Equatons: Are x and y both positive ?

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Equatons: Are x and y both positive ?

by II » Sat Aug 23, 2008 6:21 am
Please illustrate your logic behind the answer.

Thanks.
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by eccentric » Sat Aug 23, 2008 8:11 am
Both statement together is sufficient to answer the question

1. 2x-2y = 1 ---> there are 2 possibility for this to be correct both +ve or both -ve such that their difference is 1/2 so A,D ruled out

2. x/y > 1 ---> would again give two choice where x and y both must either be +ve or -ve so B ruled out as well

if we combine both; the difference must be +ve hence it can be concluded that both x and y must be +ve

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by rishi235 » Sat Aug 23, 2008 9:23 pm
Hi...
i agree to rule out A & B...with the same logic given my eccentric

Now combining the 2 statements...
X & Y...both can be -ve...

Lets take an example...
X = -5/2 &
Y = -3

From Stmt 1)..
X-Y = 1/2
-5/2 - (-3) = -5/2+3 = 1/2 ...Satisfies the condition (1) & also X>Y..so satisfies the condition (2)...

So IMO ans is E....as we cannot tell whether both r +ve or both r -ve...

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by 001 » Sun Aug 24, 2008 1:23 am
Hi, my ans is C.

Statement 1) 2x-2y=1
This statement tells x is more than y, so we can choose any value of x and y (+, -, and 0) to complete this statement. -- Insufficient.

Statement 2) (x/y) > 1
if x>0, then y >0
if x<0, then y<0
Insufficent.

taking boht statements together;
we knows that 2x-2y=1, so x=(2y+1)/2 -------(1)
and (x/y) > 1 ---------(2)

thus, (2y+1)/2y >1

[(2y+1)/2y]-1>0

1/2y >0

so we got y>0, and x must be positive to go with (x/y) > 1

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by pepeprepa » Sun Aug 24, 2008 1:37 am
rishi you are almost totally true you just have to check that
"-5/2 - (-3) = -5/2+3 = 1/2 ...Satisfies the condition (1) & also X>Y "
but this does not satisfy the condition 1) (-5/2) / (-3) = 5/6<1

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by II » Sun Aug 24, 2008 2:28 am
Thank for the input guys !

Lets ask ourselves what this question is aiming to test:
I think from the Q-stem "Are x and y both positive?" we know that this question will in some way test our knowledge of number properties (positives/negatives). so we need to bear this in mind when working through the problem.
And also from the statements we know that it will also test our ability to work with equations and inequalities.
I think this initial analysis is helpful when jumping into a problem, since it enables me to determine my approach to solve the problem.

I think it is fairly easy to rule out choices: A, B, and D. Each statement by itself is not sufficient.
So this leaves C and E.

(1) and (2) together:

(1) tells us: 2x-2y=1 --> x-y=1/2 --> x=y+1/2 (this means that x will always be greater then y)
(2) tells us that x/y > 1 --> we know from this that both x and y have to be the same sign (either pos or neg) otherwise the number will be negative.
We can also rearrange the inequality. Remember the golden rule with inequalities ... if you multiply/divide by a negative number, then the sign flips. Note: we dont know yet if y (the denominator) is neg or pos ... so we have to consider both cases.
If y is pos ... then x/y > 1 becomes x > y (we keep the same sign)
If y is neg ... then x/y > 1 becomes x < y (we flip the sign)

So in summary:
from (1) we know that x has to be greater than y (x>y) because x=y+1/2
from (2) we know that if x is greater than y, then both x and y are positive

So C is SUFF.

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by rishi235 » Sun Aug 24, 2008 6:41 am
Oh i didn't even bother to check the 2nd condition...
Thanks guys 4 all the discussions & pepeprepa thanks 4 correcting me...
:)

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by shaden » Thu Nov 06, 2008 7:31 am
I am still confused... I believe both statements are insufficent, GMAT says it is solveable with both. My logic is as follows: Statement 2 says X > Y. Therefore, we can plug #'s into statement one.

First I used X =3, Y=2.5 --> 2(3)-2(2.5) ---> 6-5 =1, X and Y positive work.

Now lets try some negative numbers. X=-2, Y=-2.5 ---> 2(-2) - 2(-2.5) ---> -4 +5 = 1

We can solve it with both x and y positive, and x and y negative, therefore we dont know if x and y are both positive.... I said E, they say C.

Can anyone explain where I went wrong?

