Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 25 May 2017, 10:58

### 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

# M18#13

Author Message
Manager
Joined: 07 Jan 2008
Posts: 87
Followers: 2

Kudos [?]: 189 [0], given: 1

### Show Tags

07 Feb 2009, 06:22
3
This post was
BOOKMARKED
Does line Ax + By + C = 0 (A is not 0) intersect the x-axis on the negative side?

1. BA < 0
2. AC > 0

[Reveal] Spoiler: OA
B

Source: GMAT Club Tests - hardest GMAT questions
Manager
Joined: 07 Jan 2008
Posts: 87
Followers: 2

Kudos [?]: 189 [0], given: 1

### Show Tags

07 Feb 2009, 06:24
I solved this as follows:

Ax+By+C=0

therefore y = -(A/B) x + (C/B)

When y = 0, x = (BC)/A

therefore you need both (1) and (2) to check if x is negative.
Intern
Joined: 10 Feb 2009
Posts: 1
Followers: 0

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

### Show Tags

10 Feb 2009, 06:02
I'm pretty sure that's wrong..... (A/B)X + (C/B) = Y
Y=0 --> X = -C/A

Then we need only (b) to answer the question...

Who is right here?
Intern
Joined: 02 Jun 2008
Posts: 4
Followers: 0

Kudos [?]: 2 [0], given: 0

### Show Tags

10 Feb 2009, 11:51
tzvister wrote:
I'm pretty sure that's wrong..... (A/B)X + (C/B) = Y
Y=0 --> X = -C/A

Then we need only (b) to answer the question...

Who is right here?

absolutely you are correct,

the general form of equn of line is x/a + y/b = 1 ( a and b are intercept of x and y axis respectively) comparing it to the given equn a = -C/A, hence from stmt 2 itself v can answer the question
SVP
Joined: 07 Nov 2007
Posts: 1806
Location: New York
Followers: 38

Kudos [?]: 931 [0], given: 5

### Show Tags

11 Feb 2009, 07:53
topmbaseeker wrote:
I solved this as follows:

Ax+By+C=0

therefore y = -(A/B) x + (C/B)

When y = 0, x = (BC)/A
therefore you need both (1) and (2) to check if x is negative.

your approach is correct.. you did mistake in the equation.
Y= -(A/B)x+(-C/B) =0

--> -Ax-C =0 -->x intercept = -C/A

_________________

Smiling wins more friends than frowning

Manager
Joined: 20 Oct 2009
Posts: 111
Followers: 8

Kudos [?]: 43 [0], given: 0

### Show Tags

24 Dec 2009, 07:17
X=(-By-C)/A
when y=0, X=-C/A, B is sufficient.

This question is not hard, just be careful x-intersection means y=0.

topmbaseeker wrote:
Does line Ax + By + C = 0 (A is not 0) intersect the x-axis on the negative side?

1. BA < 0
2. AC > 0

* Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
* Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
* BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
* EACH statement ALONE is sufficient
* Statements (1) and (2) TOGETHER are NOT sufficient

[Reveal] Spoiler: OA
B

Source: GMAT Club Tests - hardest GMAT questions

_________________

Dream the impossible and do the incredible.

Live. Love. Laugh.

Senior Manager
Joined: 01 Nov 2010
Posts: 289
Location: India
Concentration: Technology, Marketing
GMAT Date: 08-27-2012
GPA: 3.8
WE: Marketing (Manufacturing)
Followers: 10

Kudos [?]: 80 [0], given: 44

### Show Tags

01 Jan 2011, 03:28
Hey all,

my answer is different from all.
i ll go with E.
here is my explanation:
ax+by+c=0;
or, y= -(c+ax)/b; or, y= -(c^2+acx) / bc ; (on multiplying C )
statement 1 : BA <0 ;not applicable
statement 2 : AC >0 ; applicable;
in the eq, c^2 is +ve
ac is +ve
bc not known;
on comparing with standard (y=mx+c)
m = -ac/bc ; since bc is unknown, you cant predict the slope of line ,
hence you cant say, whether it will cut Y-Axis or not.

next;

ax+by+c=0;
or, y= -(c+ax)/b; or, y= -(bc+bax) / b^2 ; (on multiplying B )
statement 1 : BA <0 ; applicable
statement 2 : AC >0 ; not applicable;
in the eq, b^2 is +ve
ab is -ve
bc not known;
on comparing with standard (y=mx+c)
m = -ba/b^2 ; since b^2 is +ve, the slope of line is +ve .
constant (i.e c in standard eq) = bc; not known
since, bc is not known, whether it will be on +ve or -ve Y-axis.
so, you cant say that whether this line will cut the X-axis or not.

please make me correct if m wrong.

hence you cant say, whether it will cut Y-Axis or not.
_________________

kudos me if you like my post.

