Two pumps, A and B, deliver water at constant rates of 30 and 50 gallons per minute, respectively. Together they...

1. TRANSLATE the problem information

• Given information:
  • Pump A rate: 30 gallons per minute
  • Pump B rate: 50 gallons per minute
  • Total gallons delivered: 9,000
  • Pump B ran 60 minutes longer than Pump A
• What we need to find: How many gallons did Pump B deliver?

2. INFER the solution strategy

• Since we have two unknowns (time for each pump), we need to create a relationship between them
• Key insight: Use one variable for Pump A's time, then express Pump B's time in terms of that variable
• Strategy: Set up one equation using Rate × Time = Quantity for both pumps

3. TRANSLATE into mathematical expressions

• Let \(\mathrm{t}\) = minutes that Pump A ran
• Then Pump B ran for \(\mathrm{(t + 60)}\) minutes
• Total gallons equation: \(\mathrm{30t + 50(t + 60) = 9,000}\)

4. SIMPLIFY the equation to solve for t

• Distribute: \(\mathrm{30t + 50(t + 60) = 9,000}\)
• \(\mathrm{30t + 50t + 3,000 = 9,000}\)
• Combine like terms: \(\mathrm{80t + 3,000 = 9,000}\)
• Subtract 3,000: \(\mathrm{80t = 6,000}\)
• Divide by 80: \(\mathrm{t = 75}\)

5. INFER the final answer

• Pump A ran for 75 minutes
• Pump B ran for \(\mathrm{75 + 60 = 135}\) minutes
• Pump B delivered: \(\mathrm{50 \times 135 = 6,750}\) gallons

Answer: D (6,750)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students set up the wrong relationship between the pump times. They might write Pump A time as \(\mathrm{(t + 60)}\) and Pump B time as \(\mathrm{t}\), reversing which pump ran longer.

This leads to the equation: \(\mathrm{30(t + 60) + 50t = 9,000}\), which gives \(\mathrm{t = 75}\) for Pump B's time and \(\mathrm{t + 60 = 135}\) for Pump A's time. Then they calculate Pump B's gallons as \(\mathrm{50 \times 75 = 3,750}\), which doesn't match any answer choice. This leads to confusion and guessing.

Second Most Common Error:

Poor SIMPLIFY execution: Students correctly set up \(\mathrm{30t + 50(t + 60) = 9,000}\) but make distribution errors. They might forget to distribute the 50 to the 60, writing \(\mathrm{30t + 50t + 60 = 9,000}\) instead of \(\mathrm{30t + 50t + 3,000 = 9,000}\).

This gives them \(\mathrm{80t = 8,940}\), so \(\mathrm{t = 111.75}\). Then Pump B time becomes 171.75 minutes, and they calculate \(\mathrm{50 \times 171.75 = 8,587.5}\) gallons, which doesn't match any choice. This causes them to get stuck and randomly select an answer.

The Bottom Line:

This problem requires careful attention to which pump ran longer and systematic algebraic manipulation. Students who rush through the setup or make careless algebra errors will struggle to reach the correct answer.