Decay of a Chemical ConcentrationHourConcentration per milliliter at end of hour16.4 * 10^723.2 * 10^731.6 * 10^7A chemical breaks down...
GMAT Advanced Math : (Adv_Math) Questions
Decay of a Chemical Concentration
| Hour | Concentration per milliliter at end of hour |
|---|---|
| 1 | \(6.4 \times 10^7\) |
| 2 | \(3.2 \times 10^7\) |
| 3 | \(1.6 \times 10^7\) |
A chemical breaks down at an exponential rate, as shown in the table above. At this rate, at the end of which hour will the concentration first be less than \(1.25 \times 10^5\) molecules per milliliter?
Hour 8
Hour 9
Hour 10
Hour 11
1. TRANSLATE the problem information
- Given information:
- Hour 1: \(\mathrm{6.4×10^7}\) molecules/mL
- Hour 2: \(\mathrm{3.2×10^7}\) molecules/mL
- Hour 3: \(\mathrm{1.6×10^7}\) molecules/mL
- Need to find when concentration first drops below \(\mathrm{1.25×10^5}\) molecules/mL
2. INFER the mathematical pattern
- Looking at the data: \(\mathrm{6.4×10^7 → 3.2×10^7 → 1.6×10^7}\)
- Each hour, the concentration becomes exactly half of the previous hour
- This is exponential decay with a decay factor of 1/2
- Mathematical model: \(\mathrm{N(n) = 6.4×10^7 × (1/2)^{(n-1)}}\)
3. TRANSLATE the question into mathematical inequality
- 'First time concentration is less than \(\mathrm{1.25×10^5}\)'
- Set up: \(\mathrm{N(n) \lt 1.25×10^5}\)
- Substitute model: \(\mathrm{6.4×10^7 × (1/2)^{(n-1)} \lt 1.25×10^5}\)
4. SIMPLIFY the inequality through algebraic steps
- Divide both sides by \(\mathrm{6.4×10^7}\):
\(\mathrm{(1/2)^{(n-1)} \lt \frac{1.25×10^5}{6.4×10^7}}\)
- Calculate the right side (use calculator):
\(\mathrm{\frac{1.25×10^5}{6.4×10^7} = \frac{1.25}{640} = \frac{1}{512}}\)
- So we have: \(\mathrm{(1/2)^{(n-1)} \lt \frac{1}{512}}\)
5. SIMPLIFY further by converting to positive exponents
- Rewrite as: \(\mathrm{2^{(n-1)} \gt 512}\)
- Since \(\mathrm{512 = 2^9}\), we have: \(\mathrm{2^{(n-1)} \gt 2^9}\)
- Therefore: \(\mathrm{n-1 \gt 9}\), which gives us \(\mathrm{n \gt 10}\)
6. APPLY CONSTRAINTS to find the specific answer
- We need the smallest integer n where \(\mathrm{n \gt 10}\)
- This gives us \(\mathrm{n = 11}\)
- But let's verify by checking both Hour 10 and Hour 11:
- Hour 10: \(\mathrm{N(10) = \frac{6.4×10^7}{2^9} = 1.25×10^5}\) (equal to threshold, not less than)
- Hour 11: \(\mathrm{N(11) = \frac{6.4×10^7}{2^{10}} = 6.25×10^4 \lt 1.25×10^5}\) ✓
Answer: D. Hour 11
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak INFER skill: Students may recognize there's a pattern but fail to identify it as exponential decay with base 1/2. Instead, they might try to find a linear pattern or get confused about how exponential functions work with fractional bases. This leads to setting up incorrect equations and getting stuck, causing them to abandon systematic solution and guess.
Second Most Common Error:
Poor APPLY CONSTRAINTS reasoning: Students correctly solve the inequality and find \(\mathrm{n \gt 10}\), but then select Choice C (Hour 10) because they don't carefully check that Hour 10 gives exactly \(\mathrm{1.25×10^5}\), which equals (but is not less than) the threshold. The problem specifically asks for 'first be less than,' requiring strict inequality.
The Bottom Line:
This problem requires students to bridge pattern recognition with exponential modeling, then carefully execute multi-step algebraic manipulations while paying attention to inequality constraints. The combination of scientific notation, exponential equations, and precision in mathematical language makes it challenging.
Hour 8
Hour 9
Hour 10
Hour 11