Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 19 Jun 2013, 05:31

# If x and y are positive, is3x>7y? (1) x>y+4 (2)

Author Message
TAGS:
Senior Manager
Joined: 03 Mar 2010
Posts: 460
Followers: 3

Kudos [?]: 76 [0], given: 21

If x and y are positive, is3x>7y? (1) x>y+4 (2) [#permalink]  16 Jun 2011, 01:45
00:00

Question Stats:

8% (00:00) correct 91% (01:35) wrong based on 12 sessions
If x and y are positive, is3x>7y?
(1) x>y+4
(2) -5x<-14y

How to solve algebraically?
[Reveal] Spoiler: OA

_________________

My dad once said to me: Son, nothing succeeds like success.

Current Student
Joined: 26 May 2005
Posts: 575
Followers: 18

Kudos [?]: 79 [0], given: 13

Re: Inequality Approach [#permalink]  16 Jun 2011, 03:21
If x and y are positive, is3x>7y?
(1) x>y+4
(2) -5x<-14y

How to solve algebraically?

I think the OA is B
the question can be rephrases as x/y>7/3
X>y+4
pick up number
y = 1 x = 5
5/1>7/3 yes
now Y= 100
X = 106

X/Y is not greater than 7/3

hence insufficient

st2: -5x<-14y
questions can be rephrases as x>2.33Y

-5x<-14y( multiply -ve on both sides and flip the sign)
5x>14Y
X>14/5Y
X>2.8Y.. hence sufficient
CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2762
Location: Malaysia
Concentration: Marketing, Entrepreneurship
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Followers: 126

Kudos [?]: 644 [0], given: 222

Re: Inequality Approach [#permalink]  16 Jun 2011, 03:44
the OA should be B

to solve it graphically just draw all the three lines. you would notice that the 3x=7y and 5x=14y would intersect at 0. in the first quadrant the 5x=14y would be more steeper.

since 5x>14 y => the area between x axis and 5x=14y.

to check whether it satisfies 3x>7y : since the 5x=14y lies below 3x=7y => 3x>7y would always be true.
Hence 2nd statement is sufficient.

1st statement is not sufficient. x>y+4 intersects at 7,3 with 3x=7y....after the intersection it the answer is NO.
before the intersection the answer is Yes.
Hence not sufficient.

_________________

Fight for your dreams :For all those who fear from Verbal- lets give it a fight

Money Saved is the Money Earned

Jo Bole So Nihaal , Sat Shri Akaal

Find out what's new at GMAT Club - latest features and updates

Gmat test review :
670-to-710-a-long-journey-without-destination-still-happy-141642.html

Manager
Joined: 03 Jun 2010
Posts: 140
Location: Dubai, UAE
Schools: IE Business School, Manchester Business School, HEC Paris, Rotterdam School of Management, Babson College
Followers: 2

Kudos [?]: 4 [0], given: 4

Re: Inequality Approach [#permalink]  16 Jun 2011, 03:58
If this was a problem solving question, I would have gotten a wrong reply, having said that since this is DS we don't need to solve we just need to know we can solve. Let's look at the question now

We know x>7y/3

Statement 1
Sufficient because we put in a value of x and get a value of y. Let's assume from an understanding point of view x=4
x>y+4
y<0
hence x>0
Sufficient
Statement 2
x>14y/5
Assume x=5
y=25/14

In this case 3x not greater than 7y
Sufficient

D

Posted from GMAT ToolKit

Posted from GMAT ToolKit
VP
Status: There is always something new !!
Affiliations: PMI,QAI Global,eXampleCG
Joined: 08 May 2009
Posts: 1395
Followers: 8

Kudos [?]: 86 [0], given: 10

Re: Inequality Approach [#permalink]  16 Jun 2011, 04:08
3x>7y
critical points are x = 7 and y = 3. Considering first quadrant.
for 0< > x<7 and 0<y<3 is the region for which 3x > 7y.

a x=6,y= 1 gives 3x > 7y and for x=10, y= 5 the sign changes.
Not sufficient.

b 5x > 14y

2<x < 14 and 0<y<5 satisfies.

checking for region 2<x<7 and 0<y<3 gives 3x > 7y always.

