Rate of inlet pipe = 1/4
rate of outlet pipe = 1/6

effective rate = 1/4-1/6 = 1/12.
time for total job = 12 hrs but 1/4 of the job already done when pipe was on at 8am,
hence time required to fill the tank = (3/4)*12 = 9hrs.
Option C 18:00

Assume the total volume of the pool = 24 lts.

Rate of pipe A = 6 lts/hour
Rate of pipe B = 4 lts/hour

Water in the pool at 9 AM = 6 lts.
Rate of filling after 9 AM = 2 lts/hour

Hence the time needed = (24 - 6) /2 = 9 more hours
The pool will be full after 10 hours or at 18:00 hours.

Correct Option: C

IMO C

Time to fill = Total Job/Combined Rate

Combined Rate = 1/4 - 1/6

Total Job = 3/4 ( As 1/4 is already filled in 1 hour gap)

Thus Time = 9 hours, and ans C.

Please correct if the method is wrong.

Pipe A fills the pool in 4 hrs.
1 hour's work : 1/4

Pipe B empties the pool in 6 hrs.
1 hour's work : 1/6

Together if they work, 1 hour's work = 1/4 -1/6 = 1/12

Given : Pipe A started at 8:00 a.m and Pipe B at 9:00 a.m
Pool filled after 1 hour by Pipe A : 1/4 or 3/12

After 9:00 a.m
Pool filled after 1 hour with both the pipes on : 1/12
Pool filled after 9 hours with both pipes on : 9/12

Pool filled in 1 hour + Pool filled in 9 hours = 3/12 +9/12 =1

Therefore, it takes 10 hrs to fill the pool

As Pipe A started at 8:00 a.m, pool is full at 18:00 hrs

Answer is C.
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Let capacity of the tank be 12 units..

Efficiency of A = +3 units/hr
Efficiency of B = -2 units/hr
Combined efficiency of A + B is 1 unit/hr



In 1 hour pipe A fills up 3 units and 9 units is left
Time required is \(\frac{9}{1} = 9\) hours more

Thus, time required to fill the take is 9 hours after 9 AM , ie 18:00 hours , answer will be (C)
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This is an easy question if done carefully.

so in the first hour, 1/4 of pool is full.
from the next hour (9 am onwards), let it take x hours to fill the pool.

So,

1/4 + (1/4 - 1/6)*x = 1
solving x we get x = 9.

Hence after 9 am it takes 9 hours ie 6 pm or 1800 hours.

let t=time to fill pool
at 9am pool is 1/6 full
rate of A-B=1/12
t(1/12)=5/6
t=9 hours
9am+9 hours=18:00
C

We are given that pipe A fills a swimming pool in 4 hours and pipe B empties the pool in 6 hours. Thus, the fill rate of pipe A is 1/4 and the empty rate of pipe B is 1/6.

Since pipe A was opened at 8 a.m. and pipe B was at 9 a.m., we can let the time of pipe A = t + 1 and the time of pipe B = t. Also, we will subtract the work of pipe B (emptying) from the work done by pipe A (filling) to get the net amount of filling:

(1/4)(t + 1) - (1/6)(t) = 1

t/4 + 1/4 - t/6 = 1

Multiplying by 12, we have:

3t + 3 - 2t = 12

t = 9

So, the pool will be filled at 9 a.m. + 9 hours = 18:00.

Answer: C
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\(Rate of A = \frac{1}{4}\) and \(Rate of B = \frac{1}{6}\)

Let the W = total work = 1
Pipe A fills 1/4 of the pool in 1 hour. The remaining unfilled portion at 9 am is \(1-\frac{1}{4} = \frac{3}{4}\)

At 9 am pipe B starts emptying the pool. Hence the combined rate is that of \(A-B = \frac{1}{12}\)

We know \(W=\frac{3}{4}\), \(R=\frac{1}{12}\), hence T = 9

Answer is 9 hours from 9 am, i.e 6 PM, C

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