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A tank has two water pumps Alpha and Beta and one drain Gamma. Pump

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A tank has two water pumps Alpha and Beta and one drain Gamma. Pump [#permalink]

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A tank has two water pumps Alpha and Beta and one drain Gamma. Pump Alpha alone can fill the whole tank in x hours, and pump Beta alone can fill the whole tank in y hours. The drain can empty the whole tank in z hours such that z>x. When the tank was empty, pumps Alpha and Beta started pumping water in the tank and the drain Gamma simultaneously was draining water from the tank. When finally the tank is full, which of the following represents the amount of water in terms of the fraction of the tank which pump Alpha pumped into the tank?

(A) yz/(x+y+z)
(B) yz/(yz + xz – xy)
(C) yz/(yz + xz + xy)
(D) xyz/(yz + xz – xy)
(E) (yz + xz – xy)/yz

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[Reveal] Spoiler: OA

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Re: A tank has two water pumps Alpha and Beta and one drain Gamma. Pump [#permalink]

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New post 09 Sep 2015, 06:36
Solution : Let n be no. of the hours taken to fill the whole tank when all three pumps are running.
Amount of water in the tank after its full = n[1/x + 1/y - 1/z]
Amount of water in terms of the fraction of the tank which pump Alpha pumped into the tank = (n/x)/n[1/x + 1/y - 1/z] ==> yz/(xz+yz-xy)

Option B.

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A tank has two water pumps Alpha and Beta and one drain Gamma. Pump [#permalink]

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New post 09 Sep 2015, 10:56
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Bunuel wrote:
A tank has two water pumps Alpha and Beta and one drain Gamma. Pump Alpha alone can fill the whole tank in x hours, and pump Beta alone can fill the whole tank in y hours. The drain can empty the whole tank in z hours such that z>x. When the tank was empty, pumps Alpha and Beta started pumping water in the tank and the drain Gamma simultaneously was draining water from the tank. When finally the tank is full, which of the following represents the amount of water in terms of the fraction of the tank which pump Alpha pumped into the tank?

(A) yz/(x+y+z)
(B) yz/(yz + xz – xy)
(C) yz/(yz + xz + xy)
(D) xyz/(yz + xz – xy)
(E) (yz + xz – xy)/yz

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SUGGESTION: Always calculate one hour work done by each machine/individual first

Alpha Takes x hours to fill the Tank alone
Beta Takes y hours to fill the Tank alone
Gamma Takes z hours to Empty the Tank alone

Consider filling the tank as POSITIVE WORK and Emptying the Tank as NEGATIVE WORK

i.e. Fraction of tank filled by Alpha in 1 hour = 1/x
i.e. Fraction of tank filled by Alpha in 1 hour = 1/y
i.e. Fraction of tank emptied by Gamma in 1 hour = -1/z

1 hour work of all three Alpha, Beta and Gamma together = (1/x)+(1/y)-(1/z)
1 hour work of Alpha = 1/x

Fraction of work done by Alpha = (1/x) / [(1/x)+(1/y)-(1/z)] = yz / (yz + xz - xy)

Answer: option B

One may also solve such question by substituting values of x=2, y=3 and z=4 as well and then check options by substituting values
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Re: A tank has two water pumps Alpha and Beta and one drain Gamma. Pump [#permalink]

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Bunuel wrote:
A tank has two water pumps Alpha and Beta and one drain Gamma. Pump Alpha alone can fill the whole tank in x hours, and pump Beta alone can fill the whole tank in y hours. The drain can empty the whole tank in z hours such that z>x. When the tank was empty, pumps Alpha and Beta started pumping water in the tank and the drain Gamma simultaneously was draining water from the tank. When finally the tank is full, which of the following represents the amount of water in terms of the fraction of the tank which pump Alpha pumped into the tank?

