A sump is filled by three tankers with uniform flow. The first two tankers operating simultaneously fill the sump in the same time during which the sump is filled by the third tanker alone. The second tanker fills the sump 5 hours faster than the first tanker and 4 hours slower than the third tanker. The time required by the first tanker is:
15 hours
Solving the Tanker Filling Time Problem
This problem involves understanding the concept of work rates. When tankers fill a sump, their work rate is the reciprocal of the time they take to fill it alone. If a tanker takes \( t \) hours to fill the sump, its rate is \( \frac{1}{t} \) of the sump per hour.
Let's define the time taken by each tanker to fill the sump alone:
- Time taken by the first tanker = \( t_1 \) hours
- Time taken by the second tanker = \( t_2 \) hours
- Time taken by the third tanker = \( t_3 \) hours
Now, let's translate the given information into equations based on the relationships between their filling times:
- The second tanker fills the sump 5 hours faster than the first tanker:
This means the second tanker takes 5 hours less than the first.
\[ t_2 = t_1 - 5 \] - The second tanker fills the sump 4 hours slower than the third tanker:
This means the second tanker takes 4 hours more than the third.
\[ t_2 = t_3 + 4 \]
From these two equations, we can also relate \( t_1 \) and \( t_3 \):
\[ t_1 - 5 = t_3 + 4 \] \[ t_1 = t_3 + 9 \]Also, we can express \( t_3 \) in terms of \( t_2 \):
\[ t_3 = t_2 - 4 \]The main condition is that the first two tankers operating simultaneously fill the sump in the same time as the third tanker alone. In terms of rates, the sum of the rates of the first two tankers is equal to the rate of the third tanker.
- Rate of the first tanker = \( \frac{1}{t_1} \)
- Rate of the second tanker = \( \frac{1}{t_2} \)
- Rate of the third tanker = \( \frac{1}{t_3} \)
The combined rate equation is:
\[ \frac{1}{t_1} + \frac{1}{t_2} = \frac{1}{t_3} \]We want to find \( t_1 \). Let's express \( t_2 \) and \( t_3 \) in terms of \( t_1 \). We already have \( t_2 = t_1 - 5 \). From \( t_1 = t_3 + 9 \), we get \( t_3 = t_1 - 9 \).
Substitute these into the combined rate equation:
\[ \frac{1}{t_1} + \frac{1}{t_1 - 5} = \frac{1}{t_1 - 9} \]To solve this equation, find a common denominator on the left side:
\[ \frac{(t_1 - 5) + t_1}{t_1(t_1 - 5)} = \frac{1}{t_1 - 9} \] \[ \frac{2t_1 - 5}{t_1^2 - 5t_1} = \frac{1}{t_1 - 9} \]Now, cross-multiply:
\[ (2t_1 - 5)(t_1 - 9) = 1 \times (t_1^2 - 5t_1) \] \[ 2t_1(t_1 - 9) - 5(t_1 - 9) = t_1^2 - 5t_1 \] \[ 2t_1^2 - 18t_1 - 5t_1 + 45 = t_1^2 - 5t_1 \] \[ 2t_1^2 - 23t_1 + 45 = t_1^2 - 5t_1 \]Move all terms to one side to form a quadratic equation:
\[ 2t_1^2 - t_1^2 - 23t_1 + 5t_1 + 45 = 0 \] \[ t_1^2 - 18t_1 + 45 = 0 \]Now we solve the quadratic equation \( t_1^2 - 18t_1 + 45 = 0 \). We can factor this equation. We need two numbers that multiply to 45 and add up to -18. These numbers are -3 and -15.
\[ (t_1 - 3)(t_1 - 15) = 0 \]This gives two possible solutions for \( t_1 \):
\[ t_1 = 3 \quad \text{or} \quad t_1 = 15 \]We must check these solutions in the context of the problem. The times taken by the tankers must be positive. Also, for \( t_2 = t_1 - 5 \) and \( t_3 = t_1 - 9 \) to be positive, \( t_1 \) must be greater than 5 and greater than 9.
- If \( t_1 = 3 \) hours:
\( t_2 = 3 - 5 = -2 \) hours. Time cannot be negative, so \( t_1 = 3 \) is not a valid solution.
- If \( t_1 = 15 \) hours:
\( t_2 = 15 - 5 = 10 \) hours. This is positive and valid.
\( t_3 = 15 - 9 = 6 \) hours. This is positive and valid.
Let's verify if \( t_1 = 15, t_2 = 10, t_3 = 6 \) satisfies the original combined rate equation \( \frac{1}{t_1} + \frac{1}{t_2} = \frac{1}{t_3} \):
\[ \frac{1}{15} + \frac{1}{10} = \frac{2}{30} + \frac{3}{30} = \frac{5}{30} = \frac{1}{6} \] \[ \frac{1}{t_3} = \frac{1}{6} \]Since \( \frac{1}{6} = \frac{1}{6} \), the values \( t_1 = 15, t_2 = 10, t_3 = 6 \) satisfy all conditions of the problem.
Therefore, the time required by the first tanker is 15 hours.
Revision Table: Key Concepts
| Concept | Explanation | Formula/Relationship |
|---|---|---|
| Work Rate | The amount of work done per unit of time. For filling a sump, it's the fraction of the sump filled per hour. | Rate = \( \frac{1}{\text{Time taken}} \) |
| Combined Rate (Working Together) | When multiple entities work together, their rates add up to find the total rate. | RateTotal = Rate1 + Rate2 + ... |
| Solving Word Problems | Translate the given information into algebraic equations and solve for the unknown variable(s). Always check solutions in the context of the problem. | Set up equations, substitute, solve (often leads to linear or quadratic equations). |
Additional Information: Work and Time Problems
Work and time problems are common in quantitative aptitude. They often involve individuals or machines working at different rates to complete a task. The key idea is that the total work done is equal to the rate of work multiplied by the time taken.
If a person completes a task in \( T \) hours, their rate of work is \( \frac{1}{T} \) of the task per hour. In this problem, the 'task' is filling the sump.
When people or machines work together, their individual rates are added to find the combined rate. If person A has rate \( R_A \) and person B has rate \( R_B \), their combined rate is \( R_A + R_B \). If they work together for time \( t \), the total work done is \( (R_A + R_B) \times t \).
In filling problems like this one, the 'work' is filling 1 whole sump. So, if the combined rate is \( R_{combined} \) and the time taken to fill the sump together is \( T_{combined} \), then \( R_{combined} \times T_{combined} = 1 \).
Quadratic equations often arise in work and time problems when relationships between times lead to squared terms. Always remember to check if the solutions to the quadratic equation are meaningful in the context of the problem (e.g., time must be positive).
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