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  # If x, y, and k are positive numbers such that
  
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  Manager
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  If x, y, and k are positive numbers such that [#permalink]  19 Nov 2006, 13:13
  If x, y, and k are positive numbers such that (10x+20y)/(x+y) = k and if x<y, which of the following could be the value of k?
  
  a) 10
  b) 12
  c) 15
  d) 18
  e) 30
  Director
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  Just by substitution I get 18.
  
  If all 3 are positive, and y >x, the minimum value of x is 1. then y can be 2, 3, 4 etc. When y = 4, k =18. For y=2, and y=3, k is not a positive integer.
  SVP
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  (D) as well
  
  (10x+20y)/(x+y) = k
  
  <=> (10x+20y) = k*(x+y)
  <=> (k-10)*x = (20-k)*y
  
  To make x and y positive, k have to be such that 10 < k < 20.
  
  This rules out A and E.
  
  Then, since x < y, k must be bigger than 15 (we must multiply a bigger number in front of x than in front of y).
  
  Hence, k = 18.
  Senior Manager
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  Fig,
  
  Help me out a little. Firstly, I've done the first part of the question two ways and come up with two different answers:
  
  First:
  
  10x + 20y = k(x +y) ---> 10x - kx = ky - 20y ---> x (10 - k) = y (k - 20)
  
  For x and y to be positive k<10 or k<20???? Where'd I go wrong????
  
  The other way I arranged it was the same as you:
  
  <=> (10x+20y) = k*(x+y)
  <=> (k-10)*x = (20-k)*y
  
  How did you go from here to working out it's 18?
  
  I think I'm up too late!!
  SVP
  Joined: 01 May 2006
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  Fig,
  
  Help me out a little. Firstly, I've done the first part of the question two ways and come up with two different answers:
  
  First:
  
  10x + 20y = k(x +y) ---> 10x - kx = ky - 20y ---> x (10 - k) = y (k - 20)
  
  For x and y to be positive k<10 or k<20???? Where'd I go wrong????
  
  The other way I arranged it was the same as you:
  
  <=> (10x+20y) = k*(x+y)
  <=> (k-10)*x = (20-k)*y
  
  How did you go from here to working out it's 18?
  
  I think I'm up too late!!
  
  By your first way, it's a little harder to view the elements ... It makes us deal with an equation with negative factors in front. But it's possible.
  
  x (10 - k) = y (k - 20)
  
  Imagine now:
  o If k = 21 the equation is turned to be -11*x = y than x and y cannot both be positive. >>>> k < 20.
  o If k = 9 the equation is turned to be x = -11*y than x and y cannot both be positive. >>>> k > 10.
  
  Then, 10-k and k-20 suggest a sort of symetry with a center at 15. To confirm this feeling, we can take k=15.
  
  x (10 - 15) = y (15 - 20) <=> x = y : Bingo there is a symetrical behaviour at 15.
  
  So now we can try k = 12,
  
  x (10 - 12) = y (12 - 20) <=> 2 x = 8 y >>>> x > y : the opposite of what we are looking for.
  
  So k = 18
  
  By the second way, we are using positive factors in front of x & y.
  
  (k-10)*x = (20-k)*y
  
  Here, we can observe without calculation that:
  > at left side of the equation, k must be greater than 10 (otherwise we have a negative factor)
  > at right side, k must be fewer than 20 (otherwise we have a negative factor)
  
  One more time, k-10 and 20-k suggest a symetry at 15. Then, similar to the first way but without calculation, we can see that greater k is fewer x is.
  
  Thus, as we want x < y and as k = 15 is a symetry point, k must be equal to 18.
  Senior Manager
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  Thanks a lot Fig and hats off to you.
  GMAT Club Legend
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  (10x+20y)/(x+y) = 10(x+2y)/(x+y) = 10[(x+y)+y]/x+y = 10 + 10y/(x+y) = 10(1+y/x+y)
  
  10(1+y/x+y) = k
  1 + y/(x+y) = k/10
  y/(x+y) = k/10 - 1
  
  So k cannot be 10 or 30 since y/x+y is a proper fraction.
  
  Testing:
  k = 12, y/x+y = 1/5 --> would require x > y. Out.
  k = 15, y/x+y = 1/2 --> would mean x = y. Out.
  k = 18, y/x+y = 4/5 --> possible since y can be greater than x for this result.
  
  Ans D
  VP
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  plug in, D is right.
  Senior Manager
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  (10x + 20y)/(x + y) = 10 + 10 * y / (x + y)
  as y > x
  1 > y / (x + y) > 1/2
  
  Only possible ans is 18
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