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A tank can be filled by pipe A in 2 hours and pipe B in 6 hours. At 10 A.M. pipe A was opened. At what time will the tank be filled if pipe B is opened at 11 A.M. ?
Answer & Solution
Correct Answer:
Option
C
Pipe A will fill 3 units till 11 A.M. Remaining capacity
= 6 - 3
= 3 units
Now both pipes will fill the tank in
$$\frac{{{\text{Total Capacity}}}}{{{\text{Efficiency }}}} = \frac{3}{{\left( {3 + 1} \right)}} = \frac{3}{4}{\text{ hours}}$$
So, $$\left( {11 + \frac{3}{4}} \right)$$ A.M., tank will be filled = 11.45 A.M.
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LoginA can fill in 1 hr=1/2 part of the tank
So remaining 1/2 part
2/3 part filled by (A+B) in 1 hr
So, remaining 1/2 part filled by (A+B)= (3/2)*(1/2)
=3/4 hr or 45 mins
So total 1 hr 45 mins.Starting from 10 am the tank will be filled within 11:45 am