Question

Four pipes \(A,B,C,D\) can fill a tank in \(15, 20, 30,\) and \(60\) hours respectively. Pipe \(A\) is opened at 4 a.m., \(B\) at 5 a.m., \(C\) at 6 a.m., and \(D\) at 7 a.m. When is the tank completely filled?

• A. 9:30 a.m.
• B. 10:00 a.m.
• C. 10:30 a.m.
• D. 11:00 a.m.

Hint:

Handle staggered openings by adding work done over each interval, then use the final combined rate for the remainder.

Answer 1

The Correct Option is D

Step 1: Write hourly rates.
\(A=\frac{1}{15},\ B=\frac{1}{20},\ C=\frac{1}{30},\ D=\frac{1}{60}\) tank/hour.
Step 2: Work done before 7 a.m.
From 4–5 a.m.: only \(A\Rightarrow \frac{1}{15}\).
From 5–6 a.m.: \(A+B\Rightarrow \frac{1}{15}+\frac{1}{20}=\frac{7}{60}\).
From 6–7 a.m.: \(A+B+C\Rightarrow \frac{1}{15}+\frac{1}{20}+\frac{1}{30}=\frac{9}{60}=\frac{3}{20}\).
Total by 7 a.m.: \[ \frac{1}{15}+\frac{7}{60}+\frac{9}{60}=\frac{4+7+9}{60}=\frac{20}{60}=\frac{1}{3}. \] Step 3: Work after 7 a.m. with all four pipes.
Combined rate \(A+B+C+D=\frac{1}{15}+\frac{1}{20}+\frac{1}{30}+\frac{1}{60}=\frac{10}{60}=\frac{1}{6}\) tank/hour.
Remaining work \(=1-\frac{1}{3}=\frac{2}{3}\).
Time needed \(=\dfrac{\frac{2}{3}}{\frac{1}{6}}=\frac{2}{3}\times 6=4\) hours.
Step 4: Finish time.
7 a.m. \(+\,4\) hours \(=\) \(\boxed{\text{11:00 a.m.}}\)