A printer produces between 22 and 28 pages per minute, inclusive, at a steady rate. It is used to print...
GMAT Algebra : (Alg) Questions
A printer produces between \(22\) and \(28\) pages per minute, inclusive, at a steady rate. It is used to print a document containing \(660\) pages. Based on this information, which inequality represents the possible number of minutes, m, the printing could take?
\(22 + 660 \leq \mathrm{m} \leq 28 + 660\)
\(\frac{660}{28} \leq \mathrm{m} \leq \frac{660}{22}\)
\((22)(660) \leq \mathrm{m} \leq (28)(660)\)
\(\frac{660}{22} \leq \mathrm{m} \leq \frac{660}{28}\)
1. TRANSLATE the problem information
- Given information:
- Printer rate: between 22 and 28 pages per minute (inclusive)
- Document size: 660 pages
- Find: inequality for possible printing time m (in minutes)
- What this tells us: We need to connect rate, total pages, and time
2. INFER the fundamental relationship
- The key relationship is: Time = Total Pages ÷ Rate
- So our equation becomes: \(\mathrm{m = 660/r}\), where r is the printing rate
- We know that \(\mathrm{22 \leq r \leq 28}\)
3. INFER how rate affects time
- Since \(\mathrm{m = 660/r}\), rate and time have an inverse relationship
- When rate increases, time decreases (and vice versa)
- This means:
- Maximum rate (28 pages/min) → Minimum time = \(\mathrm{660/28}\)
- Minimum rate (22 pages/min) → Maximum time = \(\mathrm{660/22}\)
4. Write the final inequality
- Since time ranges from minimum to maximum:
- \(\mathrm{660/28 \leq m \leq 660/22}\)
Answer: B
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak TRANSLATE skill: Students may incorrectly set up the relationship between rate, pages, and time. Instead of recognizing that \(\mathrm{time = pages/rate}\), they might think \(\mathrm{time = rate \times pages}\) or \(\mathrm{rate = pages \times time}\), leading them to formulas like \(\mathrm{(22)(660) \leq m \leq (28)(660)}\).
This may lead them to select Choice C (\(\mathrm{(22)(660) \leq m \leq (28)(660)}\)).
Second Most Common Error:
Poor INFER reasoning about inverse relationships: Students correctly establish \(\mathrm{m = 660/r}\) but get confused about which rate gives the minimum time and which gives the maximum time. They might incorrectly think that higher rate means more time, setting up the inequality backwards as \(\mathrm{660/22 \leq m \leq 660/28}\).
This may lead them to select Choice D (\(\mathrm{660/22 \leq m \leq 660/28}\)).
The Bottom Line:
This problem challenges students to both translate a word problem into mathematical relationships AND understand how inverse relationships affect inequalities. The key insight is recognizing that when rate increases, time decreases, so the inequality bounds flip compared to the rate bounds.
\(22 + 660 \leq \mathrm{m} \leq 28 + 660\)
\(\frac{660}{28} \leq \mathrm{m} \leq \frac{660}{22}\)
\((22)(660) \leq \mathrm{m} \leq (28)(660)\)
\(\frac{660}{22} \leq \mathrm{m} \leq \frac{660}{28}\)