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A printer produces pages at a constant rate. The table shows the number of pages p the printer produces in...

GMAT Algebra : (Alg) Questions

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Algebra
Linear functions
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A printer produces pages at a constant rate. The table shows the number of pages \(\mathrm{p}\) the printer produces in \(\mathrm{t}\) minutes for three runs.

Time \(\mathrm{t}\) (minutes)Pages \(\mathrm{p}\)
0.618
1.030
1.751

Which equation expresses \(\mathrm{t}\) as a function of \(\mathrm{p}\)?

A
\(\mathrm{t = \frac{1}{60}p}\)
B
\(\mathrm{t = \frac{1}{30}p}\)
C
\(\mathrm{t = \frac{1}{18}p}\)
D
\(\mathrm{t = 30p}\)
Solution

1. TRANSLATE the problem information

  • Given information:
    • Printer produces pages at constant rate
    • Three data points: (0.6 min, 18 pages), (1.0 min, 30 pages), (1.7 min, 51 pages)
    • Need equation for t as function of p

2. INFER the mathematical relationship

  • Constant rate means the pages-per-minute ratio stays the same
  • I can calculate this rate using any data pair: \(\mathrm{rate = pages ÷ time}\)
  • Once I have the rate, I can write \(\mathrm{p = (rate) × t}\), then solve for t

3. Calculate the constant rate

  • Using first data pair: \(\mathrm{18 ÷ 0.6 = 30}\) pages per minute
  • Verify with second pair: \(\mathrm{30 ÷ 1.0 = 30}\) pages per minute ✓
  • Verify with third pair: \(\mathrm{51 ÷ 1.7 = 30}\) pages per minute ✓

4. INFER the direct relationship

  • Since rate = 30 pages per minute: \(\mathrm{p = 30t}\)
  • But the question asks for t as a function of p, so I need t = something involving p

5. SIMPLIFY to solve for t

  • Starting with: \(\mathrm{p = 30t}\)
  • Divide both sides by 30: \(\mathrm{p/30 = t}\)
  • Rewrite: \(\mathrm{t = p/30 = (1/30)p}\)

Answer: (B) \(\mathrm{t = (1/30)p}\)




Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students misread which variable should be the function of which. They see the table with t values and p values, and incorrectly assume they need p as a function of t (since that's how the data is arranged).

This leads them to write \(\mathrm{p = 30t}\) and stop there, selecting Choice (D) \(\mathrm{t = 30p}\) because they flip the relationship incorrectly.

Second Most Common Error:

Poor rate calculation (SIMPLIFY): Students make arithmetic errors when calculating \(\mathrm{18 ÷ 0.6}\), getting rates like \(\mathrm{18 × 0.6 = 10.8}\) instead of \(\mathrm{18 ÷ 0.6 = 30}\).

With an incorrect rate, they might get relationships like \(\mathrm{t = (1/10.8)p}\), leading them to guess among the fractional choices or select Choice (C) \(\mathrm{t = (1/18)p}\) if they confuse 18 pages with the rate.

The Bottom Line:

This problem tests whether students can correctly interpret 'constant rate' data and manipulate the resulting equation to express the requested variable as a function of the other. The key insight is recognizing that the question specifically asks for t as a function of p, not the other way around.

Answer Choices Explained
A
\(\mathrm{t = \frac{1}{60}p}\)
B
\(\mathrm{t = \frac{1}{30}p}\)
C
\(\mathrm{t = \frac{1}{18}p}\)
D
\(\mathrm{t = 30p}\)
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