A research team is monitoring the population of a certain species of frog in a protected wetland. The population was...

1. TRANSLATE the problem information

• Given information:
  • Initial population: 1,200 frogs
  • Population is decreasing (exponential decay model)
  • After 8 years: population ≈ 616
  • Need to find equation for P(t)

• What this tells us: We need \(\mathrm{P(t) = P_0 \times b^t}\) where \(\mathrm{P_0 = 1{,}200}\) and b represents decay

2. INFER which characteristics eliminate answer choices

• Since \(\mathrm{P_0 = 1{,}200}\), we can immediately eliminate any choice without 1,200 as the coefficient
• This eliminates choice (B) which starts with 584

• Since population is decreasing, we need exponential decay where \(\mathrm{0 \lt b \lt 1}\)
• This eliminates choice (D) which has base \(\mathrm{1.08 \gt 1}\) (that's growth, not decay)

3. INFER that verification with data point is needed

• We're left with choices (A) and (C), both starting with 1200
• Choice (A): \(\mathrm{P(t) = 1200(0.08)^t}\)
• Choice (C): \(\mathrm{P(t) = 1200(0.92)^t}\)
• Both have decay bases, so we need to check which matches \(\mathrm{P(8) = 616}\)

4. SIMPLIFY the verification calculations

• For choice (A): \(\mathrm{P(8) = 1200(0.08)^8}\)
  • Calculate \(\mathrm{0.08^8 \approx 0.0000168}\) (use calculator)
  • \(\mathrm{1200 \times 0.0000168 \approx 0.02}\)
  • This is nowhere near 616!

• For choice (C): \(\mathrm{P(8) = 1200(0.92)^8}\)
  • Calculate \(\mathrm{0.92^8 \approx 0.5132}\) (use calculator)
  • \(\mathrm{1200 \times 0.5132 \approx 616}\) ✓

Answer: C

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't systematically eliminate choices based on mathematical constraints. They might jump straight to testing all four choices with the data point \(\mathrm{P(8) = 616}\), making the problem much longer and more prone to calculation errors.

Some students might not recognize that choice (B) can be eliminated immediately due to wrong initial value, or that choice (D) represents growth instead of decay. This leads to testing inappropriate models and potentially getting confused about which calculation errors are causing wrong answers.

This may lead them to select Choice (A) (\(\mathrm{1200(0.08)^t}\)) if they make a calculation error or Choice (D) (\(\mathrm{1200(1.08)^t}\)) if they don't understand the difference between growth and decay models.

Second Most Common Error:

Poor SIMPLIFY execution: Students correctly identify that choice (C) is the answer but make calculator errors when computing \(\mathrm{0.92^8}\), getting a result that doesn't verify properly with \(\mathrm{P(8) = 616}\).

This causes them to doubt their systematic elimination process and potentially select a wrong choice like Choice (A) or abandon their work and guess.

The Bottom Line:

This problem rewards systematic thinking over computational brute force. The key insight is using the mathematical meaning of "initial value" and "decreasing" to eliminate wrong choices before doing any complex calculations.