A meteorologist studying the relationship between altitude and atmospheric pressure found that the relationship is linear within a certain range....

1. TRANSLATE the problem information

• Given information:
  • At altitude 1,000 ft: pressure = 14.7 psia
  • At altitude 4,000 ft: pressure = 13.2 psia
  • Relationship is linear
  • Find: altitude when pressure = 14.0 psia

• This gives us two coordinate points: \(\mathrm{(1000, 14.7)}\) and \(\mathrm{(4000, 13.2)}\)

2. INFER the approach needed

• Since we have a linear relationship, we need to find the equation of the line first
• Linear equations have the form \(\mathrm{P = mh + b}\) (pressure = slope × altitude + y-intercept)
• Once we have the equation, we can substitute \(\mathrm{P = 14.0}\) to find the altitude

3. SIMPLIFY to find the slope

• Using slope formula: \(\mathrm{m = \frac{P_2 - P_1}{h_2 - h_1}}\)
• \(\mathrm{m = \frac{13.2 - 14.7}{4000 - 1000}}\)
• \(\mathrm{m = \frac{-1.5}{3000}}\)
• \(\mathrm{m = -0.0005}\)

4. SIMPLIFY to establish the linear equation

• Using point-slope form with \(\mathrm{(1000, 14.7)}\):
• \(\mathrm{P - 14.7 = -0.0005(h - 1000)}\)
• \(\mathrm{P - 14.7 = -0.0005h + 0.5}\)
• \(\mathrm{P = -0.0005h + 15.2}\)

5. SIMPLIFY to solve for the target altitude

• Substitute \(\mathrm{P = 14.0}\):
• \(\mathrm{14.0 = -0.0005h + 15.2}\)
• \(\mathrm{-0.0005h = 14.0 - 15.2 = -1.2}\)
• \(\mathrm{h = \frac{1.2}{0.0005} = 2,400\ feet}\)

Answer: (B) 2,400

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Not recognizing that you need to establish the complete linear equation first before finding the specific altitude.

Students might try to use proportional reasoning directly: "The pressure dropped 1.5 psia over 3,000 feet, so to drop 0.7 psia (from 14.7 to 14.0) would take..." This approach gets messy and often leads to setup errors. This may lead them to select Choice (A) (1,400) or causes confusion and guessing.

Second Most Common Error:

Poor SIMPLIFY execution: Making algebraic errors when solving the final equation \(\mathrm{-0.0005h = -1.2}\).

Students might incorrectly divide, getting \(\mathrm{h = 1.2 \times 0.0005 = 0.0006}\) instead of \(\mathrm{h = \frac{1.2}{0.0005} = 2,400}\). Or they might lose track of negative signs during the manipulation. This leads to wildly incorrect answers and typically results in guessing.

The Bottom Line:

This problem tests whether students can systematically work with linear relationships rather than trying shortcuts. The key insight is that real-world linear problems require the same mathematical approach as abstract ones: establish the equation, then use it.