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# If x and y are positive, is 3x > 7y? (1) x > y + 4 (2)

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Manager
Joined: 21 Jul 2009
Posts: 249
Location: New York, NY

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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.
Manager
Joined: 21 Jul 2009
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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!
Manager
Joined: 04 Aug 2009
Posts: 54
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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.
Senior Manager
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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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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.

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.
Manager
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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...
Senior Manager
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15 Oct 2009, 19:09
1
KUDOS
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 ***

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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Manager
Joined: 21 Jul 2009
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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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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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keep uppp...ing the tempo...

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Manager
Joined: 14 Mar 2011
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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.

Manager
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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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Senior Manager
Joined: 16 Feb 2011
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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
Manager
Joined: 13 Apr 2010
Posts: 166
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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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Re: 3x > 7y?   [#permalink] 05 Sep 2011, 23:30

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