Attitude determine everything.
all the best and God bless you.

Math Expert
Joined: 02 Sep 2009
Posts: 38870
Followers: 7731

Kudos [?]: 106111 [3] , given: 11607

### Show Tags

01 Jan 2011, 08:16
3
KUDOS
Expert's post
321kumarsushant wrote:
Hey all,

my answer is different from all.
i ll go with E.
here is my explanation:
ax+by+c=0;
or, y= -(c+ax)/b; or, y= -(c^2+acx) / bc ; (on multiplying C )
statement 1 : BA <0 ;not applicable
statement 2 : AC >0 ; applicable;
in the eq, c^2 is +ve
ac is +ve
bc not known;
on comparing with standard (y=mx+c)
m = -ac/bc ; since bc is unknown, you cant predict the slope of line ,
hence you cant say, whether it will cut Y-Axis or not.

next;

ax+by+c=0;
or, y= -(c+ax)/b; or, y= -(bc+bax) / b^2 ; (on multiplying B )
statement 1 : BA <0 ; applicable
statement 2 : AC >0 ; not applicable;
in the eq, b^2 is +ve
ab is -ve
bc not known;
on comparing with standard (y=mx+c)
m = -ba/b^2 ; since b^2 is +ve, the slope of line is +ve .
constant (i.e c in standard eq) = bc; not known
since, bc is not known, whether it will be on +ve or -ve Y-axis.
so, you cant say that whether this line will cut the X-axis or not.

please make me correct if m wrong.

hence you cant say, whether it will cut Y-Axis or not.

Official answer is B, not E.

Does line Ax + By + C = 0 (A is not 0) intersect the x-axis on the negative side?

$$ax+by+c=0$$ is equation of a line. Note that the line won't have interception with x-axis when $$a=0$$ (and $$c\neq{0}$$): in this case the line will be $$y=-\frac{c}{b}$$ and will be parallel to x -axis.

Now, in other cases (when $$a\neq{0}$$) x-intercept of a line will be the value of $$x$$ when $$y=0$$, so the value of $$x=-\frac{c}{a}$$. Question basically asks whether this value is negative, so question asks is $$-\frac{c}{a}<0$$? --> is $$\frac{c}{a}>0$$? --> do $$c$$ and $$a$$ have the same sign?

(1) BA < 0. Not sufficient as we can not answer whether $$c$$ and $$a$$ have the same sign.
(2) AC > 0 --> $$c$$ and $$a$$ have the same sign. Sufficient.

Check more on this topic here: math-coordinate-geometry-87652.html

Hope it helps.
_________________
Manager
Joined: 21 Nov 2010
Posts: 130
Followers: 0

Kudos [?]: 5 [0], given: 12

### Show Tags

07 Jan 2012, 17:40
I missed step 1 which is put it in slope form...messed me up.
Intern
Joined: 30 Aug 2012
Posts: 8
Followers: 0

Kudos [?]: 3 [0], given: 12

### Show Tags

06 Jan 2013, 15:55
I hope this helps
The question is if x intercept is negative, which can be found by solving for x, assuming y=0, x= -(C/A)
situation 1 says BA<0 - no relevance to whether x intercept is -ve or =ve - not sufficient
2) AC>0, would mean both a and c are either =ve or -ve in either situation X intercept is -ve therefore B is sufficient,
also A not equal to zero would mean C cannot be zero as well. so this covers all grounds i guess.

Re: M18#13   [#permalink] 06 Jan 2013, 15:55
Display posts from previous: Sort by

# M18#13

Moderator: Bunuel

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.