(A) yz/(x+y+z)
(B) yz/(yz + xz – xy)
(C) yz/(yz + xz + xy)
(D) xyz/(yz + xz – xy)
(E) (yz + xz – xy)/yz

Kudos for a correct solution.




amount of water in tank is proportional to time taken to fill
so ratio can be taken with respect to times
total time to fill the tank (with A and B filling and G draining) = xyz/(yz + xz – xy)
dividing the time taken for A to fill the tank

xyz/x(yz + xz – xy) = yz/(yz + xz – xy)

so B

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Re: A tank has two water pumps Alpha and Beta and one drain Gamma. Pump [#permalink]

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New post 09 Sep 2015, 18:28
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the answer is option B by taking the work done by third drain as negative. so combined rate is 1/x+1/y-1/z.
now taking rate of 1/x*1/ (1/x+1/y-1/z) gives option b

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Re: A tank has two water pumps Alpha and Beta and one drain Gamma. Pump [#permalink]

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Bunuel wrote:
A tank has two water pumps Alpha and Beta and one drain Gamma. Pump Alpha alone can fill the whole tank in x hours, and pump Beta alone can fill the whole tank in y hours. The drain can empty the whole tank in z hours such that z>x. When the tank was empty, pumps Alpha and Beta started pumping water in the tank and the drain Gamma simultaneously was draining water from the tank. When finally the tank is full, which of the following represents the amount of water in terms of the fraction of the tank which pump Alpha pumped into the tank?

(A) yz/(x+y+z)
(B) yz/(yz + xz – xy)
(C) yz/(yz + xz + xy)
(D) xyz/(yz + xz – xy)
(E) (yz + xz – xy)/yz

Kudos for a correct solution.


Work done by Alpha in 1 hour = \(\frac{1}{x}\)
Work done by beta in 1 hour = \(\frac{1}{y}\)
Work done by Gamma in 1 hour = \(\frac{1}{z}\)
Total work done by 3 pumps in 1 hour=

\(\frac{1}{x}\) + \(\frac{1}{y}\) - \(\frac{1}{z}\) = \(\frac{zy + xz - xy}{xyz}\)

Alpha's contribution in this work will be

\(\frac{\frac{1}{x}}{\frac{zy+xz-xy}{xyz}}\) = \(\frac{yz}{zy + xz - xy}\)

Answer:- B

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Re: A tank has two water pumps Alpha and Beta and one drain Gamma. Pump [#permalink]

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New post 11 Sep 2015, 22:59
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A tank has two water pumps Alpha and Beta and one drain Gamma. Pump Alpha alone can fill the whole tank in x hours, and pump Beta alone can fill the whole tank in y hours. The drain can empty the whole tank in z hours such that z>x. When the tank was empty, pumps Alpha and Beta started pumping water in the tank and the drain Gamma simultaneously was draining water from the tank. When finally the tank is full, which of the following represents the amount of water in terms of the fraction of the tank which pump Alpha pumped into the tank?

(A) yz/(x+y+z)
(B) yz/(yz + xz – xy)
(C) yz/(yz + xz + xy)
(D) xyz/(yz + xz – xy)
(E) (yz + xz – xy)/yz

Work done by Alpha in 1 hour = 1/x
Work done by beta in 1 hour = 1/y
Work done by Gamma in 1 hour = 1/z
Total work done by 3 pumps in 1 hour=

1/x + 1/y - 1/z = (zy+xz−xy)/xyz ------> when the tank is consider to be full


Alpha's part as a fraction of the total work done = (1/x)/(zy+xz−xy)/xyz

On rearranging we get yz/(yz + xz – xy)
hence B ans .
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Re: A tank has two water pumps Alpha and Beta and one drain Gamma. Pump [#permalink]

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New post 13 Sep 2015, 08:37
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Bunuel wrote:
A tank has two water pumps Alpha and Beta and one drain Gamma. Pump Alpha alone can fill the whole tank in x hours, and pump Beta alone can fill the whole tank in y hours. The drain can empty the whole tank in z hours such that z>x. When the tank was empty, pumps Alpha and Beta started pumping water in the tank and the drain Gamma simultaneously was draining water from the tank. When finally the tank is full, which of the following represents the amount of water in terms of the fraction of the tank which pump Alpha pumped into the tank?

(A) yz/(x+y+z)
(B) yz/(yz + xz – xy)
(C) yz/(yz + xz + xy)
(D) xyz/(yz + xz – xy)
(E) (yz + xz – xy)/yz

Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION:

Note that you have variables in the question and the options. Since we are looking for a lazy solution, making equations out of the variables is not acceptable. So then, should we plug in numbers? With three variables to take care of, that might involve a lot of calculations too. Then what else?

Here is our minimum-work-solution to this problem; try to think one of your own and don’t forget to share it with us.

Plugging in numbers for the variables can be troublesome but you can give some very convenient values to the variables so that the effect of a pump and a drain will cancel off.

