More Fun with Weighted Averages

By - Aug 23, 11:18 AM Comments [2]

This article was written by Rich Zwelling. For more GMAT prep advice, check out the Knewton GMAT blog.

In last week's post, I discussed how weighted averages are often tested conceptually and thus show up on Data Sufficiency problems. However, even when weighted average problems appear on the more results-oriented Problem Solving questions, this conceptual understanding can be mighty handy. As I said in a previous post, the GMAT is not a speed calculation test, but that doesn't mean the test won't reward you if you can find conceptual shortcuts!



(A) $752
(B) $755
(C) $765
(D) $773
(E) $775

You could certainly solve this problem the standard way: his first 6 paychecks average $750, meaning that they total $750*6 = $4500, then you'd calculate the total of the remaining paychecks and find an overall average. But instead, let’s have some fun and try to short-circuit the problem using weighted averages:





where the X marks our final average. Again, think of the relative sizes of the groups as "weights" that pull the overall average away from the average of 750 and 780. Since there are three times as many entries in the 780 group, the overall average is pulled towards 780 and ends up three times closer to 780 than to 750.

Now, notice that the distance of our whole number line is 30. That distance of 30 is divided into 4 equal segments; thus, each marking is 30/4 = 7.5 apart. That means our "fudged" final average (X) is 780-7.5 = 772.5.



TAKEAWAY: When you see a problem involving multiple sets of numbers, each of which has a unique average and each of which has a different number of entries ("weight"), you should immediately think of weighted averages and see if this conceptual approach can get you to the answer faster.



A) 28
B) 32
C) 35
D) 40
E) 42

[2] Comments to this Article

  1. Joy February 11, 11:32 PM

    Hey, what’s the answer?

    1. Grant August 18, 8:40 PM

      Joy, here’s how I solved it.

      I first found the current number of small cases and large cases.


      There’s 48 between the smallest and the average, and 12 between the largest and the average. So the weight of the largest:smallest is 48:12 or 4:1. Since you know that there is 70 total cases, 4x + x = 70. x = 14, and there are currently 14 small cases and 56 large cases.

      Now, for the average to be 70, the ratio of small to large cases must be equal.


      Ratio is 30:30, or 1:1. Since there are 14 small cases and 56 large cases, and the question asks how many large cases must be removed, you must remove enough large cases to make the ration 14:14. 56-x=14. x = 42.