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If two pipes function simultaneously, a tank is filled in 12 hours. One pipe fills the tank 10 hours faster than the other. How many hours does the faster pipe alone take to fill the tank?
- a)12 hrs
- b)20 hrs
- c)18 hrs
- d)15 hrs
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
If two pipes function simultaneously, a tank is filled in 12 hours. On...
Using Rule 1,
If the slower pipe fills the tank in x hours, then
Required time = 30 - 10 = 20 hours
Required time = 30 - 10 = 20 hours
Most Upvoted Answer
If two pipes function simultaneously, a tank is filled in 12 hours. On...
Understanding the Problem
To tackle the problem, let's define the variables and the relationships between the pipes.
- Let the time taken by the slower pipe to fill the tank be \( x \) hours.
- Then, the faster pipe will take \( x - 10 \) hours to fill the tank.
Working Together
When both pipes work together, they fill the tank in 12 hours. The rates of filling for each pipe are:
- Slower pipe: \( \frac{1}{x} \) (tank per hour)
- Faster pipe: \( \frac{1}{x - 10} \) (tank per hour)
The combined rate of both pipes working together is:
\[ \frac{1}{x} + \frac{1}{x - 10} = \frac{1}{12} \]
Setting Up the Equation
To find \( x \), we can combine the fractions:
\[ \frac{(x - 10) + x}{x(x - 10)} = \frac{1}{12} \]
This simplifies to:
\[ \frac{2x - 10}{x^2 - 10x} = \frac{1}{12} \]
Cross-multiplying gives:
\[ 12(2x - 10) = x^2 - 10x \]
Expanding and rearranging results in:
\[ x^2 - 34x + 120 = 0 \]
Solving the Quadratic Equation
Now, we can solve this quadratic equation using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \( a = 1, b = -34, c = 120 \):
- Discriminant: \( b^2 - 4ac = 1156 - 480 = 676 \)
- Roots:
\[ x = \frac{34 \pm 26}{2} \]
Calculating gives:
- \( x = 30 \) (slower pipe)
- \( x = 4 \) (not valid since it must be greater than 10)
Therefore, the slower pipe takes 30 hours, and the faster pipe takes:
\[ x - 10 = 30 - 10 = 20 \text{ hours} \]
Conclusion
The faster pipe alone takes 20 hours to fill the tank, confirming the correct answer is option B.
To tackle the problem, let's define the variables and the relationships between the pipes.
- Let the time taken by the slower pipe to fill the tank be \( x \) hours.
- Then, the faster pipe will take \( x - 10 \) hours to fill the tank.
Working Together
When both pipes work together, they fill the tank in 12 hours. The rates of filling for each pipe are:
- Slower pipe: \( \frac{1}{x} \) (tank per hour)
- Faster pipe: \( \frac{1}{x - 10} \) (tank per hour)
The combined rate of both pipes working together is:
\[ \frac{1}{x} + \frac{1}{x - 10} = \frac{1}{12} \]
Setting Up the Equation
To find \( x \), we can combine the fractions:
\[ \frac{(x - 10) + x}{x(x - 10)} = \frac{1}{12} \]
This simplifies to:
\[ \frac{2x - 10}{x^2 - 10x} = \frac{1}{12} \]
Cross-multiplying gives:
\[ 12(2x - 10) = x^2 - 10x \]
Expanding and rearranging results in:
\[ x^2 - 34x + 120 = 0 \]
Solving the Quadratic Equation
Now, we can solve this quadratic equation using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \( a = 1, b = -34, c = 120 \):
- Discriminant: \( b^2 - 4ac = 1156 - 480 = 676 \)
- Roots:
\[ x = \frac{34 \pm 26}{2} \]
Calculating gives:
- \( x = 30 \) (slower pipe)
- \( x = 4 \) (not valid since it must be greater than 10)
Therefore, the slower pipe takes 30 hours, and the faster pipe takes:
\[ x - 10 = 30 - 10 = 20 \text{ hours} \]
Conclusion
The faster pipe alone takes 20 hours to fill the tank, confirming the correct answer is option B.
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