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

Video solution from Quant Reasoning starts at 8:56 here:

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.

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

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

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 .

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.

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)

(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

Let x = 2 hours, y = 3 hours and z = 4 hours.
Let the tank = 12 gallons.

Since Alpha takes 2 hours to fill the 12-gallon tank, Alpha's rate \(= \frac{w}{t} = \frac{12}{2} = 6\) gallons per hour.
Since Beta takes 3 hours to fill the 12-gallon tank, Beta's rate \(= \frac{w}{t} = \frac{12}{3} = 4\) gallons per hour.
Since Gamma takes 4 hours to empty the 12-gallon tank, Gamma's rate \(= \frac{w}{t} = \frac{12}{4} = 3\) gallons per hour.

When all 3 pumps work together -- Alpha and Beta increasing the volume by 6 gallons per hour and 4 gallons per hour, Gamma reducing the volume by 3 gallons per hour -- the net gain per hour = 6+4-3 = 7 gallons.
Since the net gain for all 3 pumps = 7 gallons per hour, and Alpha's rate alone = 6 gallons per hour, the fraction attributed to Alpha = \(\frac{6}{7}\).

The correct answer must yield a value of \(\frac{6}{7}\)when x=2, y=3 and z=4.
Only B works:
\(\frac{yz}{yz + xz - xy} = \frac{3*4}{3*4 + 2*4 - 2*3} = \frac{12}{14} = \frac{6}{7}\).

Approach under timed condition: I got bewildered as the question looked too lengthy and over that choices too would leave anyone breathless. However, as i read situation got clear. We have respective times for different pumps - two filling(A and B) and one draining(G) - and we have to calculate the fraction of water filled by A.
Fraction of water filled by A = \(\frac{Work A}{ Work A, B and G}\)
So, our target is to calculate the total water(work) that the three can do where one(G) is doing a negative work. Since we have time taken by each of them, we can find the rate of work i.e. \(\frac{1}{x}\), \(\frac{1}{y}\) and \(\frac{1}{z}\).

But before we do write anything on paper we can observe that one rate would be negative and while solving for work we would have one of the multipliers(in the numerator) with a negative sign before, which would be 'xy' in this case. On seeing the options we can get rid of A and C immediately as there is no '-xy' term. Since we are looking for ratio to get fraction of work by A, we would get rid of 'x' from work we get from the three together. Hence we are left with B and E only, giving us 50-50% probability of answering correctly.

Now, here's when things get tricky for which we need to be careful, where i spent half of the time figuring out which is correct. And to answer correctly we need to keep in mind that rates would give us the entity 'xyz' as denominator(while calculating for work). Finally, calculating fraction(ratio of rates) would give us 'xyz' as numerator wherein 'x' gets canceled out later. B is the answer.

Refer here:
Rate of A = \(\frac{1}{x}\)
Rate of B = \(\frac{1}{y}\)
Rate of G = \(\frac{1}{z}\)
Total work done by 3 = \(\frac{1}{x}\) + \(\frac{1}{y}\) - \(\frac{1}{z}\) = \(\frac{zy + xz - xy}{xyz}\)

Fraction of water filled by A = \(\frac{\frac{1}{x}}{\frac{zy+xz-xy}{xyz}}\) = \(\frac{yz}{zy + xz - xy}\)

Answer B
[Note: z > x is required only if one substitutes some numbers. Though number picking works, it sometimes screws up - for example see for yourself when x, y and z are 3,3,4 and 3,5,4 respectively]

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