The temperature T of an object h hours after it is placed in a room is modeled by \(\mathrm{T =...
GMAT Advanced Math : (Adv_Math) Questions
The temperature \(\mathrm{T}\) of an object \(\mathrm{h}\) hours after it is placed in a room is modeled by \(\mathrm{T = A + (T_0 - A)k^h}\), where \(\mathrm{A}\) is the constant room temperature, \(\mathrm{T_0}\) is the object's initial temperature (with \(\mathrm{T_0 \gt A}\)), and \(\mathrm{k}\) is a constant. If the object approaches room temperature such that each hour the temperature difference decreases by the same percentage, which of the following must be true about \(\mathrm{k}\)?
\(\mathrm{k \leq 0}\)
\(\mathrm{0 \lt k \lt 1}\)
\(\mathrm{k = 1}\)
\(\mathrm{1 \lt k \lt 2}\)
\(\mathrm{k \geq 2}\)
1. TRANSLATE the problem information
- Given temperature model: \(\mathrm{T = A + (T_0 - A)k^h}\)
- Key constraint: "each hour the temperature difference decreases by the same percentage"
- We need to find what this tells us about k
2. INFER what the percentage decrease means
The temperature difference from room temperature is \(\mathrm{T - A = (T_0 - A)k^h}\)
When something decreases by the same percentage each time period, it means we multiply by the same fraction each time. For example:
- If it decreases by 25% each hour → multiply by 0.75 each hour
- If it decreases by 40% each hour → multiply by 0.60 each hour
This fraction is always positive and less than 1.
3. INFER the constraint on k
Since each hour the difference \(\mathrm{(T_0 - A)k^h}\) is multiplied by k to get \(\mathrm{(T_0 - A)k^{(h+1)}}\), the value k itself must be this positive fraction less than 1.
Therefore: \(\mathrm{0 \lt k \lt 1}\)
4. INFER by checking other possibilities
Let's verify this makes sense:
- If \(\mathrm{k = 1}\): The difference stays \(\mathrm{(T_0 - A) \times 1^h = (T_0 - A)}\), no change
- If \(\mathrm{k \gt 1}\): The difference grows as \(\mathrm{k^h}\) increases, not decreasing
- If \(\mathrm{k \leq 0}\): We get negative differences or alternating signs, not a steady decrease
Answer: B (\(\mathrm{0 \lt k \lt 1}\))
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak INFER skill: Students misinterpret what "decreases by the same percentage" means mathematically. They might think this just means k needs to be less than some value, or they might confuse percentage decrease with absolute decrease.
This conceptual confusion leads them to incorrectly analyze the constraints, possibly selecting Choice A (\(\mathrm{k \leq 0}\)) if they think any decrease means k must be non-positive, or Choice C (\(\mathrm{k = 1}\)) if they think constant percentage means k stays at 1.
Second Most Common Error:
Insufficient INFER reasoning: Students correctly identify that we need \(\mathrm{0 \lt k \lt 1}\) but then second-guess themselves by not fully analyzing what happens in other cases. They might worry that k could equal 1 and still represent some kind of change.
This incomplete reasoning causes them to get stuck between choices B and C, often leading to guessing.
The Bottom Line:
This problem tests whether students can translate a real-world description of exponential decay into mathematical constraints. The key insight is recognizing that "same percentage decrease" means multiplication by a constant positive fraction less than 1.
\(\mathrm{k \leq 0}\)
\(\mathrm{0 \lt k \lt 1}\)
\(\mathrm{k = 1}\)
\(\mathrm{1 \lt k \lt 2}\)
\(\mathrm{k \geq 2}\)