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

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Senior Manager
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If x and y are positive, is 3x > 7y? (1) x > y + 4 (2) [#permalink]

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New post 23 Sep 2008, 20:28
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If x and y are positive, is 3x > 7y?
(1) x > y + 4
(2) -5x < -14y
Intern
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Re: DS - Help pl [#permalink]

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New post 24 Sep 2008, 06:43
I think B is sufficient .

St2: -5x < -14y.

By trial and error method,
x = 1 and 2 is not possible as violating 2nd condition.

x = 3 and y = 1.
3x = 9 and 7y = 7 so 9 > 7.

Same, x = 4 and 5 not possible.

for x = 6 and y = 2
3x = 18 and 7y = 14 so 18 > 14.

and so on.. so for any value of x and y .. 3x > 7y. Hence, B is sufficient.
Senior Manager
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Re: DS - Help pl [#permalink]

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New post 24 Sep 2008, 08:42
I agree on B. I solved by:

-5x < -14y
x > (-14/-5)y
x > 2.8y
3(x > 2.8y)
3x > 8.4y
Director
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Re: DS - Help pl [#permalink]

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New post 24 Sep 2008, 10:07
Why not D?
x & y>0

(1)
x>y+4
Let, y = 1, x = 6
6>5
3x > 7y?
18>7

y = 1/2, x has to be greater than 4.5, let x = 6
18 >3.5
suff

(2) agree with the explanations.
Intern
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Re: DS - Help pl [#permalink]

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New post 24 Sep 2008, 10:33
1) NOT sufficient:

if X = 6
if Y = 1
X>y+4
and 3x>7y......18>7

but

if X = 15
if Y = 10
x>y+4
however, 3(15) is not > 7(10)
1 KUDOS received
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Re: DS - Help pl [#permalink]

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New post 24 Sep 2008, 14:03
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Is there any other way to solve it without plugin-in numbers,

I did it like this,
From 2,
(2) -5x < -14y
5x >14y
2.5x > 7y compare it with one in the Q , 3x > 7y?

Is this approach fast?
Senior Manager
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Re: DS - Help pl [#permalink]

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New post 25 Sep 2008, 06:26
IMO, usually, solving algebraically (when you know how to) will be faster than plugging numbers.
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Re: DS - Help pl [#permalink]

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New post 25 Sep 2008, 09:47
jackychamp wrote:
Is there any other way to solve it without plugin-in numbers,

I did it like this,
From 2,
(2) -5x < -14y
5x >14y
2.5x > 7y compare it with one in the Q , 3x > 7y?

Is this approach fast?



For the first statement, my approach will be
x > y + 4
Hence, 3x > 3y + 12
Now, the above sentence will always satisfy 3x > 7y provided 12 is always more than 4y. But, that will not always be true. For y = 1 and 2, this will be true. But, for y = 4, 12 will become less than 4y. Hence, stmt 1 is not sufficient.
Senior Manager
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Posts: 304
Re: DS - Help pl [#permalink]

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New post 25 Sep 2008, 09:51
scthakur wrote:
jackychamp wrote:
Is there any other way to solve it without plugin-in numbers,

I did it like this,
From 2,
(2) -5x < -14y
5x >14y
2.5x > 7y compare it with one in the Q , 3x > 7y?

Is this approach fast?



For the first statement, my approach will be
x > y + 4
Hence, 3x > 3y + 12
Now, the above sentence will always satisfy 3x > 7y provided 12 is always more than 4y. But, that will not always be true. For y = 1 and 2, this will be true. But, for y = 4, 12 will become less than 4y. Hence, stmt 1 is not sufficient.


Good way to simplify....thanks
Re: DS - Help pl   [#permalink] 25 Sep 2008, 09:51
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