A certain radioactive substance decays such that the amount remaining halves every 4 hours. At the beginning of an experiment,...

1. TRANSLATE the problem information

• Given information:
  • Initial amount: 1,280 grams
  • The substance 'halves every 4 hours'
  • Need to find y (grams remaining) after t hours

• What 'halves every 4 hours' means mathematically:
  • Every 4 hours, the amount becomes 1/2 of what it was
  • The decay factor is 1/2 per 4-hour period

2. INFER the mathematical model

• This is exponential decay following the pattern:
  \(\mathrm{y = (initial\,amount) \times (decay\,factor)^{(number\,of\,periods)}}\)

• Key insight: We need to count how many 4-hour periods occur in t hours
  \(\mathrm{Number\,of\,4-hour\,periods = \frac{t}{4}}\)

3. INFER the complete function

• Putting it together:
  \(\mathrm{y = 1{,}280 \times \left(\frac{1}{2}\right)^{\frac{t}{4}}}\)

• This matches the exponential decay form where:
  • 1,280 is the initial amount
  • 1/2 is the decay factor
  • t/4 is the number of decay periods

4. Verify with test values

• At \(\mathrm{t = 0}\):
  \(\mathrm{y = 1{,}280 \times \left(\frac{1}{2}\right)^{0}}\)
  \(\mathrm{= 1{,}280 \times 1}\)
  \(\mathrm{= 1{,}280\,grams}\) ✓
• At \(\mathrm{t = 4}\):
  \(\mathrm{y = 1{,}280 \times \left(\frac{1}{2}\right)^{1}}\)
  \(\mathrm{= 1{,}280 \times 0.5}\)
  \(\mathrm{= 640\,grams}\) ✓
• At \(\mathrm{t = 8}\):
  \(\mathrm{y = 1{,}280 \times \left(\frac{1}{2}\right)^{2}}\)
  \(\mathrm{= 1{,}280 \times 0.25}\)
  \(\mathrm{= 320\,grams}\) ✓

Answer: C

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students struggle to convert 'halves every 4 hours' into the mathematical structure \(\left(\frac{1}{2}\right)^{\frac{t}{4}}\). They might incorrectly think the base should be 2 instead of 1/2, reasoning that 'halving' involves the number 2.

This leads them to consider options with base 2, making them select Choice D (\(\mathrm{y = 1{,}280(2)^{\frac{t}{4}}}\)) - which would represent growth, not decay.

Second Most Common Error:

Poor INFER reasoning: Students recognize this is exponential decay but place the components incorrectly in the formula. They might put the initial amount (1,280) as the base of the exponential rather than as the coefficient.

This confusion about the exponential decay structure may lead them to select Choice A (\(\mathrm{y = \left(\frac{1}{2}\right)(1{,}280)^{\frac{t}{4}}}\)) - where 1,280 becomes the base instead of the initial amount.

The Bottom Line:

This problem requires recognizing that 'halving' creates a decay factor of 1/2 (not 2) and understanding that the time variable t must be divided by the period length (4 hours) to count the number of decay cycles. The exponential decay model has a specific structure that students must apply correctly.