There are no constraints on the values of x, y and z except z > x (drain Gamma empties slower than pipe Alpha fills)

Let’s say, x = 2 hrs, y = 4 hrs, z = 4 hrs

What did we do here? We made the rate of Beta same as the rate of Gamma i.e. 1/4 of the tank each. This means, whenever both of them are working together, drain Gamma cancels out the work of pump Beta. Every hour, pump Beta fills 1/4th of the tank and every hour drain Gamma empties 1/4th of the tank. So the entire tank will be filled by pump Alpha alone. Hence, if y = z, pump Alpha fills the entire tank i.e. the amount of water in terms of fraction of the tank pumped by Alpha will be 1.

In the options, put y = z and see which option gives you 1. Note that you don’t have to put in the values of 2, 4 and 4. We gave those values only for illustration purpose.

If y = z, xy = xz.

So in option (B), xz cancels xy in the denominator giving yz/yz = 1

Again, in option (E), xz cancels xy in the numerator giving yz/yz = 1

The other options will not simplify to 1 even though when we put y = z, the answer should be 1 irrespective of the value of x, y and z. The other options will depend on the values of x and/or y. Hence the only possible options are (B) and (E). But we still need to pick one out of these two.

Now let’s say, x = 4, y = 2, z = 4.00001 ( z should be greater than x but let’s assume it is infinitesimally greater than x such that we can approximate it to 4 only)

Rate of work of Gamma (1/4th of the tank per hour) is half the rate of work of Beta (1/2 the tank per hour). Rate of work of Gamma is same as rate of work of Alpha. Half the work done by pump Beta is removed by drain Gamma. So if pump Beta fills the tank, drain Gamma empties half of it in that time – the other half would be filled by pump Alpha i.e. the amount of water in terms of fraction of the tank pumped by Alpha will be 1/2.

Put x = z in the options (B) and (E). The one that gives you 1/2 with these values should be the answer. Again, you don’t need to plug in the actual values till the end.

If x = z, yx = yz

(B) \(\frac{yz}{(yz + xz - xy)}\)
yz cancels xy in the denominator giving us \(\frac{yz}{xz} = \frac{y}{x}= \frac{2}{4} = \frac{1}{2}\)

(E) \(\frac{(yz + xz - xy)}{yz}\)
yz cancels xy in the numerator giving us \(\frac{xz}{yz} = \frac{x}{y} = \frac{4}{2} = 2\)

Only option (B) gives 1/2.
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Re: A tank has two water pumps Alpha and Beta and one drain Gamma. Pump [#permalink]

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New post 30 Sep 2016, 07:39
Bunuel wrote:
A tank has two water pumps Alpha and Beta and one drain Gamma. Pump Alpha alone can fill the whole tank in x hours, and pump Beta alone can fill the whole tank in y hours. The drain can empty the whole tank in z hours such that z>x. When the tank was empty, pumps Alpha and Beta started pumping water in the tank and the drain Gamma simultaneously was draining water from the tank. When finally the tank is full, which of the following represents the amount of water in terms of the fraction of the tank which pump Alpha pumped into the tank?

(A) yz/(x+y+z)
(B) yz/(yz + xz – xy)
(C) yz/(yz + xz + xy)
(D) xyz/(yz + xz – xy)
(E) (yz + xz – xy)/yz

Kudos for a correct solution.


oi oi oi...3 variables...i just remembered the question I got at the real GMAT...with 5 variables...

this one has wording kind of confusing, but in the end, it's not that bad...
rate for A = 1/x
rate for B = 1/y
rate drains for G = 1/z

now, A and B work together, while G takes out water...
we then fill the tank at a rate of:
1/x + 1/y - 1/z.
or, after some simplifications: (zx+zy-xy)/xyz.

now...1/x multiplied by the reciprocal of the above mentioned
x cancels, and what is left yz/(xz+zy-xy)

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Re: A tank has two water pumps Alpha and Beta and one drain Gamma. Pump [#permalink]

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New post 14 Aug 2017, 11:41
(1/x+1/y-1/z)*n=1 tank
(yz+xz-xy)/xyz

1/x : (yz+xz-xy)/xyz = 1/x * xyz/(yz+xz-xy) = yz/(yz+xz-xy)
hence, B

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Re: A tank has two water pumps Alpha and Beta and one drain Gamma. Pump   [#permalink] 14 Aug 2017, 11:41
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