A lab sample has a mass of 135 milligrams at time t = 0. Each hour, the mass becomes one-third...

1. TRANSLATE the problem information

• Given information:
  • Initial mass: 135 milligrams at \(\mathrm{t} = 0\)
  • Each hour: mass becomes \(\frac{1}{3}\) of the previous hour's mass
  • Find: mass after 2 hours

• What this tells us: The mass is multiplying by \(\frac{1}{3}\) each hour, creating an exponential decay pattern.

2. INFER the mathematical approach

• This is exponential decay where the mass gets multiplied by the same factor \((\frac{1}{3})\) repeatedly
• We can either:
  • Calculate step by step for each hour
  • Use the exponential formula: \(\mathrm{M(t)} = \text{initial mass} \times (\text{decay factor})^\mathrm{t}\)

3. SIMPLIFY using the exponential formula

• Set up the formula: \(\mathrm{M(t)} = 135 \times (\frac{1}{3})^\mathrm{t}\)
• For \(\mathrm{t} = 2\) hours: \(\mathrm{M(2)} = 135 \times (\frac{1}{3})^2\)
• Calculate the exponent: \((\frac{1}{3})^2 = \frac{1}{9}\)
• Final calculation: \(135 \times \frac{1}{9} = 15\) milligrams

Answer: C (15)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students misinterpret "becomes one-third of the mass" as "loses one-third of the mass" (meaning \(\frac{2}{3}\) remains each hour instead of \(\frac{1}{3}\) remains).

Using \(\frac{2}{3}\) as the decay factor: \(\mathrm{M(2)} = 135 \times (\frac{2}{3})^2 = 135 \times \frac{4}{9} = 60\). Since 60 isn't among the choices, this leads to confusion and guessing.

Second Most Common Error:

Poor SIMPLIFY execution: Students correctly set up \(\mathrm{M(2)} = 135 \times (\frac{1}{3})^2\) but make arithmetic errors.

Common mistake: Calculating \((\frac{1}{3})^2\) as \(\frac{2}{3}\) instead of \(\frac{1}{9}\), leading to \(\mathrm{M(2)} = 135 \times \frac{2}{3} = 90\). Again, 90 isn't among the choices, causing them to second-guess their approach and potentially select Choice B (45) thinking they made an error in the time calculation.

The Bottom Line:

This problem tests whether students can correctly interpret repeated proportional change as exponential decay and execute the calculations accurately. The key insight is recognizing that "becomes one-third" means multiply by \(\frac{1}{3}\), not subtract \(\frac{1}{3}\).