Pool A and Pool B each have the same volume of water and are drained

Question

Pool A and Pool B each have the same volume of water and are drained by Pipe X and Pipe Y, respectively. The draining rate of Pipe X is double that of Pipe Y. If, after  t  hours, the remaining volume of water in Pool A is two-thirds that of Pool B, how long, in terms of  t , will it take for Pipe X to completely drain Pool A from its full volume?

A.  \frac{t}{2}  hours
B.  \frac{2t}{2}  hours
C.  \frac{3t}{2}  hours
D.  2t  hours
E.  3t  hours

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Bunuel · Accepted Answer

Official Solution:

Pool A and Pool B each have the same volume of water and are drained by Pipe X and Pipe Y, respectively. The draining rate of Pipe X is double that of Pipe Y. If, after  t  hours, the remaining volume of water in Pool A is two-thirds that of Pool B, how long, in terms of  t , will it take for Pipe X to completely drain Pool A from its full volume?

A.  \frac{t}{2}  hours
B.  \frac{2t}{2}  hours
C.  \frac{3t}{2}  hours
D.  2t  hours
E.  3t  hours

Assume Pipe X needs  x  hours to completely drain Pool A, then Pipe Y would need  2x  hours to drain Pool B.

After  t  hours, the volume of water left in Pool A is  \frac{x - t}{x}  of the initial volume. For example, if it takes Pipe X 4 hours to empty Pool A, then after 1 hour,  \frac{4 - 1}{4} = \frac{3}{4}  of the water remains.

Similarly, after  t  hours, the volume of water left in Pool B is  \frac{2x - t}{2x} . At this point, the remaining volume in Pool A is two-thirds of the volume in Pool B, which gives us the equation:
 
 \frac{x - t}{x}=\frac{2}{3}*\frac{2x - t}{2x} ;

x - t=\frac{1}{3}*(2x - t);

3(x - t)=(2x - t);

x=2t .

Answer: D

pranjalshah · Answer

Given rate of X = 2*rate of Y (X=2Y)
Let volume of pool be V.
Also, since pool A is emptied at X/hr, therefore to empty V volume, we need to V/X hours.

So we need to find V/X in terms of t.

Volume of water pumped in t hrs for pool A and B will be Xt and Yt respectively.
Volume of water remaining in pool A = V-Xt
Volume of water remaining in pool B = V-Yt

Given, V-Xt=(2/3)*(V-Yt)
Also given, X=2Y. Therefore, Y=X/2.
So, V-Xt=(2/3)*(V-Xt/2)
Solving, we get V/X=2t (D)

Purnank · Answer

Pool A and Pool B each have the same volume (assuming V) of water and are drained by Pipe X and Pipe Y, respectively. 
The draining rate of Pipe X is double that of Pipe Y. x = 2y
So at time t remaining volume in tank after drainage V - 2yt, for Pool A and V-yt for pool B
after t hours, the remaining volume of water in Pool A is two-thirds that of Pool B
V-2yt = 2/3 * (V-yt)
3V - 6yt = 2V - 2yt
V = 4yt
how long, in terms of t, will it take for Pipe X to completely drain Pool A from its full volume = Volume of pool / Rate of drainage = 4yt / 2y = 2t
Answer D

HarishD · Answer

A, drained by pipe x, let's say volume V
B, drained by pipe y, let's say volume V (same volume)

rate of B = a liters/hour
rate of A = 2a liters/hour

volume filled in time "t" by A= 2at , remaining A= V-2at
volume filled in time "t" by B= at, remaining B= V-at

the remaining volume of water in Pool A is two-thirds that of Pool B
V-2at = (2/3)(V-at)
V= 4at

time by pipe x to completely drain volume V= 4at/2a= 2t units

ANSWER:D

Nidzo · Answer

Let Pool A and Pool B =  a  when full.
Let Pipe X have a rate of  2x , and therefore Pipe Y has a rate of  x .
This means that it'll take Pipe X, alone,  \frac{a}{2x}  to empty the pool.

After  t  hours, Pipe X will have  a-2xt  to pump out and Pipe Y will have  a - xt  left to pump out.

As we are told, " the remaining volume of water in Pool A is two-thirds that of Pool B ":

a-2xt = \frac{2}{3}(a - xt)

3a - 6xt = 2a - 2xt

a = 4xt

Plugging this back into  \frac{a}{2x} :

\frac{4xt}{2x}

2t

ANSWER D

vasu1104 · Answer

say capacity for each pool is m litre
rate of pipe X= 2p
rate of pipe Y = p

now each pipe work for T hr.
so lets calculate how much water they drain out.

for pipe X= (-2pt)
for pipe Y = (-pt)

left water in pool A= m - 2pt
left in pool B = m - pt

m - 2pt = (2/3) (m-pt)
m= 4pt

we need to find total time take drain pool A
time = m / 2p
4pt / 2p
2t hrs

PeanutButter429 · Answer

Let a be the rate of drainage of Pool A, and b be the same for pool B

1-a=2/3(1-b)

Then you subsitute t/x for a, and t/2x for b - as rate of Pool A is double rate of pool B

Then you solve and get t/x = 1/2 -> which means that for a time of t, the pool is drained by 1/2 -> which means that in 2 hours it will be fully drained

Pool A and Pool B each have the same volume of water and are drained

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Pool A and Pool B each have the same volume of water and are drained

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