Given:
average (arithmetic mean) of x, y, and 20 is 10 greater than the average of x, y, 20, and 30
Writing it in equation form
(x+y+20)/3 = 10 + (x+y+20+30)/4
Solving it:-
(x+y)/3 + (20/3) = 10 + (x+y)/4 + 50/4
(x+y)/3 - (x+y)/4 = 10 + 50/4 - 20/3
(x+y)(1/3-1/4) = 10 + 25/2 - 20/3
(x+y)(1/12) = 95/6
Multiply with 6 on both sides
(x+y)/2 = 95
Ans is E
average problem
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punit.kaur.mba
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VP_Jim
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Another way of doing it, if you're like me and you don't like equations, is to plug in the answer choices for X and Y.
If the average of X and Y is, say, 95, it doesn't really matter what those numbers are individually, since we know that they will add up to 190 (95*2). By plugging in 190 for X and Y in the answer choices, we find:
The average of X, Y, and 20 is 70, or (190+20)/3
The average of X, Y, 20, and 30 is 60 (190+20+30)/4
Thus, since the average of the first set is 10 more than the average of the second set, 95 must be the correct answer. If you were to plug in other answer choices, the numbers would not work out such that the average of the first set is 10 more than the average of the second.
Doing it this way vs. using equations might take some more calculations, but I think it's harder to make a mistake when you're plugging in. I could easily see a test taker messing up either setting up those equations or not solving them correctly.
If the average of X and Y is, say, 95, it doesn't really matter what those numbers are individually, since we know that they will add up to 190 (95*2). By plugging in 190 for X and Y in the answer choices, we find:
The average of X, Y, and 20 is 70, or (190+20)/3
The average of X, Y, 20, and 30 is 60 (190+20+30)/4
Thus, since the average of the first set is 10 more than the average of the second set, 95 must be the correct answer. If you were to plug in other answer choices, the numbers would not work out such that the average of the first set is 10 more than the average of the second.
Doing it this way vs. using equations might take some more calculations, but I think it's harder to make a mistake when you're plugging in. I could easily see a test taker messing up either setting up those equations or not solving them correctly.
Jim S. | GMAT Instructor | Veritas Prep
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AleksandrM
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I have said it before and will say it again: the chances of you making a mistake with algebra are much greater than with arithmetic. However, for this problem, and many like it, it is better to use algebra.
x + y + 20/3 = (x + y + 20 + 30)/4 + 10
Then you just multiply both sides by 12 and the rest is pretty much arithmetic. The only trick to this question is NOT to include the 10 along with x, y, 20, and 30 over the 4. Under time constraints, this problem will flow much better and quicker than it would with plugging in answer choices.
However, if you are going to use plug-in the answer method, then do the following (in my opinion this has been the most useful approach):
Try D, then B, after that it will be pretty clear whether to go higher, lower. Otherwise, you have eliminated enough answer choices to choose C.
x + y + 20/3 = (x + y + 20 + 30)/4 + 10
Then you just multiply both sides by 12 and the rest is pretty much arithmetic. The only trick to this question is NOT to include the 10 along with x, y, 20, and 30 over the 4. Under time constraints, this problem will flow much better and quicker than it would with plugging in answer choices.
However, if you are going to use plug-in the answer method, then do the following (in my opinion this has been the most useful approach):
Try D, then B, after that it will be pretty clear whether to go higher, lower. Otherwise, you have eliminated enough answer choices to choose C.
