Math Problem Statement

Solve the following problems using two variables : Two taps A and B can together fill a swimming pool in 15 days . Taps A and B are kept open for 12 days and then tap B is closed . It takes another 8 days for the pool to be filled . How many days does each tap require to fill the pool ?

Solution

Let the rate at which tap A fills the pool be 1x\frac{1}{x} pools per day, and the rate at which tap B fills the pool be 1y\frac{1}{y} pools per day, where xx and yy are the number of days it takes for taps A and B, respectively, to fill the pool individually.

Step 1: Using the information that both taps together fill the pool in 15 days:

The combined rate of both taps is 1x+1y\frac{1}{x} + \frac{1}{y}, and they fill the pool in 15 days, so: 1x+1y=115\frac{1}{x} + \frac{1}{y} = \frac{1}{15} This is the first equation.

Step 2: Using the information that both taps are open for 12 days and then only tap A is open for another 8 days:

  • For the first 12 days, both taps are open, so they fill (1x+1y)×12\left( \frac{1}{x} + \frac{1}{y} \right) \times 12 of the pool.
  • For the remaining 8 days, only tap A is open, so it fills 1x×8\frac{1}{x} \times 8 of the pool.

Together, the pool is fully filled, so: 12(1x+1y)+8(1x)=112\left(\frac{1}{x} + \frac{1}{y}\right) + 8\left(\frac{1}{x}\right) = 1 This is the second equation.

Step 3: Solving the system of equations:

From the first equation: 1x+1y=115\frac{1}{x} + \frac{1}{y} = \frac{1}{15} Substitute 1x+1y=115\frac{1}{x} + \frac{1}{y} = \frac{1}{15} into the second equation: 12×115+8×1x=112 \times \frac{1}{15} + 8 \times \frac{1}{x} = 1 Simplifying: 1215+8x=1\frac{12}{15} + \frac{8}{x} = 1 45+8x=1\frac{4}{5} + \frac{8}{x} = 1 Subtract 45\frac{4}{5} from both sides: 8x=145=15\frac{8}{x} = 1 - \frac{4}{5} = \frac{1}{5} Solve for xx: x=815=40x = \frac{8}{\frac{1}{5}} = 40

Step 4: Solving for yy:

Substitute x=40x = 40 into the first equation: 140+1y=115\frac{1}{40} + \frac{1}{y} = \frac{1}{15} Subtract 140\frac{1}{40} from both sides: 1y=115140\frac{1}{y} = \frac{1}{15} - \frac{1}{40} Find a common denominator: 1y=81203120=5120=124\frac{1}{y} = \frac{8}{120} - \frac{3}{120} = \frac{5}{120} = \frac{1}{24} Solve for yy: y=24y = 24

Final Answer:

  • Tap A can fill the pool in 40 days.
  • Tap B can fill the pool in 24 days.

Would you like a more detailed explanation or any clarifications?

Here are 5 related questions to expand on this problem:

  1. What happens if the rates of taps A and B are altered? How will it affect the solution?
  2. How would the solution change if the taps were left open for different time intervals?
  3. What is the general formula for solving such problems with two variables?
  4. How can we modify this problem to include a third tap?
  5. How does solving with two variables compare to solving similar problems with just one variable?

Tip: For rate problems, always break down the work done by each part and use algebraic equations to track their contributions over time.

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Math Problem Analysis

Mathematical Concepts

Work Problems
Algebra
Systems of Equations

Formulas

Rate of work = 1/time
Combined rate: 1/x + 1/y = 1/total time
Work completed = rate × time

Theorems

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Suitable Grade Level

Grades 8-10