What helps in tests which need you to have speed and accuracy both is hacks - the innovative shortcuts for various questions – the way very smart people would solve the question – without putting in too much effort! Mind you, these so called 'shortcuts' are based on a fabulous understanding on concepts. Think of it this way - put in the effort before the test so that it's easy going during the test.

Do minimum work on paper. This means making equations and solving them is a big no-no and doing too many calculations is cumbersome.

**Question**: 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

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) yz/(yz + xz – xy)

yz cancels xy in the denominator giving us yz/xz = y/x = 2/4 = 1/2

(E) (yz + xz – xy)/yz

yz cancels xy in the numerator giving us xz/yz = x/y = 4/2 = 2

Only option (B) gives 1/2. Answer (B)

Even if you end up feeling that this method is complicated, try and wrap your head around it. If you do, you are on your way to becoming a lazy genius yourself!

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