Question

A cistern has two pipes A and B which can fill the cistern independently in 18 hours and 24 hours respectively. If they are opened alternately with one hour duration starting with A, then the time required to fill the empty cistern (in hours) is:

• A. \(18\)
• B. \(20\frac{1}{2}\)
• C. \(22\frac{1}{4}\)
• D. \(23\frac{1}{2}\)

Hint:

For alternate pipe problems, first calculate the work done in one complete cycle and then proceed cycle by cycle.

Answer 1

The Correct Option is B

Step 1: Find hourly filling rates. Pipe A: \[ \frac1{18} \] Pipe B: \[ \frac1{24} \]

Step 2: Work completed in 2 hours. One cycle consists of: \[ A+B \] Work done: \[ \frac1{18}+\frac1{24} = \frac{7}{72} \]

Step 3: Determine work after 20 hours. 20 hours means 10 complete cycles. \[ 10\times\frac7{72} = \frac{70}{72} = \frac{35}{36} \] Remaining work: \[ 1-\frac{35}{36} = \frac1{36} \]

Step 4: Use the next pipe A. A fills at \[ \frac1{18} \] Time required for remaining work: \[ \frac{\frac1{36}}{\frac1{18}} = \frac12 \] hour. Hence total time: \[ 20+\frac12 = 20\frac12 \] Therefore, \[ \boxed{20\frac12\text{ hours}} \]