A research lab begins with a 100-gram sample of a radioactive isotope, Bismuth-210. The mass of the Bismuth-210 is observed...
GMAT Advanced Math : (Adv_Math) Questions
A research lab begins with a \(100\)-gram sample of a radioactive isotope, Bismuth-210. The mass of the Bismuth-210 is observed to decrease by \(13\%\) each day. Over a \(5\)-day period, the mass will decrease from \(100\) grams to approximately \(50\) grams, which corresponds to the isotope's half-life. Which of the following functions, \(\mathrm{m(t)}\), best models the mass, in grams, of the Bismuth-210 remaining after \(\mathrm{t}\) days?
1. TRANSLATE the problem information
- Given information:
- Initial mass: 100 grams of Bismuth-210
- Mass decreases by 13% each day
- After 5 days: mass is approximately 50 grams
- What this tells us: We need an exponential decay function where the amount gets smaller over time.
2. INFER the appropriate formula
- This is exponential decay, so we need: \(\mathrm{m(t) = initial\_amount \times (decay\_factor)^t}\)
- The key insight: if something decreases by 13%, then 87% remains each day
- We need to find the decay factor, not use the decay rate directly
3. TRANSLATE the decay information
- "Decreases by 13% each day" means:
- Decay rate = \(\mathrm{13\% = 0.13}\)
- Remaining each day = \(\mathrm{100\% - 13\% = 87\% = 0.87}\)
- So \(\mathrm{decay\ factor = 0.87}\)
4. Build the function
- \(\mathrm{m(t) = 100 \times (0.87)^t}\)
- This matches choice (C)
5. INFER verification using half-life
- Check: After 5 days, \(\mathrm{m(5) = 100(0.87)^5}\)
- Calculate: \(\mathrm{(0.87)^5 \approx 0.498}\) (use calculator)
- So \(\mathrm{m(5) \approx 100 \times 0.498 = 49.8\ grams \approx 50\ grams}\) ✓
Answer: C
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak TRANSLATE skill: Students confuse the decay rate (13%) with the decay factor and incorrectly use 0.13 as the base in their exponential function.
Their reasoning: "It decreases by 13%, so I'll use 0.13 in the formula."
This leads them to create \(\mathrm{m(t) = 100(0.13)^t}\), causing them to select Choice A \(\mathrm{(100(0.13)^t)}\)
Second Most Common Error:
Poor INFER reasoning about exponential models: Students mix up growth and decay, thinking that since there's a percentage involved, they should add it to 1 instead of subtract it.
Their reasoning: "There's 13% change, so the factor should be 1.13."
This may lead them to select Choice D \(\mathrm{(100(1.13)^t)}\), which models exponential growth instead of decay.
The Bottom Line:
The critical insight is that "decreases by x%" means you keep \(\mathrm{(100-x)\%}\) of the original amount. Students who master this translation between percentage language and mathematical factors will confidently identify the decay factor as 0.87, not 0.13.