Question : Two taps A and B can fill a tank in 10 hours and 12 hours respectively. If the two taps are opened at 10 AM, at what time (in PM) should tap A be closed to fill the tank at exactly 4 PM?
Option 1: 2:00
Option 2: 3:00
Option 3: 1:00
Option 4: 1:30
Correct Answer: 3:00
Solution : Tap A's hourly work = $\frac{1}{10}$ Tap B's hourly work = $\frac{1}{12}$ (Tap A + Tap B)'s hourly work $=\frac{1}{10}+\frac{1}{12}=\frac{11}{60}$ Let tap A be closed after $x$ hours. Then, part filled by (A + B) in $x$ hours + part filled by B in $(6 -x)$ hours $= 1$ ⇒ $\frac{11x}{60}+\frac{(6–x)}{12}=1$ ⇒ $\frac{11x+30–5x}{60}=1$ ⇒ $6x+30=60$ ⇒ $6x=30$ $\therefore x=5$ After 5 hours, Tap A should be closed i.e, 10:00 AM + 5 hours = 3:00 PM Hence, the correct answer is 3:00.
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