The function \(\mathrm{S(h) = 2.4(h - 18.5) + 76}\) gives the predicted score on a standardized test for a student...

1. TRANSLATE the problem information

• Given information:
  • Function: \(\mathrm{S(h) = 2.4(h - 18.5) + 76}\)
  • Preparation time increases by 8.75 hours
• What we need to find: How much the predicted test score increases

2. INFER the key relationship

• This function is linear (in the form \(\mathrm{y = mx + b}\))
• In linear functions, the coefficient of the variable (here 2.4) tells us the rate of change
• Rate of change = 2.4 points per hour of preparation
• Since we want the change in score (not the actual score), we use this rate directly

3. SIMPLIFY the calculation

• Change in score = Rate of change × Change in input
• Change in score = \(\mathrm{2.4 \times 8.75}\)
• Change in score = 21.0 (use calculator)

Answer: A (21.0)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't recognize this as a rate of change problem and instead try to calculate actual scores by substituting values into the function.

They might calculate \(\mathrm{S(h + 8.75) - S(h)}\) by picking a specific value for h, like \(\mathrm{h = 20}\):

• \(\mathrm{S(28.75) = 2.4(28.75 - 18.5) + 76}\)
• \(\mathrm{= 2.4(10.25) + 76}\)
• \(\mathrm{= 100.6}\)
• \(\mathrm{S(20) = 2.4(20 - 18.5) + 76}\)
• \(\mathrm{= 2.4(1.5) + 76}\)
• \(\mathrm{= 79.6}\)
• Difference = \(\mathrm{100.6 - 79.6 = 21.0}\)

While this actually gives the right answer, it's unnecessarily complicated and increases chances for arithmetic errors.

Second Most Common Error:

Poor SIMPLIFY execution: Students understand the rate of change concept but make arithmetic errors when calculating \(\mathrm{2.4 \times 8.75}\).

Common mistakes include:

• \(\mathrm{2.4 \times 8.75 = 18.0}\) (forgetting decimal placement)
• \(\mathrm{2.4 \times 8.75 = 210}\) (decimal error)

This may lead them to select Choice B (53.0) or other incorrect options.

The Bottom Line:

Success depends on recognizing that linear functions have constant rates of change, making this a simple multiplication problem rather than a complex substitution exercise.