It is currently 22 Feb 2018, 20:43

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
Senior Manager
Joined: 05 Jun 2008
Posts: 304
If x and y are positive, is 3x > 7y? (1) x > y + 4 (2) [#permalink]

### Show Tags

23 Sep 2008, 20:28
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If x and y are positive, is 3x > 7y?
(1) x > y + 4
(2) -5x < -14y
Intern
Joined: 02 Sep 2008
Posts: 45
Re: DS - Help pl [#permalink]

### Show Tags

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
Joined: 09 Oct 2007
Posts: 463
Re: DS - Help pl [#permalink]

### Show Tags

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
Joined: 27 Jun 2008
Posts: 538
WE 1: Investment Banking - 6yrs
Re: DS - Help pl [#permalink]

### Show Tags

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
Joined: 22 Oct 2007
Posts: 3
Re: DS - Help pl [#permalink]

### Show Tags

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)
Manager
Joined: 11 Jan 2008
Posts: 54
Re: DS - Help pl [#permalink]

### Show Tags

24 Sep 2008, 14:03
1
KUDOS
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
Joined: 09 Oct 2007
Posts: 463
Re: DS - Help pl [#permalink]

### Show Tags

25 Sep 2008, 06:26
IMO, usually, solving algebraically (when you know how to) will be faster than plugging numbers.
SVP
Joined: 17 Jun 2008
Posts: 1529
Re: DS - Help pl [#permalink]

### Show Tags

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
Joined: 05 Jun 2008
Posts: 304
Re: DS - Help pl [#permalink]

### Show Tags

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
Display posts from previous: Sort by