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In 2020, a biologist estimated that there were 4,500 individuals of a certain fish species in a lake. The population...

GMAT Advanced Math : (Adv_Math) Questions

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Advanced Math
Nonlinear functions
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In 2020, a biologist estimated that there were 4,500 individuals of a certain fish species in a lake. The population is projected to decrease by \(15\%\) every 6 months. The function \(\mathrm{P}\) models the estimated population \(\mathrm{t}\) years after 2020. Which equation defines \(\mathrm{P}\)?

  1. \(\mathrm{P(t)} = 4500(0.85)^{\mathrm{t}/2}\)
  2. \(\mathrm{P(t)} = 4500(0.15)^{\mathrm{t}/2}\)
  3. \(\mathrm{P(t)} = 4500(0.85)^{2\mathrm{t}}\)
  4. \(\mathrm{P(t)} = 4500(1.15)^{2\mathrm{t}}\)
A
\(\mathrm{P(t) = 4500(0.85)^{(t/2)}}\)
B
\(\mathrm{P(t) = 4500(0.15)^{(t/2)}}\)
C
\(\mathrm{P(t) = 4500(0.85)^{2t}}\)
D
\(\mathrm{P(t) = 4500(1.15)^{2t}}\)
Solution

1. TRANSLATE the problem information

  • Given information:
    • Initial population in 2020: 4,500 fish
    • Population decreases by 15% every 6 months
    • t represents years after 2020
    • Need function P(t) for population after t years

2. INFER the mathematical model needed

  • This describes exponential decay, which follows the pattern: \(\mathrm{P(t) = a(b)^x}\)
  • Need to identify: initial value (a), decay factor (b), and number of periods (x)

3. TRANSLATE each component

  • Initial value: \(\mathrm{a = 4{,}500}\) (the starting population)
  • Decay factor: "decreases by 15%" means 85% remains each period
    • \(\mathrm{b = 1 - 0.15 = 0.85}\)
  • Time periods: This is where it gets tricky!
    • Decay happens every 6 months
    • But t is measured in years
    • In 1 year, there are 2 six-month periods
    • In t years, there are \(\mathrm{2t}\) six-month periods

4. SIMPLIFY to final form

  • Substitute into exponential model:
    \(\mathrm{P(t) = 4500(0.85)^{2t}}\)
  • This matches choice C

Answer: C. \(\mathrm{P(t) = 4500(0.85)^{2t}}\)




Why Students Usually Falter on This Problem


Most Common Error Path:

Weak TRANSLATE skill: Confusing the relationship between years and 6-month periods, leading to the wrong exponent.

Students often think: "If t is in years and decay happens every 6 months, then in t years there are \(\mathrm{t/2}\) periods." This backwards reasoning comes from thinking "6 months is half a year, so I divide by 2."

This leads them to select Choice A: \(\mathrm{P(t) = 4500(0.85)^{t/2}}\)


Second Most Common Error:

Conceptual confusion about decay factors: Using 0.15 instead of 0.85 as the decay factor.

Students mistakenly think the decay factor should be the percentage that's lost (15%) rather than the percentage that remains (85%). This fundamental misunderstanding of how exponential decay works causes them to use the wrong base.

This may lead them to select Choice B: \(\mathrm{P(t) = 4500(0.15)^{t/2}}\) (combining both errors)


The Bottom Line:

The key challenge is correctly TRANSLATING between different time units while understanding that exponential decay uses the "remaining factor" (0.85), not the "lost factor" (0.15).

Answer Choices Explained
A
\(\mathrm{P(t) = 4500(0.85)^{(t/2)}}\)
B
\(\mathrm{P(t) = 4500(0.15)^{(t/2)}}\)
C
\(\mathrm{P(t) = 4500(0.85)^{2t}}\)
D
\(\mathrm{P(t) = 4500(1.15)^{2t}}\)
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