GMAT Math : Weighted averages

Let’s start with an example :

John has 5 solutions in different containers. The first solution has 10 mg/liter of salt, the second solution has 15 mg/liter of salt, the third has 15 mg/liter of salt, the fourth has 20 mg/liter of salt and the fifth has 20 mg/liter of salt. He takes 150 liter of the first solution., 300 liter of the second solution, 450 liter of the third solution, 600 liter of the fourth solution and 750 liter of the fifth solution and mixes them together, what is the resulting salt concentration of the resulting solution?

A. 20/3

B. 74/3

C. 22/3

D. 53/3

E. 59/3

Wow, there is a ton of data in the question 🙂

If we cut to the chase, what the question is asking for is the ‘average’ concentration of the solution when we mix the individual solutions. In the case the volumes of the individual solutions were the same, we would essentially determine the simple mathematical average or specifically:

Average concentration = sum of concentrations/(# of liquids)

Average concentration = (concentration 1 + concentration 2 + concentration 3 + concentration 4 + concentration 5)/5

But in this case, since the volumes of the liquid are different we are looking for a weighted average.

Weighted average = k1 x1 + k2 x2 + k3 x3 + k4 x4 + … knxn/ (k1 + k2 + k3 + k4 + … kn)

Where k = weights or in this case the volumes of the solutions

x = concentration of solutions

Weights signify the relative importance of each solution – in this case relative importance is depicted by the volume of the solutions. Think about it? If there is one solution’s volume accounting for 99% of the combined solution’s volume we would expect the concentration of the combined solution to be quite similar to this one mega solution’s concentration.

Coming to the original question:

Weighted average (average concentration) = k1 x1 + k2 x2 + k3 x3 + k4 x4 + … knxn/ (k1 + k2 + k3 + k4 + … kn)

= (10 x 150 + 15 x 300 + 15 x 450 + 20 x 600 + 20 x 750)/(150 + 300 + 450 + 600 + 750)

= 3-4 mins of intensive calculation 🙂

This brings us to our trick:

Before plugging into the formula, weights (i.e., k1, k2, k3, k4, … kn) can be simplified by a common denominator.

So, instead of using 150, 300, 450, 600 and 750, we can use 150/150 = 1, 300/150 = 2, 450/150 = 3, 600/150 = 4 and 750/150 = 5 i.e 1, 2, 3, 4 and 5 as the respective weights

So our weighted average gets simplified to a more palatable version we can solve in 1-2 mins :

= (10 x 1 + 15 x 2+ 15 x 3 + 20 x 4 + 20 x 5)/(1 + 2 + 3 + 4 + 5)

= 53/3

We go with option (D) 🙂