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Simplifying algebraic expressions

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Master | Next Rank: 500 Posts
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by jeffedwards » Wed Nov 11, 2009 6:57 pm
Is there a better way to solve it than what the book says? It makes sense, but I doubt I could duplicate that on the test. Any tips?

If x, y, and k are positive numbers such that ( (x/(x+y) (10) + (x/(x+y) (20) ) = k and if x < y, which of the following could be the value of k?
  • 1
    10
    12
    15
    18
    30
OA: D
Source: OG 12th E; PS #148
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by gmatv09 » Wed Nov 11, 2009 8:48 pm
Is there a mistake in the problem...
there eqn should be [x/(x+y)]10+[y/(x+y)]20=k
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by palvarez » Wed Nov 11, 2009 11:12 pm
(10x+20y)/(x+y) = k, which is an integer

10 + (10y/(x+y)) = k

k > 10

10y/(x+y) = 10/(1+(y/x))

y/x's max value 1.

10/(1+(y/x))'s min value = 5

therefore, k > 15

20 - (10x/(x+y)) = k

here, k < 20


We got one value in that range. 18

10y = 8x+8y
y = 4x, which is consistent with x < y.
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by KICKGMATASS123 » Thu Nov 12, 2009 4:43 pm
palvarez wrote:(10x+20y)/(x+y) = k, which is an integer

10 + (10y/(x+y)) = k

k > 10

10y/(x+y) = 10/(1+(y/x))

y/x's max value 1.

10/(1+(y/x))'s min value = 5

therefore, k > 15

20 - (10x/(x+y)) = k

here, k < 20


We got one value in that range. 18

10y = 8x+8y
y = 4x, which is consistent with x < y.

dont get 10 + (10y/(x+y)) = k step!
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by palvarez » Thu Nov 12, 2009 5:33 pm
10x+20y = 10(x+y)+10y

divide by (x+y), you are left with 10 + 10y/(x+y)
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by okigbo » Thu Nov 12, 2009 11:31 pm
This is just a weighted average problem. No need to do calculations - average of 10 and 20 is 15 - equation is weigthed on the 20 side so must be greater than 15 - pick 18 and move on.
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