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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pranjalshah · Accepted 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)

Bunuel · 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

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

Vivek1707 · Answer

Assuming some numbers can help your reduce the number of variables and solve this faster

Assume the  volume  of water is say  300 units  in both pools

Assume that Pipe x does  2 units  per hour and  y does 1 units  per hour

So as per the question: 300 - 2t = 2/3(300-t)

Hence t = 75

So time taken to drain pool x = 300 / 2 = 150 hours

Hence in termf of t, 150 would be 2t hence answer D

RSaurastri · Answer

Solution that avoids variables:

Assume the following:
 
 A and B have a capacity of 10L each 
 Rate of X = 2L/hr 
 Rate of Y = 1L/hr  
 Since rate of x is double of y

After t hours, remaining capacity is
A' = 10 - 2t
B' = 10 - t

It's given that A'=2/3 B'
10 - 2t = 2/3 (10 - t)
Upon solving, t = 2.5 hrs

If after 2.5 hrs, 10-2x2.5 = 5L remain in A.
Therefore, it will take another 2.5 hours to drain the remaining 5L.
Total time to drain A = 5 hours

Substitute t=2.5 in answer choices to see which equals to 5.

Answer = (d) 2t hours

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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