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Re: 3x > 7y? [#permalink]
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15 Oct 2009, 16:06
windycu wrote: I think put it in simple
To define statement (1) is enough or not  > it can say yes or no or Both. If it can say both > then it is not sufficient right?
First if y = 1 and x =5 then 15 > 7 yes Second if y = 10 and x = 15 then 45 > 70 no
So for the positive number x and y, we cannot sure that this Q 3x>7y or not? So statement (1) alone is not enough
For statement (2) 5X < 14Y  > Can rewrite to get 5X>14Y or 2.5X>7Y. Since x and y are positive and 2.5x>7y, so 3x>7y for sure.
So I prefer B krub.
Please correct me if I am worng krub. This cannot be true, because according to (1) x > y+4. In that scenario x = y+4.



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Re: 3x > 7y? [#permalink]
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15 Oct 2009, 16:09
Those who claim that (1) in insufficient should provide a case of x and y where 3x is not greater than 7y!



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Re: 3x > 7y? [#permalink]
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15 Oct 2009, 16:19
use this 2 sets of data and then think what the answer ...
1) use y = 20 so x is 24 ..
2) has to be true .....
so answer is D , d as Denver.



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Re: 3x > 7y? [#permalink]
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15 Oct 2009, 16:28
S1 is insufficient and OA has to be wrong. correct ans is B
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Re: 3x > 7y? [#permalink]
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15 Oct 2009, 16:30
pittgreek wrote: use this 2 sets of data and then think what the answer ...
1) use y = 20 so x is 24 ..
2) has to be true .....
so answer is D , d as Denver. what about y=1 x=6 3x=18, 7y=7 3x > 7y y=20 x=25 3x=75, 7y=140 3x < 7y
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Last edited by srini123 on 15 Oct 2009, 19:12, edited 1 time in total.



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Re: 3x > 7y? [#permalink]
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15 Oct 2009, 16:58
srini123 wrote: S1 is insufficient and OA has to be wrong.
correct ans is B Prove statement 1 is insufficient by showing two cases of x and y that are not consistent...



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Re: 3x > 7y? [#permalink]
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15 Oct 2009, 19:09
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mendelay wrote: srini123 wrote: S1 is insufficient and OA has to be wrong.
correct ans is B Prove statement 1 is insufficient by showing two cases of x and y that are not consistent... mendeley see my reply to pittgreek post. I am reposting here what I posted there: *** edited by srini123 *** what about y=1 x=6 3x=18, 7y=7 3x > 7y y=20 x=25 3x=75, 7y=140 3x < 7y *** edited by srini123 *** I am thinking S1 is not giving us one single solution, unless I am seriously missing something here
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Re: 3x > 7y? [#permalink]
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15 Oct 2009, 20:26
You're right. Kudos! Thanks! I took a practice test and answered B, but the OA said D, and I could not fully understand the explanation some gave on this topic trying to prove 1 as sufficient...



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Re: 3x > 7y? [#permalink]
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16 Oct 2009, 07:01
mendelay wrote: You're right. Kudos! Thanks! I took a practice test and answered B, but the OA said D, and I could not fully understand the explanation some gave on this topic trying to prove 1 as sufficient... Thanks for the Kudos
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Re: 3x > 7y? [#permalink]
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03 Sep 2011, 10:26
I am getting B as answer.
If x and y are positive, is 3x > 7y? (1) x > y + 4 (2) 5x < 14y
let y= 1 then x could be 6, in this case 3x > 7y as 18 is > 7. let y = 10 then x could be 15. in this case 3x < 7y as 45 is less than 70. so statement 1 is insufficient.
statement 2 says 5x < 14 y, so 5x > 14 y or x > 14/5 Y. now whatever positive value you put for x and y 3x always will be greater than 7y.
so answer should be B



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Re: 3x > 7y? [#permalink]
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04 Sep 2011, 11:31
I vote for B OA is A??!!! for statment 1: x > y +4 3x > 3y+12 for 3x >7y 3y+12 > 7y y<4............so only for +ve and y < 4 we get 3x > 7y, for y > 4 , we get 3x<=7y Nw Stment 2: 5x < 14y 5x > 14y (inverse the sign) 2.5x > 7y (divide by 2) add 0.5 x on both sides 3x > 7y + 0.5x which implies 3x >7y (since 0.5x is +ve) So Stment 2 is sufficient ...Choice B
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Re: 3x > 7y? [#permalink]
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04 Sep 2011, 11:40
I got B too...am i improving or is it just an easy question..hehehe..lol



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Re: 3x > 7y? [#permalink]
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05 Sep 2011, 23:30
for 1, Plug in numbers for x,y are 7,2 for 2, Plug in numbers for x,y are 6,2 +1 for D
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