The table shows the population of a certain bacteria colony over time, where t is the time in hours since...
GMAT Advanced Math : (Adv_Math) Questions
The table shows the population of a certain bacteria colony over time, where \(\mathrm{t}\) is the time in hours since observation began, and \(\mathrm{p}\) is the population in hundreds.
| Time (hours) | Population (hundreds) |
|---|---|
| 0 | 1.00 |
| 2 | 1.04 |
| 4 | 1.0816 |
The population grows exponentially. If no external factors affect the growth, which of the following equations best represents the relationship between \(\mathrm{t}\) and \(\mathrm{p}\)?
\(\mathrm{p = (1 + 1)^{t/2}}\)
\(\mathrm{p = (1 + 0.02)^t}\)
\(\mathrm{p = 1(1 + 0.04)^{t/2}}\)
\(\mathrm{p = 0.04(1 + 1)^{t/2}}\)
1. TRANSLATE the table data into growth information
- Given information:
- At \(\mathrm{t=0}\) hours: \(\mathrm{p=1.00}\) hundreds
- At \(\mathrm{t=2}\) hours: \(\mathrm{p=1.04}\) hundreds
- At \(\mathrm{t=4}\) hours: \(\mathrm{p=1.0816}\) hundreds
- Population grows exponentially
2. INFER the growth pattern from consecutive data points
- Calculate growth factors between time periods:
- From \(\mathrm{t=0}\) to \(\mathrm{t=2}\): \(\mathrm{1.04 \div 1.00 = 1.04}\)
- From \(\mathrm{t=2}\) to \(\mathrm{t=4}\): \(\mathrm{1.0816 \div 1.04 = 1.04}\)
- Key insight: The population multiplies by 1.04 every 2 hours, not every hour
3. INFER the correct time variable for the exponential model
- Since growth occurs every 2 hours, after t total hours there have been \(\mathrm{t/2}\) growth periods
- Standard form: \(\mathrm{p = initial\ value \times (growth\ factor)^{(number\ of\ periods)}}\)
- Therefore: \(\mathrm{p = 1.00 \times (1.04)^{(t/2)}}\)
4. TRANSLATE into the answer choice format
- Since \(\mathrm{1.04 = 1 + 0.04}\), we can write:
\(\mathrm{p = 1(1 + 0.04)^{(t/2)}}\) - This matches choice C exactly
Answer: C
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak INFER skill: Not recognizing that growth occurs every 2 hours rather than every hour
Students see exponential growth and automatically think the time variable should be t (every hour). They calculate a growth factor per hour instead of recognizing the 2-hour growth periods shown in the table. This leads them to try finding a model like \(\mathrm{p = 1(1 + r)^t}\) where \(\mathrm{r \approx 0.02}\).
This may lead them to select Choice B (\(\mathrm{p = (1 + 0.02)^t}\)) which represents hourly growth rather than the correct 2-hour growth periods.
The Bottom Line:
The key challenge is recognizing that exponential models must match the time intervals in the given data. When data points are spaced 2 hours apart, you're seeing growth every 2 hours, not every hour.
\(\mathrm{p = (1 + 1)^{t/2}}\)
\(\mathrm{p = (1 + 0.02)^t}\)
\(\mathrm{p = 1(1 + 0.04)^{t/2}}\)
\(\mathrm{p = 0.04(1 + 1)^{t/2}}\)