Math Problem Statement

6. A swimming pool was being refilled after being drained. Water flowed in from a garden hose at a rate of 30.3 liters per minute. How much does the water depth change in 12 hours? In 16 hours? 7. A gas station owner finds that for every penny increase in the price of gas, she sells 120 fewer gallons of gas a day. a. How much more or less gas will she sell if she raises the price by $0.10/gallon? Explain your reasoning or show work. b. How much more or less gas will she sell if she decreases the price by 25 cents per gallon? Explain your reasoning or show work.

Solution

Let's solve these problems one by one.

Problem 6

The pool is being refilled with water at a rate of 30.3 liters per minute. We need to find the change in water depth after 12 hours and 16 hours.

  1. Convert hours to minutes:

    • 12hours=12×60=720minutes12 \, \text{hours} = 12 \times 60 = 720 \, \text{minutes}
    • 16hours=16×60=960minutes16 \, \text{hours} = 16 \times 60 = 960 \, \text{minutes}
  2. Calculate total water added:

    • For 12 hours: 30.3liters/minute×720minutes=21,816liters30.3 \, \text{liters/minute} \times 720 \, \text{minutes} = 21,816 \, \text{liters}
    • For 16 hours: 30.3liters/minute×960minutes=29,088liters30.3 \, \text{liters/minute} \times 960 \, \text{minutes} = 29,088 \, \text{liters}
  3. The problem asks about the change in water depth. However, without knowing the area of the pool, we cannot convert this directly to depth. Assuming the question expects just the total volume, the answers would be:

    • 12 hours: 21,816 liters
    • 16 hours: 29,088 liters

Problem 7

This problem involves the effect of gas price changes on sales.

  • Given: For every 1 cent (or $0.01) increase in price, the station sells 120 fewer gallons.

Part (a)

  1. Increase the price by $0.10 (or 10 cents):
    • Decrease in gallons = 120×10=1,200gallons120 \times 10 = 1,200 \, \text{gallons}
    • Answer: She will sell 1,200 gallons less if she raises the price by $0.10.

Part (b)

  1. Decrease the price by $0.25 (or 25 cents):
    • Increase in gallons = 120×25=3,000gallons120 \times 25 = 3,000 \, \text{gallons}
    • Answer: She will sell 3,000 gallons more if she decreases the price by $0.25.

Would you like further details on any of these solutions, or have additional questions?

Related Questions

  1. How would the answer to Problem 6 change if we knew the pool's area?
  2. What would happen in Problem 7 if the price changes were different for every gallon sold?
  3. How can we calculate the change in water depth in Problem 6 using the pool dimensions?
  4. What other factors could affect gas sales besides price?
  5. Can we use similar methods to calculate the effect of discounts in retail settings?

Tip

When solving rate problems, always double-check your units (like hours to minutes) to avoid small errors that can significantly impact the final answer.

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Math Problem Analysis

Mathematical Concepts

Rate and Time Calculation
Proportion

Formulas

Total water added = Flow rate × Time
Gallons sold change = Rate of decrease × Price change

Theorems

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Suitable Grade Level

Grade 8-10