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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

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Advanced Math
Nonlinear functions
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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 \(\mathrm{k}\)?

A

\(\mathrm{k \leq 0}\)

B

\(\mathrm{0 \lt k \lt 1}\)

C

\(\mathrm{k = 1}\)

D

\(\mathrm{1 \lt k \lt 2}\)

E

\(\mathrm{k \geq 2}\)

Solution

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.

Answer Choices Explained
A

\(\mathrm{k \leq 0}\)

B

\(\mathrm{0 \lt k \lt 1}\)

C

\(\mathrm{k = 1}\)

D

\(\mathrm{1 \lt k \lt 2}\)

E

\(\mathrm{k \geq 2}\)

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