Math Problem Statement

Two rectangular tanks, P and Q, are shown. Tank P is filled with water while Tank Q is empty. Both taps were turned on at the same time, and water flowed into both tanks at the same rate of 2.1 liters per minute. Find the time taken for the height of the water level in both tanks to be the same.

Solution

The problem involves two rectangular tanks, Tank P and Tank Q, with different initial water levels and identical water flow rates from taps. Here's the detailed setup and solution:

Problem Statement:

  • Tank P (30 cm x 30 cm base, 75 cm height) is initially filled with water.
  • Tank Q (50 cm x 30 cm base, 75 cm height) is initially empty.
  • Water flows into both tanks at the same rate of 2.1 liters per minute from taps.
  • We need to find the time taken for the water level in both tanks to be the same.

Solution:

  1. Calculate the base areas of both tanks:

    • Tank P: Base Area = 30 cm × 30 cm = 900 cm²
    • Tank Q: Base Area = 50 cm × 30 cm = 1500 cm²
  2. Convert the flow rate to cubic centimeters per minute:

    • Flow rate = 2.1 liters/min = 2100 cm³/min (since 1 liter = 1000 cm³).
  3. Determine the rate of height increase in each tank:

    • Height increase rate in Tank P: Rate=2100cm3/min900cm2=2.33cm/min\text{Rate} = \frac{2100 \, \text{cm}^3/\text{min}}{900 \, \text{cm}^2} = 2.33 \, \text{cm/min}
    • Height increase rate in Tank Q: Rate=2100cm3/min1500cm2=1.4cm/min\text{Rate} = \frac{2100 \, \text{cm}^3/\text{min}}{1500 \, \text{cm}^2} = 1.4 \, \text{cm/min}
  4. Set up equations for water heights over time:

    • Let hPh_P be the height of water in Tank P at time tt minutes.
    • Let hQh_Q be the height of water in Tank Q at time tt minutes.
    • Since Tank P was initially full, it starts at 75 cm and decreases as water is added to Tank Q: hP=752.33th_P = 75 - 2.33t
    • Tank Q starts empty, so its height is: hQ=1.4th_Q = 1.4t
  5. Set the heights equal to solve for tt: 752.33t=1.4t75 - 2.33t = 1.4t Simplify this equation: 75=3.73t75 = 3.73t t=753.7320.11minutest = \frac{75}{3.73} \approx 20.11 \, \text{minutes}

Answer:

The time taken for the height of the water level in both tanks to be the same is approximately 20.11 minutes.

Would you like any additional details or have any questions about the solution? Here are some related questions you might consider:

  1. How would the answer change if the flow rate was different for each tank?
  2. What happens if Tank Q started with some water in it?
  3. How can you adjust the calculations if the dimensions of the tanks change?
  4. What would the final water levels be if the flow continued for 30 minutes?
  5. How would evaporation or leaks affect the solution?

Tip: Always double-check units, especially when converting between liters and cubic centimeters.

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Math Problem Analysis

Mathematical Concepts

Algebra
Rate of Change
Geometry (Volume)

Formulas

Rate of height increase = Flow rate / Base area
Height equation for Tank P: h_P = Initial height - rate * t
Height equation for Tank Q: h_Q = rate * t

Theorems

Linear equations
Concept of rates and ratios

Suitable Grade Level

Grades 8-10