# How to Solve Relative Rate of Work Questions on the GMAT

Today, we look at the relative rate concept  of work, rate and time – the parallel of relative speed of distance, speed and time.

But before we do that, we will first look at one fundamental principle of work, rate and time (which has a parallel in distance, speed and time).

Say, there is a straight long track with a red flag at one end. Mr A is standing on the track 100 feet away from the flag and Mr B is standing on the track at a distance 700 feet away from the flag. So they have a distance of 600 feet between them. They start walking towards each other. Where will they meet? Is it necessary that they will meet at 400 feet from the red flag – the mid point of the distance between them? Think about it – say Mr A walks very slowly and Mr B is super fast. Of the 600 feet between them, Mr A will cover very little distance and Mr B will cover most of the distance. So where they meet depends on their rate of walking. They will not necessarily meet at the mid point. When do they meet at the mid point? When their rate of walking is the same. When they both cover equal distance.

Now imagine that you have two pools of water. Pool A has 100 gallons of water in it and the Pool B has 700 gallons. Say, water is being pumped into pool A and water is being pumped out of pool B. When will the two pools have equal water level? Is it necessary that they both have to hit the 400 gallons mark to have equal amount of water? Again, it depends on the rate of work on the two pools. If water is being pumped into pool A very slowly but water is being pumped out of pool B very fast, at some point, they both might have 200 gallons of water in them. They will both have 400 gallons at the same time only when their rate of pumping is the same. This case is exactly like the case above.

Now let’s go on to the question from the GMAT Club tests which tests this understanding and the concept of  relative rate of work:

Question: Tanks X and Y contain 500 and 200 gallons of water respectively. If water is being pumped out of tank X at a rate of K gallons per minute and water is being added to tank Y at a rate of M gallons per minute, how many hours will elapse before the two tanks contain equal amounts of water?

(A) 5/(M+K) hours

(B) 6/(M+K) hours

(C) 300/(M+K) hours

(D) 300/(M−K) hours

(E) 60/(M−K) hours

Solution: There are two tanks with different water levels. Note that the rate of pumping is given as K gallons per min and M gallons per min i.e. they are different. So we cannot say that they both will have equal amount of water when they have 350 gallons. They could very well have equal amount of water at 300 gallons or 400 gallons etc. So when one expects that water in both tanks will be at 350 gallon level, one is making a mistake. The two tanks are working for the same time to get their level equal but their rates are different. So the work done is different. Note here that equal level does not imply equal work done. The equal level could be achieved at 300 gallons when work done would be different – 200 gallons removed from tank X and 100 gallons added to tank Y. The equal level could be achieved at 400 gallons when work done would be different again – 100 gallons removed from tank X and 200 gallons added to tank Y.

To achieve the ‘equal level,’ tank Y needs to gain water and tank X needs to lose water. Total 300 gallons (500 gallons – 200 gallons) of work needs to be done. Which tank will do how much depends on their respective rates.

Work to be done together = 300 gallons

Relative rate of work = (K + M) gallons/minute

The rates get added because they are working in opposite directions – one is removing water and the other is adding water. So we get relative rate (which is same as relative speed) by adding the individual rates.

Note here that rate is given in gallons per minute. But the options have hours so we must convert the rate to gallons per hour.

Relative rate of work = (K + M) gallons/minute = (K + M) gallons/(1/60) hour = 60*(K + M)  gallons/hour

Time taken to complete the work = 300/60(K+M) hours = 5/(K+M) hours