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GMATprep DS: Number theory

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GMATprep DS: Number theory [#permalink] New post 09 Feb 2007, 02:23
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

38% (01:02) correct 63% (00:53) wrong based on 6 sessions
Please discuss.
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 [#permalink] New post 09 Feb 2007, 03:16
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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 [#permalink] New post 09 Feb 2007, 04:49
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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 [#permalink] New post 09 Feb 2007, 05:08
(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.
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 [#permalink] New post 09 Feb 2007, 14:31
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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 [#permalink] New post 10 Feb 2007, 03:30
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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Re: GMATprep DS: Number theory [#permalink] New post 13 Jun 2011, 06:28
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] New post 13 Jun 2011, 07:42
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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
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] New post 21 Jun 2011, 10:20
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
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   [#permalink] 21 Jun 2011, 10:20
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