It is currently 23 Mar 2018, 00:06

### 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

# ps:m24 #21

Author Message
Manager
Joined: 05 Jan 2009
Posts: 76

### Show Tags

25 Jul 2009, 14:14
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 1 sessions

### HideShow timer Statistics

How to solve it?
GMATclub test24

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.

Attachments

m24.JPG [ 39.5 KiB | Viewed 1390 times ]

Senior Manager
Joined: 25 Jun 2009
Posts: 291

### Show Tags

25 Jul 2009, 14:50
pmal04 wrote:
How to solve it?
GMATclub test24

IMO its should be C. These are equations for parabola, now trying substituting diff values of 'a' and try to make a figure out of it ( if a= +ve then you will get a figure some thing like this which is attached and if a=-ve then its mirror image)

I hope this helped else let me know I will elaborate.

Cheers
Attachments

untitled.JPG [ 7.66 KiB | Viewed 1369 times ]

GMAT Tutor
Joined: 24 Jun 2008
Posts: 1346

### Show Tags

26 Jul 2009, 08:16
Recall how we find intersection points - we just solve the two equations together. So really you're just trying to determine if ax^2 + b = cx^2 + d has any solutions, or equivalently, if (a - c)x^2 = d - b has any solutions, or perhaps most simply, if x^2 = (d-b)/(a-c) has any solutions. If the right side of this equation is zero or positive, we'll be able to find one or two values of x, respectively. If the right side is negative, there will be no solutions. So we can rephrase the question: is (d-b)/(a-c) > 0?

Since we need to know about both the numerator and denominator of (d-b)/(a-c), neither statement is sufficient alone. Combining the statements, since c = -a, our fraction becomes (d-b)/(2a). From Statement 2, d-b is negative. Still, we don't know if a is positive or negative; if a is positive, the fraction is negative, and if a is negative, the fraction is positive. Insufficient.

Of course, if you recognize that these are equations of parabolas, you can use coordinate geometry principles to solve. S1 tells us that one parabola is rising, one falling. Statement 2 tells us that the first parabola has a higher vertex than the second. Still, we have two possibilities; if the first parabola is rising, and the second falling, they don't intersect. If the first parabola is falling and the second rising, they do. Insufficient.

All that said, the wording of the question is very bad - the equations in the question clearly do not define 'lines'; they define 'curves'.
_________________

GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

Director
Joined: 05 Jun 2009
Posts: 725
WE 1: 7years (Financial Services - Consultant, BA)

### Show Tags

27 Jul 2009, 05:38
nitishmahajan wrote:
pmal04 wrote:
How to solve it?
GMATclub test24

IMO its should be C. These are equations for parabola, now trying substituting diff values of 'a' and try to make a figure out of it ( if a= +ve then you will get a figure some thing like this which is attached and if a=-ve then its mirror image)

I hope this helped else let me know I will elaborate.

Cheers

HI Nitish,

Right approach, but I think you missed the point when a is negative, it'll not be mirror image, but there will be two intersection(because Y intercepts didn't change)

_________________

Consider kudos for the good post ...
My debrief : http://gmatclub.com/forum/journey-670-to-720-q50-v36-long-85083.html

Senior Manager
Joined: 25 Jun 2009
Posts: 291

### Show Tags

27 Jul 2009, 05:51
sudeep wrote:
nitishmahajan wrote:
pmal04 wrote:
How to solve it?
GMATclub test24

IMO its should be C. These are equations for parabola, now trying substituting diff values of 'a' and try to make a figure out of it ( if a= +ve then you will get a figure some thing like this which is attached and if a=-ve then its mirror image)

I hope this helped else let me know I will elaborate.

Cheers

HI Nitish,

Right approach, but I think you missed the point when a is negative, it'll not be mirror image, but there will be two intersection(because Y intercepts didn't change)

Hi Sudeep,

I agree that when a is -ve it wont be the mirror image but I guess in that case also it won't intersect as d< b. So, I think the figure would be some thing like this. In this case the second parabola will be inside the first parabola but I think it won't intersect.

Cheers
Attachments

untitled.JPG [ 8.86 KiB | Viewed 1323 times ]

Director
Joined: 05 Jun 2009
Posts: 725
WE 1: 7years (Financial Services - Consultant, BA)

### Show Tags

27 Jul 2009, 06:56
Nitish,

See this:
Attachments

quad1.jpg [ 19.16 KiB | Viewed 1314 times ]

_________________

Consider kudos for the good post ...
My debrief : http://gmatclub.com/forum/journey-670-to-720-q50-v36-long-85083.html

Senior Manager
Joined: 25 Jun 2009
Posts: 291

### Show Tags

27 Jul 2009, 10:13
sudeep wrote:
Nitish,

See this:

Thanks Sudeep, I got it now

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
Re: ps:m24 #21   [#permalink] 27 Jul 2009, 10:13
Display posts from previous: Sort by