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1 st)
2x-2y=1
x-y=1/2 insuff since x and y can be anything
x=1 and y=1/2 or x= -1/2 and y=-1
2) x/y >1 x and y may be both pos and both negative
-2/-1>1
E for me _________________

1 st) 2x-2y=1 x-y=1/2 insuff since x and y can be anything x=1 and y=1/2 or x= -1/2 and y=-1 2) x/y >1 x and y may be both pos and both negative -2/-1>1 E for me

E for me too ........ used the exact methodology and numbers

Re: DS - x and y positive [#permalink]
11 Jun 2006, 10:58

gidimba wrote:

Are x and y both positive?

1. 2x-2y=1 2. x/y >1

(1) 2x-2y = 1
x = y+ 1/2. x or y, both, could be either - or +.

(2) x/y >1.
x>y but both should be +ves.
x<y but both should be -ves.

togather:
if y = 0.5, x = 1 (and we have x/y = 1/0.5 = 2 which is greater than1).
ok, if y = -0.5, x = 0 (and we have x/y = 0 which isn't greater than1).

Re: DS - x and y positive [#permalink]
11 Jun 2006, 13:55

Professor wrote:

gidimba wrote:

Are x and y both positive?

1. 2x-2y=1 2. x/y >1

(1) 2x-2y = 1 x = y+ 1/2. x or y, both, could be either - or +.

(2) x/y >1. x>y but both should be +ves. x<y but both should be -ves.

togather: if y = 0.5, x = 1 (and we have x/y = 1/0.5 = 2 which is greater than1). ok, if y = -0.5, x = 0 (and we have x/y = 0 which isn't greater than1).

So E is it.

Prof, we cannot take this exemple simply because this exemple is not respecting statment 2. It respects the line equation but not x/y > 1. So it cannot be use to say answer (E) as well.

Re: DS - x and y positive [#permalink]
11 Jun 2006, 14:08

Fig wrote:

Prof, we cannot take this exemple simply because this exemple is not respecting statment 2. It respects the line equation but not x/y > 1. So it cannot be use to say answer (E) as well.

thankx fig, you are correct and i also corrected myself as well....

(1) 2x-2y = 1
x = y+ 1/2. x or y, both, could be either - or +.

(2) x/y >1.
x>y but both should be +ves.
x<y but both should be -ves.

togather:
if y = 0.5, x = 0.5+0.5 = 1 (and we have x/y = 1/0.5 = 2 which is greater than 1).
if y = -1, x = -0.5 (and we have x/y = 1/2 which is incorrect.) therefore, only x>y and only +ve values for x and y work.

2(x-y) = 1 => (x-y) = 1/2 ....... (b) From (a) (x-y) > 0 & from (b) (x-y) = 1/2

i.e 1/2 > 0 => 1 > 2 Not true...

Therefore C cannot be the answer.

your assertion in a is not correct.

if x/y > 1 then you cannot always say that x > y because x and y could be negative too

consider x = -6 and y = -3 x/y > 1 and x < y

You are missing statement 1 in which x>y
1=> x = y+1/2
For any value of y, x will always be greater than y.
Taking your example of y =-3, x = -3+1/2
x = -2.5
Hence x/y>1.
Therefore C.

2(x-y) = 1 => (x-y) = 1/2 ....... (b) From (a) (x-y) > 0 & from (b) (x-y) = 1/2

i.e 1/2 > 0 => 1 > 2 Not true...

Therefore C cannot be the answer.

your assertion in a is not correct.

if x/y > 1 then you cannot always say that x > y because x and y could be negative too

consider x = -6 and y = -3 x/y > 1 and x < y

You are missing statement 1 in which x>y 1=> x = y+1/2 For any value of y, x will always be greater than y. Taking your example of y =-3, x = -3+1/2 x = -2.5 Hence x/y>1. Therefore C.

you are missing the point

I am not disputing the answer (ie C). I was just pointing out that what haas_mba07 asserted (in red font above) is not correct .

ie just considering x/y > 1 , one cannot conclude that x > y.

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