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  # GMATprep DS: Number theory
  
  Author Message
  VP
  Joined: 21 Mar 2006
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  09 Feb 2007, 03:23
  00:00
  
  Difficulty:
  
  (N/A)
  
  Question Stats:
  
  38% (00:02) correct 63% (00:53) wrong based on 6 sessions
  
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  Attachments
  
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  Director
  Joined: 12 Jun 2006
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  09 Feb 2007, 04:16
  C
  (1) 2x - 2y = 1 ---> 2(x-y) = 1 ---> x-y = 1/2
  x and y can be +ves or -ves and still sum to .5. Ins
  
  (2) x and y can be neg or pos and be greater than 1
  -4/-2 = 2 or 4/2 = 2. Ins
  
  1 and 2
  In order for x/y to be greater than 1 amd x - y to equal 1/2 x and y must both be positive.
  
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  VP
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  09 Feb 2007, 05:49
  Should be C
  
  1 is st forward
  
  (2)
  means x>y if both are positive
  x<y if both are -ve
  
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  SVP
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  09 Feb 2007, 06:08
  (C) for me
  
  Stat1
  It's the equation of a line. Apart the line x=0 and y=0, all lines are passing through 2 cadran (a cadran changes the sign of another cadran on either x, y or x and y).
  
  Line : y = x - 1/2
  
  INSUFF.
  
  Stat2
  x/y > 1
  
  Here too, it's under the line y=x when x is potive and y positive, and it's above the line y=x when x is negative and y is negative. Thus, x and y could represent a point in the cadran I partially, in the cadran III partially.
  
  INSUFF.
  
  Both 1 and 2
  The line y = x - 1/2 cut since x=1/2 and x increasing represents the solution of the 2 statments.
  
  In other words, we are always in cadran I.
  
  SUFF.
  Attachments
  
  Fig1_y equal x.gif [ 2.3 KiB | Viewed 2006 times ]
  
  Fig2_x div y sup 1.gif [ 2.87 KiB | Viewed 2056 times ]
  
  Fig3_Stats 1 and 2.gif [ 3.05 KiB | Viewed 2005 times ]
  
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  Retired Moderator
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  09 Feb 2007, 15:31
  from one
  
  x-y = 1/2
  
  this is possible in 3 scenarios
  
  1) both +ve
  
  2)x +ve and y -ve , x,y are fractions
  
  3)both -ve , /y/>/x/
  
  insuff
  
  from 2
  
  x/y > 1
  
  x/y - y/y>0
  
  (x-y) / y >0.............insuff
  
  both together
  
  x-y = 1/2 , substitute in 2
  
  (1/2)/y >0 thus y sure is positive and this is matching with scenario no: one only
  
  thus both are +ve and C is my answer
  
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  VP
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  10 Feb 2007, 04:30
  ggarr wrote:
  C
  (1) 2x - 2y = 1 ---> 2(x-y) = 1 ---> x-y = 1/2
  x and y can be +ves or -ves and still sum to .5. Ins
  
  (2) x and y can be neg or pos and be greater than 1
  -4/-2 = 2 or 4/2 = 2. Ins
  
  1 and 2
  In order for x/y to be greater than 1 amd x - y to equal 1/2 x and y must both be positive.
  
  Here's where I made the mistake. I rearranged statement (2) as x > y.
  
  Now, if you take x = 0 and y = -1/2, the equation from (1) is still satisfied but x and y are not positive. So I chose E. Obviously this is a trap and I fell for it by rearranging (2). When should you rearrange the terms in an inequality and when shouldn't you ('cause sometimes rearranging terms in a inequality does help)? Btw, OA is C.
  
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  Manager
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  Re: GMATprep DS: Number theory [#permalink]
  
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  13 Jun 2011, 07:28
  I am still getting E and don't understand the explainations for answer C. Can someone please clarify?
  
  S1.
  
  rearranged to
  
  y = x + 1/2
  
  not suff, it's a line that contains both positive values and negative values for x and y
  
  S2.
  
  rearranged to
  
  x>y and x<y or another way of looking at it is (x/y pos/pos) or (x/y neg/neg)
  
  not suff, contains both positive and negative values for x and y
  
  S1 and S2
  
  not suff,
  
  y=x+1/2
  x>y or x<y
  
  we can still get neg/neg and pos/pos
  
  where did I go wrong?
  
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  Math Forum Moderator
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  Re: GMATprep DS: Number theory [#permalink]
  
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  13 Jun 2011, 08:42
  1
  KUDOS
  kripalkavi wrote:
  Are $$x$$ and $$y$$ both correct?
  
  1. $$2x-2y=1$$
  2. $$\frac{x}{y}>1$$
  
  1.
  $$2x-2y=1$$
  $$x-y=\frac{1}{2}$$
  This statement just tells us that x is $$\frac{1}{2}$$ units to the right of y on the number line.
  
  So,
  x=1, y=0.5
  x=-0.5, y=-1.0
  x=0, y=-0.5
  x=0.5, y=0
  
  Not Sufficient.
  
  2.
  $$\frac{x}{y}>1$$
  $$\frac{x}{y}-1>0$$
  $$\frac{x-y}{y}>0$$
  
  For a fraction to be greater than 0, both numerator and denominator must have the same sign
  We have 2 scenarios here:
  if y>0, x-y>0 or x>y
  y=1; x=1.5
  
  if y<0, x-y<0 or x<y
  y=-1, x=-1.5
  
  Not Sufficient.
  
  Combining both;
  
  from statement 1, we know x-y=0.5, which is greater than 0.
  
  Thus, for expression
  $$\frac{x-y}{y}>0$$
  We definitely know that the numerator is +ve. And thus the denominator must also be positive.
  
  x-y>0
  y>0
  x>0
  Sufficient. x and y are both indeed positive.
  
  Ans: "C"
  _________________
  
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  Manager
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  Re: GMATprep DS: Number theory [#permalink]
  
  ### Show Tags
  
  21 Jun 2011, 11:20
  fluke wrote:
  kripalkavi wrote:
  Are $$x$$ and $$y$$ both correct?
  
  1. $$2x-2y=1$$
  2. $$\frac{x}{y}>1$$
  
  1.
  $$2x-2y=1$$
  $$x-y=\frac{1}{2}$$
  This statement just tells us that x is $$\frac{1}{2}$$ units to the right of y on the number line.
  
  So,
  x=1, y=0.5
  x=-0.5, y=-1.0
  x=0, y=-0.5
  x=0.5, y=0
  
  Not Sufficient.
  
  2.
  $$\frac{x}{y}>1$$
  $$\frac{x}{y}-1>0$$
  $$\frac{x-y}{y}>0$$
  
  For a fraction to be greater than 0, both numerator and denominator must have the same sign
  We have 2 scenarios here:
  if y>0, x-y>0 or x>y
  y=1; x=1.5
  
  if y<0, x-y<0 or x<y
  y=-1, x=-1.5
  
  Not Sufficient.
  
  Combining both;
  
  from statement 1, we know x-y=0.5, which is greater than 0.
  
  Thus, for expression
  $$\frac{x-y}{y}>0$$
  We definitely know that the numerator is +ve. And thus the denominator must also be positive.
  
  x-y>0
  y>0
  x>0
  Sufficient. x and y are both indeed positive.
  
  Ans: "C"
  
  thank you - great explanation!
  
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  Re: GMATprep DS: Number theory   [#permalink] 21 Jun 2011, 11:20
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