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!
If you think I have another official GMAT problem handy, you are right . Feel free to give it a shot before reading my explanation.
Last year Manfred received 26 paychecks. Each of his first 6 paychecks was $750; each of his remaining paychecks was $30 more than each of his first 6 paychecks. To the nearest dollar, what was the average (arithmetic mean) amount of his paychecks for the year?
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:
According to the problem, 6 of Manfred’s paychecks averaged $750 and the other 20 averaged $780. First of all, notice that the $780 group carries much more weight, because there are 20 paychecks versus only 6 in the other group. This means the overall average will be closer to $780 than to $750. Guess what? That means right away, we say goodbye to answers A, B, and C. So if you end up having to guess, you’ve got a 50-50 shot. Not bad!
Next, we’re relating 6 paychecks to 20 paychecks. That’s not an easy ratio to use for weighted averages. But let’s fudge things a bit and pretend that there are 6 paychecks in the $750 group and 18 in the $780 group.
In that case, we would have an 18:6 = 3:1 ratio in favor of the $780 group, which is a much cleaner ratio for dealing with weighted averages. Here’s the cool thing: numerically, that means that the overall average would be 3 times closer to $780 than to $750. Represented on a number line, it would look something like this,
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.
Of course, now we have to readjust for the actual problem. There are 20 entries in the $780 group, not 18. But those extra two entries mean the average is bumped up just a tiny smidge from 772.5, and thus we know that D is the final answer!
In addition, you could eliminate E, because an average of $775 would be too neat for the numbers involved. $780-$775 = $5, and $775-$750 = $25, which means that there would be 5 times as many entries in the $780 group. Obviously, it’s not that neat, since the actual ratio is 20:6 = 10:3.
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.
Okay, now I have a tricky problem for you. But this time, try it on your own! And remember, try to use the strategies we just talked about in this post. Feel free to list your step-by-step solutions in the comments, and be sure to apply the concepts behind weighted averages!
A certain balloon is packaged only in small cases of 40 balloons or in large cases of 100 balloons. A certain stock room has a total of 70 cases of these balloons, and the average number of balloons for each case is 88. A supervisor plans to remove some of the larger cases from the stock room without removing any of the smaller cases. How many of these larger cases must be removed such that the average number of balloons for the remaining boxes is reduced to 70?
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