Question

Three pipes A, B, and C can fill a tank in 6 hrs. After working at it together for 2 hrs, C is closed and A and B can fill the remaining part in 7 hrs. The total number of hours taken by C alone to fill the tank is

• A. 14
• B. 12
• C. 11
• D. 10
• E. 13

Hint:

Use the work-rate equation to solve time-based problems. Divide the problem into parts and solve for each.

Answer 1

The Correct Option is A

1. Define Rates:
   • Let \(R_A\), \(R_B\), and \(R_C\) be the rates at which pipes A, B, and C fill the tank, respectively (measured in fraction of tank filled per hour).
   • The time taken for a pipe alone is the reciprocal of its rate (e.g., time for C alone is \(1/R_C\)).
2. Information from the First Statement: "Three pipes A, B, and C can fill a tank in 6 hrs."
   • This means their combined rate fills 1 tank in 6 hours.
   • Combined rate = \(R_A + R_B + R_C\).
   • Equation: \((R_A + R_B + R_C) \times 6 = 1\) (where 1 represents the whole tank).
   • Therefore, \(R_A + R_B + R_C = \frac{1}{6}\) (Equation 1).
3. Information from the Second Statement: "After working at it together for 2 hrs, C is closed and A and B can fill the remaining part in 7 hrs."
   • Work done in the first 2 hours: Pipes A, B, and C work together.

     \[ \text{Work} = \text{Combined Rate} \times \text{Time} = (R_A + R_B + R_C) \times 2 \]

     Using Equation 1, Work = \(\frac{1}{6} \times 2 = \frac{2}{6} = \frac{1}{3}\) of the tank.

   • Remaining work:

     \[ \text{Fraction of tank remaining} = 1 - \frac{1}{3} = \frac{2}{3} \]

   • Work done by A and B: Pipes A and B fill the remaining \(\frac{2}{3}\) of the tank in 7 hours.

     Combined rate of A and B = \(R_A + R_B\).

     \[ \text{Equation:} (R_A + R_B) \times 7 = \frac{2}{3} \]

   • Therefore, \(R_A + R_B = \frac{2}{3 \times 7} = \frac{2}{21}\) (Equation 2).
4. Find the Rate of C (\(R_C\)):
   • We have Equation 1: \(R_A + R_B + R_C = \frac{1}{6}\)
   • We have Equation 2: \(R_A + R_B = \frac{2}{21}\)
   • Substitute the value of \((R_A + R_B)\) from Equation 2 into Equation 1:

     \[ \frac{2}{21} + R_C = \frac{1}{6} \]

   • Solve for \(R_C\):

     \[ R_C = \frac{1}{6} - \frac{2}{21} \]

   • Find a common denominator (42):

     \[ R_C = \frac{7}{42} - \frac{4}{42} = \frac{3}{42} = \frac{1}{14} \]

     So, pipe C fills \(\frac{1}{14}\) of the tank per hour.

5. Calculate Time for C Alone:
   • The time taken by C alone is the reciprocal of its rate:

     \[ \text{Time} = \frac{1}{R_C} = \frac{1}{(1/14)} = 14 \text{ hours.} \]

6. Compare with Options: The calculated time is 14 hours.

The correct option is (a) 14.

Answer 2

Let the total work (filling the tank) be 1 unit.

Let the rates of A, B, and C be $a$, $b$, and $c$ respectively.
Given:

• A + B + C together can fill the tank in 6 hours: $$ a + b + c = \frac{1}{6} $$
• They all work together for 2 hours: $$ \text{Work done in 2 hrs} = 2(a + b + c) = 2 \cdot \frac{1}{6} = \frac{1}{3} $$
• The remaining $\frac{2}{3}$ of the tank is filled by A and B in 7 hours: $$ 7(a + b) = \frac{2}{3} \Rightarrow a + b = \frac{2}{21} $$

Now substitute into the first equation: $$ a + b + c = \frac{1}{6} $$ $$ \Rightarrow \frac{2}{21} + c = \frac{1}{6} $$ $$ \Rightarrow c = \frac{1}{6} - \frac{2}{21} = \frac{7 - 4}{42} = \frac{3}{42} = \frac{1}{14} $$ Therefore, C alone can fill the tank in 14 hours.
Answer: (a) 14

Answer 3

Let the rate at which A, B, and C fill the tank be \( A \), \( B \), and \( C \) respectively. We know that: \[ \text{Time taken by A and B together to fill the tank} = 7 \text{ hrs} \] Total work done is 1 tank, and the combined rate of A and B can be calculated from the work done in 7 hrs. After working together for 2 hrs, the fraction of the tank filled is: \[ \text{Fraction filled in 2 hrs} = 2(A + B + C) \] After closing C, the remaining part is filled by A and B alone in 7 hrs. Solving for C: \[ C = 14 \]