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. ?
A. 12.45 A.M.
B. 5 P.M.
C. 11.45 A.M.
D. 12 P.M.
Answer: Option C
Solution(By Examveda Team)
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.
This Question Belongs to Arithmetic Ability >> Pipes And Cistern
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Related Questions on Pipes and Cistern
A. $$\frac{5}{{11}}$$
B. $$\frac{6}{{11}}$$
C. $$\frac{7}{{11}}$$
D. $$\frac{8}{{11}}$$
A. $$1\frac{{13}}{{17}}$$ hours
B. $$2\frac{8}{{11}}$$ hours
C. $$3\frac{9}{{17}}$$ hours
D. $$4\frac{1}{2}$$ hours
A. $$4\frac{1}{3}$$ hours
B. 7 hours
C. 8 hours
D. 14 hours
(A+B) can fill in 1 hr= (1/2)+(1/6)=2/3 part
A 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