Question : There are two inlets A and B connected to a tank. A and B can fill the tank in 32 h and 28 h, respectively. If both the pipes are opened alternately for 1 h, starting with A, then in how much time (in hours, to the nearest integer) will the tank be filled?
Option 1: 22
Option 2: 30
Option 3: 36
Option 4: 24
Correct Answer: 30
Solution :
Given,
Pipe A can fill the tank in 32 hours
⇒ Part of the tank filled by pipe A in 1 hour = $\frac{1}{32}$
Pipe B can fill a tank in 28 hours
⇒ Part of the tank filled by B in 1 hour = $\frac{1}{28}$
The part filled by (A + B) in 2 hours = $\frac{1}{32} + \frac{1}{28}=\frac{7+8}{224}=\frac{15}{224}$
Now, $14 × \frac{15}{224} = \frac{210}{224}$
⇒ In 14 × 2 = 28 hours tank filled by $\frac{210}{224}$ part.
⇒ Remaining = $1 - \frac{210}{224}=\frac{224-210}{224}=\frac{14}{224}$ part
After completion of 28 hours pipe A will be open
To fill $\frac{1}{32}$ part pipe A takes time 1 hour
Remaining for 30th hours = $\frac{14}{224}-\frac{1}{32} = \frac{7}{224}$
⇒ To fill $\frac{7}{224}$ part pipe B will take time
⇒ $\frac{7}{224}\times28$ $\approx$ 1 hours
⇒ Total time taken to fill the tank = (28 + 1 + 1) hours = 30 hours
Hence, the correct answer is 30 hours.
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