Growth of a Culture of BacteriaDayNumber of bacteria per milliliter at end of day12.5 * 10^525.0 * 10^531.0 * 10^6A...
GMAT Advanced Math : (Adv_Math) Questions
Growth of a Culture of Bacteria
| Day | Number of bacteria per milliliter at end of day |
|---|---|
| 1 | \(2.5 \times 10^5\) |
| 2 | \(5.0 \times 10^5\) |
| 3 | \(1.0 \times 10^6\) |
A culture of bacteria is growing at an exponential rate, as shown in the table above. At this rate, on which day would the number of bacteria per milliliter reach \(5.12 \times 10^8\)?
Day 5
Day 9
Day 11
Day 12
1. TRANSLATE the problem information
- Given information:
- Day 1: \(2.5\times10^5\) bacteria/mL
- Day 2: \(5.0\times10^5\) bacteria/mL
- Day 3: \(1.0\times10^6\) bacteria/mL
- Target: Find day when bacteria reaches \(5.12\times10^8\)
2. INFER the growth pattern
- Looking at the data ratios:
- Day 1 to Day 2: \(5.0\times10^5 \div 2.5\times10^5 = 2\)
- Day 2 to Day 3: \(1.0\times10^6 \div 5.0\times10^5 = 2\)
- The bacteria doubles every day - this is exponential growth with base 2
3. TRANSLATE into mathematical model
- Using Day 3 as reference point: \(\mathrm{N(d)} = 1.0\times10^6 \times 2^{(\mathrm{d}-3)}\)
- Set equal to target: \(1.0\times10^6 \times 2^{(\mathrm{d}-3)} = 5.12\times10^8\)
4. SIMPLIFY the exponential equation
- Divide both sides by \(1.0\times10^6\):
\(2^{(\mathrm{d}-3)} = 5.12\times10^8 \div 1.0\times10^6\)
\(2^{(\mathrm{d}-3)} = 5.12\times10^2\)
\(2^{(\mathrm{d}-3)} = 512\)
5. INFER the power relationship
- Need to solve: \(2^{(\mathrm{d}-3)} = 512\)
- Recognize that \(512 = 2^9\) (use calculator if needed to verify)
- Therefore: \(\mathrm{d}-3 = 9\), so \(\mathrm{d} = 12\)
Answer: D. Day 12
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak INFER skill: Students may recognize the doubling but set up the wrong equation, such as starting from Day 1 instead of Day 3, leading to \(\mathrm{N(d)} = 2.5\times10^5 \times 2^{(\mathrm{d}-1)} = 5.12\times10^8\). This gives \(2^{(\mathrm{d}-1)} = 2048\), and \(2048 = 2^{11}\), so \(\mathrm{d}-1 = 11\), making \(\mathrm{d} = 12\). Interestingly, this still gives the correct answer, but through a more complex calculation path.
Second Most Common Error:
Poor SIMPLIFY execution: Students may incorrectly handle the scientific notation conversion, calculating \(5.12\times10^8 \div 1.0\times10^6\) as 5.12 instead of 512, leading to \(2^{(\mathrm{d}-3)} = 5.12\). Since this isn't a clean power of 2, they become confused and guess.
The Bottom Line:
This problem tests whether students can recognize exponential patterns in data tables and translate them into solvable equations. The key insight is that exponential growth with doubling creates predictable powers of 2, making the algebra manageable once the pattern is identified.
Day 5
Day 9
Day 11
Day 12