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A group consists of both men and women. The average (arithmetic mean) height of the women is 66 inches, and the average (arithmetic mean) height of the men is 72 inches. If the average (arithmetic mean) height of all the people in the group is 70 inches, what is the ratio of women to men in the group?
This is a weighted average problem. There are two groups, and each group has a different average height. When the two groups are combined, the average height changes. In problems like this, you might be asked to find the size of one group, the relationship between the two groups, the average height of one of the groups, or something else entirely. If you try to memorize a mathematical formula to use, you’ll find yourself stumped as soon as the problem differs from what you’ve memorized. But with a little creativity and logical reasoning, you can solve any weighted average problem quickly. Here’s how.
Imagine the overall average height as the pivot point of a seesaw.
Figure out how each group compares to the average height. In this case, the women are on average 4 inches shorter, and the men are on average 2 inches taller. So, the women will stand 4 inches to the left of the pivot point, and the men will stand 2 inches to the right.
If you’ve ever played on a seesaw before, you know that the further you are from the pivot point, the more your weight ‘counts’. Someone who’s all the way out at the very edge will be able to balance out a much heavier person who’s closer to the center. The image should actually look like this:
The women are twice as far from the center as the men. Their weight counts for twice as much. To balance them out, you’ll need twice as many men as women.
Now, the seesaw is balanced, and the problem is solved. The ratio of women to men is 1:2.
What if you don’t know how far each group is from the center? Here’s another problem:
The average (arithmetic mean) of 13 numbers is 70. If the average of 10 of these numbers is 90, what is the average of the other 3 numbers?
Draw out your seesaw. There are 10 numbers on the right, and 3 numbers on the left. Because the average of the 10 larger numbers is 90, and the average is 20, they’ll stand 20 away from the center. You don’t know just yet how far from the center the numbers on the left are. (However, because there are fewer of them, you know they need to be more than 20 units from the center in order to balance the seesaw!)
How far away from the center are the three numbers on the left, when the seesaw is balanced? Well, there are only 3/10 as many of them. So, in order to balance everything out, each of them must count for 10/3 times as much. They have to be 10/3 times as far from the center.
10/3 * 20 is 200/3. However, that’s not the answer — it’s just how far away from the middle the three values are. If they’re 200/3 away from the center, their average value must be 70 – 200/3, or 10/3.
You can even use this strategy on some percentage problems. Here’s a final example. Try it on your own before reading onwards.
Fiber X cereal is 55% fiber. Fiber Max cereal is 70% fiber. Sheldon combines an amount of the two cereals in a single bowl of mixed cereal that is 65% fiber. If the bowl contains a total of 12 ounces of cereal, how much of the cereal, in ounces, is Fiber X?
Here’s how you’d set up the seesaw:
Fiber X is twice as far from the middle, so its weight counts for twice as much. There must be half as much of it in order for the seesaw to balance. If the bowl contains 12 ounces of cereal, 4 ounces will be Fiber X, and the remaining 8 ounces will be Fiber Max.
The lesson here is that you don’t need to memorize a bunch of mathematical formulas to succeed on the GRE. In fact, trying to memorize formulas makes you less flexible and less likely to spot your own mistakes. Work through weighted average problems like the three in this article using logical reasoning, instead. There are many more problems to try in the Averages, Weighted Averages, Median, and Mode chapter of the 5lb. Book of GRE Practice Problems! 📝
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Chelsey Cooley is a Manhattan Prep instructor based in Seattle, Washington. Chelsey always followed her heart when it came to her education. Luckily, her heart led her straight to the perfect background for GMAT and GRE teaching: she has undergraduate degrees in mathematics and history, a master’s degree in linguistics, a 790 on the GMAT, and a perfect 170/170 on the GRE. Check out Chelsey’s upcoming GRE prep offerings here.