A materials scientist studies two different alloy compositions and their relationship between temperature (in degrees Celsius) and internal pressure (...
GMAT Algebra : (Alg) Questions
A materials scientist studies two different alloy compositions and their relationship between temperature (in degrees Celsius) and internal pressure (in atmospheres). For Alloy A, the pressure \(\mathrm{P_A}\) is given by \(\mathrm{P_A = 0.35T + 8.0}\), where T is the temperature and \(\mathrm{50 \leq T \leq 200}\). For Alloy B, the pressure \(\mathrm{P_B}\) is given by \(\mathrm{P_B = 0.45T - 1.0}\), where T is the temperature and \(\mathrm{50 \leq T \leq 200}\). At 150 degrees Celsius, by how much does the pressure in Alloy B exceed the pressure in Alloy A?
5.0
6.0
8.0
15.0
1. TRANSLATE the problem requirements
- Given information:
- Alloy A: \(\mathrm{P_A = 0.35T + 8.0}\)
- Alloy B: \(\mathrm{P_B = 0.45T - 1.0}\)
- Temperature: \(\mathrm{T = 150°C}\)
- What we need: "by how much does pressure in Alloy B exceed pressure in Alloy A"
- This means we need \(\mathrm{P_B - P_A}\) at \(\mathrm{T = 150}\)
2. SIMPLIFY to find \(\mathrm{P_A}\) at \(\mathrm{T = 150}\)
- \(\mathrm{P_A = 0.35T + 8.0}\)
- \(\mathrm{P_A = 0.35(150) + 8.0}\)
- \(\mathrm{P_A = 52.5 + 8.0}\)
- \(\mathrm{P_A = 60.5}\) atmospheres
3. SIMPLIFY to find \(\mathrm{P_B}\) at \(\mathrm{T = 150}\)
- \(\mathrm{P_B = 0.45T - 1.0}\)
- \(\mathrm{P_B = 0.45(150) - 1.0}\)
- \(\mathrm{P_B = 67.5 - 1.0}\)
- \(\mathrm{P_B = 66.5}\) atmospheres
4. SIMPLIFY to find the difference
- \(\mathrm{P_B - P_A = 66.5 - 60.5}\)
- \(\mathrm{P_B - P_A = 6.0}\) atmospheres
Answer: B. 6.0
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak TRANSLATE skill: Students might misinterpret "by how much does B exceed A" and calculate \(\mathrm{P_A - P_B}\) instead of \(\mathrm{P_B - P_A}\), getting -6.0. Since this isn't an answer choice, they may select the absolute value or guess randomly.
Second Most Common Error:
Poor SIMPLIFY execution: Arithmetic errors in the decimal multiplications (\(\mathrm{0.35 × 150}\) or \(\mathrm{0.45 × 150}\)) or in the final subtraction. For example, miscalculating \(\mathrm{0.35 × 150}\) as 53.5 instead of 52.5, leading to \(\mathrm{P_A = 61.5}\) and a final difference of 5.0, causing them to select Choice A (5.0).
The Bottom Line:
This problem tests careful reading comprehension and systematic arithmetic execution. The mathematics is straightforward, but precision in both language interpretation and calculation is essential for success.
5.0
6.0
8.0
15.0