the population is estimated to increase each year by 3% from...

GMAT Advanced Math : (Adv_Math) Questions

Source: Official
Advanced Math
Nonlinear functions
HARD
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The population of a town is currently \(\mathrm{50{,}000}\), and the population is estimated to increase each year by \(\mathrm{3\%}\) from the previous year. Which of the following equations can be used to estimate the number of years, \(\mathrm{t}\), it will take for the population of the town to reach \(\mathrm{60{,}000}\)?

A

\(50,000 = 60,000(0.03)^\mathrm{t}\)

B

\(50,000 = 60,000(3)^\mathrm{t}\)

C

\(60,000 = 50,000(0.03)^\mathrm{t}\)

D

\(60,000 = 50,000(1.03)^\mathrm{t}\)

Solution

1. TRANSLATE the problem information

  • Given information:
    • Initial population: 50,000
    • Growth rate: 3% increase each year
    • Target population: 60,000
    • Need to find: equation for number of years (t)
  • What "3% increase each year" means: Each year the population becomes \(103\%\) of the previous year

2. INFER how exponential growth works

  • With exponential growth, we multiply by the same factor each time period
  • A 3% increase means we multiply by \(1.03\) each year (not \(0.03\))
  • After t years: Population = \(\mathrm{(Initial)} \times \mathrm{(Growth\ factor)}^\mathrm{t}\)

3. TRANSLATE this into our specific equation

  • Starting population: 50,000
  • Growth factor: \(1.03\)
  • After t years: \(50,000(1.03)^\mathrm{t}\)
  • We want this to equal 60,000

4. Set up the final equation

  • Target population = Initial population × \(\mathrm{(Growth\ factor)}^\mathrm{t}\)
  • \(60,000 = 50,000(1.03)^\mathrm{t}\)

Answer: D




Why Students Usually Falter on This Problem


Most Common Error Path:

Weak TRANSLATE skill: Students see "3% increase" and immediately think to use \(0.03\) as the multiplier, forgetting that a 3% increase means the new amount is \(103\%\) of the original.

With this error, they might choose Choice C (\(60,000 = 50,000(0.03)^\mathrm{t}\)), not realizing that multiplying by \(0.03\) would actually cause the population to decrease dramatically each year, not increase.


Second Most Common Error:

Poor INFER reasoning about equation setup: Students might set up the equation backwards, thinking the current population should equal some expression involving the target population.

This leads them to select Choice A (\(50,000 = 60,000(0.03)^\mathrm{t}\)) or Choice B (\(50,000 = 60,000(3)^\mathrm{t}\)), neither of which makes logical sense since the smaller current population cannot equal an expression starting with the larger target population.


The Bottom Line:

The key insight is recognizing that "3% increase" means multiplying by \(1.03\) (which is \(100\% + 3\%\)), not by \(0.03\). This fundamental translation error leads to most wrong answers on exponential growth problems.

Answer Choices Explained
A

\(50,000 = 60,000(0.03)^\mathrm{t}\)

B

\(50,000 = 60,000(3)^\mathrm{t}\)

C

\(60,000 = 50,000(0.03)^\mathrm{t}\)

D

\(60,000 = 50,000(1.03)^\mathrm{t}\)

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