In an earlier post, we tackled a medium-level GMATPrep® weighted average question; click here to read that article before reading this one. This week, we’re trying a harder GMATPrep® weighted average question in order to test whether you learned the concept as well as you thought you did. : )
As we discussed earlier, every weighted average problem I’ve seen (so far!) on GMATPrep is a Data Sufficiency question. This doesn’t mean that they’ll never give us a Problem Solving weighted average problem, but it does seem to be the case that the test-writers are more concerned with whether we understand how weighted averages work than with whether we can actually do the calculations. Last week, we focused on understanding how weighted averages work via writing some equations. We’ll try to apply that understanding to our harder problem this week, along with a more efficient solution method.
Let’s start with a sample problem. Set your timer for 2 minutes…. and… GO!
* A contractor combined x tons of gravel mixture that contained 10 percent gravel G, by weight, with y tons of a mixture that contained 2 percent gravel G, by weight, to produce z tons of a mixture that was 5 percent gravel G, by weight. What is the value of x?
(1) y = 10
(2) z = 16
There are two kinds of gravel: “10% gravel” and “2% gravel.” These are our two “sub-groups.” When the two are combined (in some unknown – for now! – amounts), we get a 3rd kind:“5% gravel.” The number of tons of “10% gravel” (x) and the number of tons of “2% gravel” (y) will add up to the number of tons of “5% gravel” (z), or x + y = z. We need to find the number of tons of “10% gravel” used in the mixture.
The problem this week throws in a new wrinkle: we’re not just trying to calculate a ratio this time. We have to have enough info to calculate the actual amount of “10% gravel” used. Last week, we never had to worry about the actual number of employees. We’ll have to keep that in mind to see how things might change.
This problem never mentions the word average. That’s annoying. – how are we supposed to tell that this is a weighted average problem? Basically, the problem should talk about 2 sub-groups that are combined in some way to make a 3rd overall group, or mixture of the original 2 sub-groups. The problem will often discuss these groups in terms of percentages (as this problem does) or ratios (as last week’s problem did). That starts to tell us that some kind of averaging is happening.
Next, we know that we’ll either have an unweighted (“normal”) average or a weighted average, so check to see whether this is a normal average. We start with 10% gravel and 2% gravel. If we mixed exactly equal amounts of each (a normal average), what would the resulting mixture be?
(10+2)/2 = 6
Does the problem say we end up with 6% gravel? No. The amount of “10% gravel” is not equal to the amount of “2% gravel.” Therefore, this is not a “normal” average; now we know we have a weighted average problem.
Let’s go back to our test of “10%” and “2%” as an unweighted average. Instead of calculating with a formula (as we did up above and last week), let’s draw it out visually. Draw a straight, horizontal line on a piece of paper. Label the left end “2” and the right end “10” to represent our two different sub-groups of gravel:
An average of any two numbers will always appear between those two numbers. If the average is unweighted (50% of each is used), then the average will be exactly halfway between the two:
If the average is weighted, then it will NOT appear exactly halfway between. It will appear closer to one end or the other, depending upon the weighting. For example, if we have all “2% gravel” and zero “10% gravel,” what’s the average? A 100% weighting of “2% gravel” will give us an “average” of 2.
In this problem, they tell us that the average of the mixture is 5, so the average is closer to the “2” end of the line than to the “10” end of the line. (Imagine a tug-of-war between 2 and 10.) Now we know that there’s more “2% gravel” in the mixture than “10% gravel” because the “2” end of the line has “pulled” the average closer to its end.
To find the weighting: calculate the difference between the two ends of the line. in this case, the difference is 10 – 2 = 8. Determine how far the dominant end (2, in this case) has “pulled” the average: calculate how far away the weighted average appears from the other end of the line. In this case, the weighted average 5 is 5 units away from the 10 sub-group.
The “dominant” end (2) therefore has the weighting 5/8, because it has “pulled” the weighted average 5/8 of the way towards its end of the line.
The “non-dominant” end (10) is just the remaining amount of the “adds up to 1” figure (from our article last week): 3/8. The “non-dominant” end (10) has “pulled” the weighted average only 3/8 of the way toward its end of the line.
Now we know that, of the total mixture, 5/8 of it will be “2% gravel” and 3/8 of it will be “10% gravel.”
Conceptually, we want to realize that if a problem tells us the two starting points (the ends of the line) and the weighted average (the middle number), then we know we will be able to calculate the relative weightings of the two sub-groups. (This is the exact same concept that we learned on last week’s problem!)
Let’s examine the statements to see whether this knowledge might be useful.
Statement 1 says “y = 10,” which tells us that there are 10 tons of the “2% gravel.” If I know I have 10 tons of the “2% gravel,” and I also know that 5/8 of all of the gravel will be this “2% gravel,” then can I calculate x, the amount of “10% gravel?” Yes! 10 tons = 5/8 of the total. Divide each side by 5: 2 tons = 1/8 of the total. Multiply by 3: 6 tons = 3/8 of the total. (I don’t actually need to do this calculation on this problem because this is Data Sufficiency, but I would if I saw this on Problem Solving.) Statement 1 is sufficient; eliminate choices B, C, and E.
Statement 2 says “z = 16,” which tells us that there are 16 tons of the “5% gravel.” I also know that z represents the sum of x and y, or 8/8. If I know that 16 is 100%, or 8/8, of the amount, then I can also calculate x: 16 tons = 8/8 of the total. Divide by 8: 2 tons= 1/8 of the total. (Is this starting to look familiar?) Multiply by 3: 6 tons = 3/8 of the total. Statement 2 is also sufficient.
The correct answer is D.
We can simplify this further (for future data sufficiency questions) by saying: if we’re told the two “ends of the line” for calculating the average, as well as the overall weighted average, then we can calculate the relative weightings, or ratio, of the sub-groups (just as we did last week). If we’re also given one of the three actual amounts, then we can calculate all of the actual amounts in the problem. In this case, if we’d realized that before examining the statements, we could have asked ourselves, “Does the statement give me one of the three real amounts?” and quickly recognized that each statement is sufficient by itself.
Key Takeaways for Data Sufficiency Weighted Average Problems:
(1) Determine that you have a weighted average problem: this occurs when an average is described (even if the word “average” is not in the problem!), but that average is not a standard 1:1 or equally weighted average.
(2) Carefully write down what you were asked to solve, then determine what you know, what you don’t know, and what you would need to know in order to solve (before you look at the statements).
(3) Check the given statements to see whether you can find a “match” (that is, a statement tells you what you had already decided you would need to know in order to solve).
* GMATPrep® questions courtesy of the Graduate Management Admissions Council. Usage of this question does not imply endorsement by GMAC.