It is currently 24 Jun 2017, 14:26

### 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

# jdeleted

Author Message
TAGS:

### Hide Tags

Intern
Joined: 18 Sep 2010
Posts: 4

### Show Tags

02 Mar 2011, 19:18
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 100% (00:22) wrong based on 9 sessions

### HideShow timer Statistics

jdeleted

Last edited by vivatran on 05 Nov 2012, 21:12, edited 1 time in total.
Director
Status: Impossible is not a fact. It's an opinion. It's a dare. Impossible is nothing.
Affiliations: University of Chicago Booth School of Business
Joined: 03 Feb 2011
Posts: 900

### Show Tags

02 Mar 2011, 19:25
1
KUDOS
This is very tricky. E. Just guessing. well it cant be B since you need to define what is line r and what is line s. And information IMO is not sufficient.
Manager
Joined: 18 Oct 2010
Posts: 90

### Show Tags

02 Mar 2011, 19:57
1
KUDOS
line r: y=ax+b
line s: g=bx+c
is b<c?
statement 1: intersection of r and s is negative or y<0 and ax+b=cx+d <=> (a-c)x=d-b
x=(d-b)/(a-c) <0
statement2: the slope of r < the slop of s or a<c ,
we dont know about b and d thus insufficient,
1+2.
(d-b)/(a-c)<0 so or d-b< o and a-C> o or (d-b)>o or (a-c)< o
from statement 2 we know that a<c thus eliminate the condision that (d-b)<0 and (a-c)>0
so sufficient
C
Director
Status: Impossible is not a fact. It's an opinion. It's a dare. Impossible is nothing.
Affiliations: University of Chicago Booth School of Business
Joined: 03 Feb 2011
Posts: 900

### Show Tags

02 Mar 2011, 20:47
1
KUDOS
I came up with 2 cases.

Case 1 intercept(r) > intercept(s)

Case II intercept(r) < intercept(s)

I am not sure if they are considering the +/- or the modulus of intercept. In case it is magnitude (modulus) of the intercept the answer is E. But consider the sign of the intercept, the answer is C.

Hope the helps!
Attachments

LINES.jpg [ 17.2 KiB | Viewed 1266 times ]

Intern
Joined: 18 Sep 2010
Posts: 4

### Show Tags

03 Mar 2011, 01:11
deleted

Last edited by vivatran on 05 Nov 2012, 23:17, edited 1 time in total.
Manager
Joined: 14 Feb 2011
Posts: 194

### Show Tags

03 Mar 2011, 03:03
1
KUDOS
shuwa wrote:
diebeatsthegmat wrote:
line r: y=ax+b
line s: g=bx+c
is b<c?
statement 1: intersection of r and s is negative or y<0 and ax+b=cx+d <=> (a-c)x=d-b
x=(d-b)/(a-c) <0
statement2: the slope of r < the slop of s or a<c ,
we dont know about b and d thus insufficient,
1+2.
(d-b)/(a-c)<0 so or d-b< o and a-C> o or (d-b)>o or (a-c)< o
from statement 2 we know that a<c thus eliminate the condision that (d-b)<0 and (a-c)>0
so sufficient
C

line r: y=ax+b
line s: g=bx+c

I think

line r: y=ax1+b
line s: g=bx2+c
if
line r: y=ax+b
line s: g=bx+c

it means that 2 lines intersect together at a common point?

Shuwa - I think diebeatsthegmat has a typo in his solution when he defines the lines the first time line r: y=ax+b, line s: g=bx+c .

In reality, he has defined the line s as y= cx+d as can be seen in his calculation for statement 1.

Also, defining lines as y=ax+b and y= cx+d, does not imply that they intersect at the same point, it is just a general form of equation for any line that has a defined slope (a and c for two lines here) and a defined y intercept (b and d for lines here).
Manager
Joined: 14 Feb 2011
Posts: 194

### Show Tags

03 Mar 2011, 03:07
1
KUDOS
gmat1220 wrote:
I came up with 2 cases.

Case 1 intercept(r) > intercept(s)

Case II intercept(r) < intercept(s)

I am not sure if they are considering the +/- or the modulus of intercept. In case it is magnitude (modulus) of the intercept the answer is E. But consider the sign of the intercept, the answer is C.

Hope the helps!

The question simply asks if y intercept of r is less than y intercept of s, so we need to take the value of the intercept and not the absolute value.

A general line would be y=mx+c where y intercept is c which can be negative or positive or zero, so we need to just compare the actual value and that would involve incorporating the sign of the intercept as well.

I think diebeatsthegmat has the best approach to solve this without any confusion whatsoever.
Director
Status: Impossible is not a fact. It's an opinion. It's a dare. Impossible is nothing.
Affiliations: University of Chicago Booth School of Business
Joined: 03 Feb 2011
Posts: 900

### Show Tags

03 Mar 2011, 06:24
1
KUDOS
Thanks beyondgmatscore. Yeah all ordered now. So I change this statement. Answer is C - whether you use visual approach or algebra.

beyondgmatscore wrote:
gmat1220 wrote:
I came up with 2 cases.

Case 1 intercept(r) > intercept(s)

Case II intercept(r) > intercept(s)

I am not sure if they are considering the +/- or the modulus of intercept. In case it is magnitude (modulus) of the intercept the answer is E. But consider the sign of the intercept, the answer is C.

Hope the helps!

The question simply asks if y intercept of r is less than y intercept of s, so we need to take the value of the intercept and not the absolute value.

A general line would be y=mx+c where y intercept is c which can be negative or positive or zero, so we need to just compare the actual value and that would involve incorporating the sign of the intercept as well.

I think diebeatsthegmat has the best approach to solve this without any confusion whatsoever.
Manager
Status: TIME FOR 700+
Joined: 06 Dec 2010
Posts: 204
Schools: Fuqua
WE 1: Research in Neurology
WE 2: MORE research in Neurology

### Show Tags

04 Mar 2011, 12:18
I'm pretty sure I marked C without doing math on this, just drawing a few sketches knowing that:

I) the slopes can be anything, the intercepts can be anything, just that their points together are in Quadrant III (-x,-y); not sufficient

II) slope R is greater than slope S, just the steepness of the line, you can put the lines anywhere in the plane or any quadrant; not sufficient

Together, intercepting in Quadrant III and bigger slope? visually it just seems that it would have to hit higher all the time on the Intercept? am i wrong to think of it this way?
_________________

Back to the grind, goal 700+

Manager
Joined: 18 Oct 2010
Posts: 90

### Show Tags

05 Mar 2011, 07:59
beyondgmatscore wrote:
gmat1220 wrote:
I came up with 2 cases.

Case 1 intercept(r) > intercept(s)

Case II intercept(r) < intercept(s)

I am not sure if they are considering the +/- or the modulus of intercept. In case it is magnitude (modulus) of the intercept the answer is E. But consider the sign of the intercept, the answer is C.

Hope the helps!

The question simply asks if y intercept of r is less than y intercept of s, so we need to take the value of the intercept and not the absolute value.

A general line would be y=mx+c where y intercept is c which can be negative or positive or zero, so we need to just compare the actual value and that would involve incorporating the sign of the intercept as well.

I think diebeatsthegmat has the best approach to solve this without any confusion whatsoever.

yep, thats also correct in some cases, anyways i am a girl! i am a her, not a him lol
Re: xy-plane- Slope   [#permalink] 05 Mar 2011, 07:59
Display posts from previous: Sort by