Question

Three pipes, A, B, and C are capable of filling a tank in 6, 8, and 12 hours, respectively. When all three pipes are opened together, they operate for 2 hours before pipe C is closed. How much additional time will it take to completely fill the tank after that?

• A. 6/7 hours
• B. 2/6 hours
• C. 3/5 hours
• D. 9/2 hours

Answer 1

Answer:

A. 6/7 hours

Work rates per hour

• $A = ( \frac{1}{6} )$

• $B = ( \frac{1}{8} )$

• $C = ( \frac{1}{12} )$

• Combined work (all three)

  $\frac{1}{6} + \frac{1}{8} + \frac{1}{12}$
  $= \frac{4 + 3 + 2}{24}$
  $= \frac{9}{24} = \frac{3}{8}$
  Work done in 2 hours

  $2 \times \frac{3}{8} = \frac{6}{8} = \frac{3}{4}$
  So, remaining work:
  $1 - \frac{3}{4} = \frac{1}{4}$
  

  After closing pipe C (only A and B)

  $\frac{1}{6} + \frac{1}{8} = \frac{4 + 3}{24} = \frac{7}{24}$
  

  Time to complete remaining work

  $\text{Time} = \frac{\frac{1}{4}}{\frac{7}{24}} = \frac{1}{4} \times \frac{24}{7} = \frac{6}{7}$
  

  Final Answer:

  $\boxed{\frac{6}{7} \text{ hours}}$