dave13
Tashin Azad
A water reservoir has two inlets and one outlet. Through the inlets water reservoir can be filled in 3 hours and 3 hours 45 minutes respectively. Water reservoir can be empties completely in 1 hour by the outlet. If the two inlets are opened at 1:00 pm and 2:00 pm respectively and the outlet at 3:00 pm then the water reservoir will be emptied at?
a) 3:55 pm
b) 5:00 pm
c) 5:20 pm
d) 5:30 pm
e) 5:45 pm
let inlets be A and B
A rate is \(\frac{1}{180}\)(i.e 3 hours is 180 minutes)
B rate is \(\frac{1}{225}\) i.e. (3 hrs and 45 min is 225 minutes)
Combined rate:\(\frac{1}{180} + \frac{1}{225} = \frac{9}{900}i.e. \frac{1}{100}\)
So total work units is 100 complete a job, BUT since A worked alone for one hour i.e 60 min, we need to subtract 100-60 =40 so 40 work units are left
let X be
total time required when B is activated while A keeps working hence x-40
\(\frac{1}{180} + \frac{1}{100} *x-40 = 1\)
\(5+9(x-40) = 900\)
\(5+9x-360 =900\)
\(9x= 1255\)
\(x = 139.4\) (round to 140)
So, 140 minutes is 2 hours and 20 minutes
if work started at 1 pm then it will be complted at 3.20 , ans by 4.20 emptied
Bunuel,
pushpitkc,
chetan2u VeritasKarishma,
generis whats wrong with my solution

p.s. I followed the method of similar problem here
https://gmatclub.com/forum/two-taps-can ... l#p2142371Hi
dave13 , that method doesn't really apply; we don't need to find your \(x\) for the first part. Time is given. In the second part, all three pipes work. A and B don't stop.
Generally, when minutes' totals are large, I switch to hours. Finally, I think this math is easier broken into stages.
(1) How much of reservoir is filled in 2 hours by A and B?
Pipe A rate: \(\frac{1R}{3hrs}=\frac{1}{3}\)
Pipe B rate* = \(\frac{4}{15}\)
Combined rate of A and B: \((\frac{1}{3}+\frac{4}{15})=\frac{9}{15}=\frac{3}{5}\)
Pipe A fills for 1 hour alone. Pipes A and B work together for 1 hour.
Total filled,
\(r*t=W\):
\(W_1=(R_{A}*t_1)+(R_{(A+B)}*t_2)=\)
\((\frac{1}{3}*1)+(\frac{3}{5}*1)=(\frac{1}{3}+\frac{3}{5})=\) \(\frac{14}{15}\) of R is filled
(2) Time needed to empty that portion?
C drains faster than A and B fill. Subtract combined rate of A and B from rate of C. The result is the rate at which the reservoir empties.
Let T = time needed to drain \(\frac{14}{15}\) of pool
Rate of C = \(\frac{1R}{1hr}\)
\((R_{C}-R_{(A+B)})*T=\frac{14}{15}\)
\((1-\frac{3}{5})*T=\frac{14}{15}\)
\(\frac{2}{5}*T=\frac{14}{15}\)
\(T=(\frac{14}{15}*\frac{5}{2})=\frac{7}{3}\) hrs
\(2\frac{1}{3}\)hrs = \(2\) hrs, \(20\) minutes
A, B, and C work simultaneously at 3 p.m.
2 hours and 20 minutes later is 5:20 p.m.
Answer C
Hope that helps
*3 hours, 45 mins =\(3\frac{3}{4}=\frac{15}{4}\) hours
B, rate:\(\frac{1R}{(\frac{15}{4}hrs)}=(1R*\frac{4}{15}hrs)=\frac{4R}{15hrs}=\frac{4}{15}\)