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GMATprep DS: Number theory [#permalink]
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09 Feb 2007, 03:23
Question Stats:
38% (01:02) correct
63% (00:53) wrong based on 6 sessions
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Please discuss.
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Joined: 12 Jun 2006
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C
(1) 2x  2y = 1 > 2(xy) = 1 > xy = 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.



VP
Joined: 28 Mar 2006
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Should be C
1 is st forward
(2)
means x>y if both are positive
x<y if both are ve



SVP
Joined: 01 May 2006
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(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.
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SVP
Joined: 05 Jul 2006
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from one
xy = 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
(xy) / y >0.............insuff
both together
xy = 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



VP
Joined: 21 Mar 2006
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ggarr wrote: C (1) 2x  2y = 1 > 2(xy) = 1 > xy = 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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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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Re: GMATprep DS: Number theory [#permalink]
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13 Jun 2011, 08:42
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kripalkavi wrote: Are \(x\) and \(y\) both correct?
1. \(2x2y=1\) 2. \(\frac{x}{y}>1\)
1. \(2x2y=1\) \(xy=\frac{1}{2}\) 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. \(\frac{x}{y}>1\) \(\frac{x}{y}1>0\) \(\frac{xy}{y}>0\) 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, xy>0 or x>y y=1; x=1.5 if y<0, xy<0 or x<y y=1, x=1.5 Not Sufficient. Combining both; from statement 1, we know xy=0.5, which is greater than 0. Thus, for expression \(\frac{xy}{y}>0\) We definitely know that the numerator is +ve. And thus the denominator must also be positive. xy>0 y>0 Adding above two expressions, x>0 Sufficient. x and y are both indeed positive. Ans: "C"
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Re: GMATprep DS: Number theory [#permalink]
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21 Jun 2011, 11:20
fluke wrote: kripalkavi wrote: Are \(x\) and \(y\) both correct?
1. \(2x2y=1\) 2. \(\frac{x}{y}>1\)
1. \(2x2y=1\) \(xy=\frac{1}{2}\) 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. \(\frac{x}{y}>1\) \(\frac{x}{y}1>0\) \(\frac{xy}{y}>0\) 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, xy>0 or x>y y=1; x=1.5 if y<0, xy<0 or x<y y=1, x=1.5 Not Sufficient. Combining both; from statement 1, we know xy=0.5, which is greater than 0. Thus, for expression \(\frac{xy}{y}>0\) We definitely know that the numerator is +ve. And thus the denominator must also be positive. xy>0 y>0 Adding above two expressions, x>0 Sufficient. x and y are both indeed positive. Ans: "C" thank you  great explanation!




Re: GMATprep DS: Number theory
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21 Jun 2011, 11:20