Thanks,

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by EricLien9122 » Thu Nov 06, 2008 10:36 am
Shaden:
x=-2, y=-2.5 will not satisfy statement 2.

-2/-2.5 > 1?

Please correct me if I made a mistake.

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by Fab » Thu Nov 06, 2008 1:58 pm
Hey guys, just one question:

What happen if X=-0.5 and Y=-1

Condition1: 2X - 2Y =1 -----> -1 - 2(-1) = 1 OK
Condition 2: x/y>1 -------> -0.5 > -1 OK

Both could be negative either, so why not E ??

THANKS.

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by shaden » Thu Nov 06, 2008 4:07 pm
Your right... I must be misunderstanding some fundamental rule of math. I thought you could go from (x/y) > 1 --> x >y (by multiplying both sides by y).

Therefore, -2 > -2.5, but your way is correct....

Why am I not allowed to convert (x/y) > 1 --> x > y?

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by eracnos » Thu Nov 06, 2008 7:35 pm
shaden wrote:Your right... I must be misunderstanding some fundamental rule of math. I thought you could go from (x/y) > 1 --> x >y (by multiplying both sides by y).

Therefore, -2 > -2.5, but your way is correct....

Why am I not allowed to convert (x/y) > 1 --> x > y?
Because when u multiply on a negative number, u should flip the sign ( <,>)
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by 4meonly » Sun Nov 09, 2008 10:48 am
Fab wrote:Hey guys, just one question:

What happen if X=-0.5 and Y=-1

Condition1: 2X - 2Y =1 -----> -1 - 2(-1) = 1 OK
Condition 2: x/y>1 -------> -0.5 > -1 OK

Both could be negative either, so why not E ??

THANKS.
You cannot take X=-0.5 and Y=-1 because (-0,5)/(-1)=0,5 which is not more that 1 according to (2)

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by rmpaes » Sun Nov 09, 2008 9:33 pm
If I can't solve an inequality with math, I try to plug in numbers such as -2,-½,0,½,2 in a table.
(1)2x - 2y = 1
(2)x/y > 1

Statement 1: 2x - 2y = 1
On the left side of the table we put the the x values and across the top we put useful test cases.
We can manipulate the equation so we can solve for y.
(2x - 1)/2 = y

x_______y_________ Are both x and y positive?
-2______-5/2_________no
-½_____-1___________no
0_______½ __________no
½_______0___________no
2_______3/2__________yes

From the table above we can see that there are times when x and y are not positive and there is a time when it is. Therefore it is not decisive.
We conclude Statement (1) is not sufficient.

Statement 2: x/y > 1

x_______y_________ Are both x and y positive?
-2______-1___________no
-½_____-1/4_________ no
0_______X ___________X
½______1/4__________no
2_______ 1__________yes

From the table above we can see that there are times when x and y are not positive and there is a time when it is. Therefore it is not decisive.
We conclude Statement (2) is not sufficient.

To combine the statements we substitute in the y value in statement 1 in statement 2 like this:
From Statement1:
y= (2x - 1)/2
Plug in y value into Statement2:
x/[(2x-1)/2] > 1=
x/[x-½] > 1
We construct another table using the values in table 1 to see if they meet the condition in statement 2.

x_______y_______x/[x-½] > 1?___ Are both x and y positive?
-2______-5/2_________4/5-no____________no
-½_____-1___________1/2-no___________no
0_______½ ___________X___________no
½_______0____________X___________no
2_______3/2_________4/3-yes__________yes

The only set of x and y's in our table that meet criteria 1 and 2 are x = 2 and y = 3/2. Both of them are positive.
We conclude that both statements together are sufficient.
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Re: Equatons: Are x and y both positive ?

by logitech » Sun Nov 09, 2008 9:49 pm
II wrote:Please illustrate your logic behind the answer.

Thanks.
1) x = y + 1/2 OKAY WE KNOW WHO IS THE DADDY! X>Y

2) X/Y > 1 HERE IS THE TRAP!

(Y+1/2) / Y > 1

OR 1 + Y/2 > 1 hmmm...so, Y/2>0 ---> Y>0

Remember X > Y

So these guys are both POSITIVE


Thing to remember:

X/Y > 1

X/Y - 1 > 0 Learn this!

(X-Y)/Y > 0

Y < 0 and X-Y < 0

OR

Y > 0 and X-Y> 0
LGTCH
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