Let x be a positive real number. A new number y is defined by y = px. The value of...

1. TRANSLATE the problem information

• Given information:
  • x is a positive real number (\(\mathrm{x \gt 0}\))
  • \(\mathrm{y = px}\) (y is defined as p times x)
  • \(\mathrm{y \gt x}\) (y is greater than x)

• What we need: Find which value of p makes this scenario possible

2. INFER the key relationship

• Since \(\mathrm{y = px}\) and we know \(\mathrm{y \gt x}\), we can substitute:
  \(\mathrm{px \gt x}\)

• The strategic insight: Since x is positive, we can divide both sides of the inequality by x without changing the inequality direction

• This gives us: \(\mathrm{p \gt 1}\)

3. APPLY CONSTRAINTS to evaluate each choice

• We need \(\mathrm{p \gt 1}\), so let's check each option:
  • Choice A: \(\mathrm{p = -2}\). Is \(\mathrm{-2 \gt 1}\)? No, negative numbers are less than 1
  • Choice B: \(\mathrm{p = \frac{3}{4}}\). Is \(\mathrm{\frac{3}{4} \gt 1}\)? No, \(\mathrm{\frac{3}{4} = 0.75 \lt 1}\)
  • Choice C: \(\mathrm{p = 1}\). Is \(\mathrm{1 \gt 1}\)? No, 1 equals 1, not greater than 1
  • Choice D: \(\mathrm{p = \frac{9}{8}}\). Is \(\mathrm{\frac{9}{8} \gt 1}\)? Yes, \(\mathrm{\frac{9}{8} = 1.125 \gt 1}\)

Answer: D (9/8)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students may not recognize they can divide both sides of \(\mathrm{px \gt x}\) by x, or they may be uncertain about whether this changes the inequality direction.

Without this key step, students might try plugging in specific values for x or get confused about how to work with the variable coefficient p. This leads to confusion and guessing.

Second Most Common Error:

Conceptual confusion about inequalities: Students might incorrectly think that dividing by x (even though \(\mathrm{x \gt 0}\)) flips the inequality sign, leading them to conclude \(\mathrm{p \lt 1}\) instead of \(\mathrm{p \gt 1}\).

This reversed thinking would make them look for values less than 1, potentially leading them to select Choice B (3/4) as it's the largest value less than 1.

The Bottom Line:

This problem tests whether students can manipulate inequalities with positive variables and recognize that the constraint naturally eliminates most answer choices through logical reasoning rather than complex calculations.