Step 1: Convert each pipe to a per–minute rate.
If a pipe fills (or drains) a tank in \(t\) minutes, its rate = \(\tfrac{1}{t}\) tank/min.
A: \(\tfrac{1}{4}\), B: \(\tfrac{1}{10}\), C: \(\tfrac{1}{12}\), D: \(\tfrac{1}{20}\).
Step 2: Check each option’s net rate and water in 30 minutes.
Take \(\mathrm{LCM}(4,10,12,20)=60\) for clean arithmetic. \[ \tfrac{1}{4}=\tfrac{15}{60},\quad \tfrac{1}{10}=\tfrac{6}{60},\quad \tfrac{1}{12}=\tfrac{5}{60},\quad \tfrac{1}{20}=\tfrac{3}{60}. \] Option A: Net rate = \(\tfrac{15}{60}-(\tfrac{6}{60}+\tfrac{5}{60}+\tfrac{3}{60})=\tfrac{15-14}{60}=\tfrac{1}{60}\) tank/min.
In 30 minutes: \(30\times\tfrac{1}{60}=\tfrac{1}{2}\) tank \(\;\Rightarrow\;\) matches. ✔
Option B: Net = \((\tfrac{6+5+3}{60})-\tfrac{15}{60}=-\tfrac{1}{60}\).
Water level decreases overall (from empty it can’t rise to \(\tfrac{1}{2}\)). ✗
Option C: Net = \(\tfrac{6+5}{60}-\tfrac{15+3}{60}=-\tfrac{7}{60}\).
Negative rate \(\;\Rightarrow\;\) not possible. ✗
Option D: Net = \(\tfrac{15+3}{60}-\tfrac{6+5}{60}=\tfrac{7}{60}\).
In 30 minutes: \(30\times\tfrac{7}{60}=\tfrac{7}{2}=3.5\) tanks \(\;\Rightarrow\;\) far more than half (would overflow). ✗
Step 3: Conclusion.
Only Option A yields exactly half a tank after 30 minutes.
Final Answer:
\[ \boxed{\text{A. Pipe A filled; B, C, D drained}} \]
A | B | C | D | Average |
---|---|---|---|---|
3 | 4 | 4 | ? | 4 |
3 | ? | 5 | ? | 4 |
? | 3 | 3 | ? | 4 |
? | ? | ? | ? | 4.25 |
4 | 4 | 4 | 4.25 |