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GRE Math Challenge #8 - If 7X + 3Y= 17 and 3X+7Y = 19,

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GRE Math Challenge #8 - If 7X + 3Y= 17 and 3X+7Y = 19, [#permalink] New post 07 Sep 2014, 06:50
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83% (01:16) correct 16% (00:00) wrong based on 6 sessions
If 7X + 3Y= 17 and 3X+7Y = 19, what is the average (arithmetic mean) of X and Y?

[Reveal] Spoiler: Answer
9/5

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Re: GRE Math Challenge #8 [#permalink] New post 10 Sep 2014, 02:38
One should not try to solve the simultaneous equations, simple addition of the given equations gives
=> 10(X+Y) = 36
=> (X+Y)/2 = 36/20 = 9/5 = average of X & Y
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Re: GRE Math Challenge #8 [#permalink] New post 15 May 2015, 13:24
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Expert's post
sandy wrote:
If 7x + 3y= 17 and 3x + 7y = 19, what is the average (arithmetic mean) of x and y?

[Reveal] Spoiler: Answer
9/5

My approach is the same as 13divyasha's, but I want to add in some middle steps (if anyone is having trouble following)

First, to find the average (arithmetic mean) of X and Y, we need to recognize that we don't need to know the individual values of x and y. All we need is the SUM of x and y, because the average of x and y = (x+y)/2

We're given two equations:
7x + 3y= 17
3x + 7y = 19

Something nice happens we when we add these equations. We get:
10x + 10y = 36
If we divide both sides by 10, we get: x + y = 36/10

Now that we know the sum of x and y, we can find the average.
Average = (x+y)/2 = = (36/10)/2 = 36/20 =
[Reveal] Spoiler:
9/5


Cheers,
Brent
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Re: GRE Math Challenge #8   [#permalink] 15 May 2015, 13:24
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GRE Math Challenge #8 - If 7X + 3Y= 17 and 3X+7Y = 19,

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