A pump removes water from a tank at a rate that can vary during the day. For any 10-minute interval...

1. TRANSLATE the problem information

• Given information:
  • Least amount removed in 10 minutes: 85 liters
  • Greatest amount removed in 10 minutes: 130 liters
  • \(\mathrm{r}\) = average rate in liters per minute during a 10-minute interval
  • Need inequality that's true for ALL possible values of \(\mathrm{r}\)

2. INFER the relationship between amounts and rates

• Since \(\mathrm{rate = amount ÷ time}\), I need to find rates for the extreme cases
• The extreme amounts will give me the extreme rates
• All other rates must fall between these extremes

3. Calculate the minimum possible rate

• Minimum rate = Least amount ÷ Time
• Minimum rate = \(\mathrm{85 ÷ 10 = 8.5}\) liters per minute

4. Calculate the maximum possible rate

• Maximum rate = Greatest amount ÷ Time
• Maximum rate = \(\mathrm{130 ÷ 10 = 13}\) liters per minute

5. APPLY CONSTRAINTS to establish the inequality

• Since 8.5 and 13 are the extreme rates, all possible values of \(\mathrm{r}\) must satisfy:
• \(\mathrm{8.5 \leq r \leq 13}\)

Answer: B. \(\mathrm{8.5 \leq r \leq 13}\)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students may misunderstand what \(\mathrm{r}\) represents or fail to connect "amount removed" with "rate of removal." They might try to work directly with the given amounts (85 and 130) without converting to rates.

This leads to confusion about what the inequality should represent, causing them to select Choice A (\(\mathrm{0 \leq r \leq 8.5}\)) by incorrectly thinking 8.5 is an upper bound rather than a lower bound.

Second Most Common Error:

Incomplete INFER reasoning: Students correctly calculate one rate (often just \(\mathrm{85 ÷ 10 = 8.5}\)) but fail to recognize they need both extreme rates to establish the complete range. They may think \(\mathrm{r}\) must be at least 8.5 but miss the upper bound.

This may lead them to select Choice D (\(\mathrm{r \geq 8.5}\)) because they only identified the lower constraint.

The Bottom Line:

This problem requires students to understand that rate problems involve dividing amount by time, and that extreme input values (amounts) produce extreme output values (rates) that define the complete range of possibilities.