dave13 wrote:

Tashin Azad wrote:

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:\(\frac{1R}{(\frac{15}{4}hrs)}=(1R*\frac{4}{15}hrs)=\frac{4R}{15hrs}=\frac{4}{15}\)
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