Hence B.
_________________

Visit -- http://www.sustainable-sphere.com/
Promote Green Business,Sustainable Living and Green Earth !!

Manager
Joined: 11 Apr 2011
Posts: 110
Followers: 0

Kudos [?]: 15 [0], given: 16

Re: Inequality Approach [#permalink]  28 Jun 2011, 09:45
1. not sufficient.
2. 5x > 14y => 2.5x > 7y => 3x > 7y, hence sufficient.

B.
Manager
Joined: 10 Jan 2010
Posts: 117
GPA: 4
WE: Programming (Computer Software)
Followers: 0

Kudos [?]: 29 [0], given: 33

Re: Inequality Approach [#permalink]  11 Jul 2011, 18:11
Can somebody please solve this algebrically?
_________________

-If you like my post, consider giving KUDOS

Manager
Joined: 04 Apr 2010
Posts: 176
Followers: 1

Kudos [?]: 22 [0], given: 31

Re: Inequality Approach [#permalink]  11 Jul 2011, 20:20
1. try following values for (x,Y)
(6, 1) -----------Yes
(106, 100)--------No

so 1 is not sufficient. B is the answer.
_________________

Consider me giving KUDOS, if you find my post helpful.
If at first you don't succeed, you're running about average. ~Anonymous

Intern
Joined: 04 May 2011
Posts: 9
Followers: 0

Kudos [?]: 1 [0], given: 7

Re: Inequality Approach [#permalink]  11 Jul 2011, 20:57
From 3x>7y
x > 7/3y

From (1) x > y+4

To prove (1) is enough, we must prove y+4 is always greater than or equal to 7/3y

y+4 >= 7/3y,
we get y<= 3
Therefore, only when y<=3 that y+4 >= 7/3y
Not every value of y make y+4 >= 7/3y

Therefore (1) is insufficient

Doing the same thing with (2), we know (2) is sufficient
Manager
Joined: 14 Apr 2011
Posts: 203
Followers: 2

Kudos [?]: 17 [0], given: 19

Re: Inequality Approach [#permalink]  14 Jul 2011, 06:26
OA must be B.

Given: x, y >0 Is 3x>7y?
Rephrase Is x/y > 7/3? (since both x and y are +ve)

St 1: x > (y+4) => x/y > 1+4/y. Insuff. as dependent on value of y. if y =1, Yes and if y=4, No.
St 2: -5x < -14y => x/y >14/5. Suff.

B. took me more than 2 mins
_________________

Looking for Kudos

Senior Manager
Joined: 08 Nov 2010
Posts: 431
Followers: 6

Kudos [?]: 25 [0], given: 161

Re: Inequality Approach [#permalink]  31 Jul 2011, 11:06
There are more ways to solve stat 1:
either multiple both sides by 3:

you will get 3x>3y+12. combining both - you will notice that you need to check whether 3y+12>7y you can see immediately that you cannot know. plugging 1 and 10 showing that quickly.

another way:
x>y+4 we can also say x=y+4+c
so the new question can be: 3(y+4+c)>7y?

again - we cannot know. we can play now with c and y for the result we want.
_________________
Re: Inequality Approach   [#permalink] 31 Jul 2011, 11:06
Similar topics Replies Last post
Similar
Topics:
If x and y are positive, is 3x > 7y? (1) x > y + 4 (2) 2 16 Aug 2004, 21:28
If x and y are positive, is 3x>7y 1. x>y+4 2. 2 23 Dec 2004, 09:58
If x and y are positive, is 3x > 7y ? (1) x > y+4 (2) 2 05 May 2006, 23:19
If x and y are positive, is 3x > 7y? (1) x > y + 4 (2) 7 07 Aug 2007, 13:23
If x and y are positive, is 3x > 7y? (1) x > y + 4 (2) 2 22 Mar 2009, 23:36
Display posts from previous: Sort by