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Veritas Geometry Challenge Problem

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Veritas Geometry Challenge Problem

by vineetbatra » Sat Apr 03, 2010 11:10 am
Line D Passes through point(-2,5), and the product of its x-intercept and y-intercept is positive. Which of the following points could be on line D

A 5,10

B -3,2

C -1,7

D -1,2

E -3,1

OA is D

I need to know a faster way to do this question. I couldn't understand the explanation provided by the book.
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by thephoenix » Sat Apr 03, 2010 11:43 am
if u luk in graph only D i.e green line only meets the condition of x-intercept*y-intercept=+ve
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by harshavardhanc » Sat Apr 03, 2010 11:59 am
vineetbatra wrote:Line D Passes through point(-2,5), and the product of its x-intercept and y-intercept is positive. Which of the following points could be on line D

A 5,10

B -3,2

C -1,7

D -1,2

E -3,1

OA is D

I need to know a faster way to do this question. I couldn't understand the explanation provided by the book.
you have to observe that the slope is -ve. This means, as you extend the line in the second quadrant (make X coordinate more -ve, like -3, -4 etc), the Y coordinate will be more than 5. So this eliminates option B and E.

Similarly, if the X coordinate is greater than -2 in the second quadrant, i.e -1, the Y coordinate will be less than 5. This eliminates C.

In the first quadrant, the Y coordinate cannot be more than 5 , so A is gone.

You are left with D. :). No choice but to pick it :)
Last edited by harshavardhanc on Wed Apr 07, 2010 4:41 am, edited 1 time in total.
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Harsha
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by eaakbari » Sat Apr 03, 2010 12:07 pm
and the product of its x-intercept and y-intercept is positive
I am not able to get past this. What exactly is its implication?
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by harshavardhanc » Sat Apr 03, 2010 12:18 pm
eaakbari wrote:
and the product of its x-intercept and y-intercept is positive
I am not able to get past this. What exactly is its implication?
it means the slope is -ve.

Y
|
|\
|-\
|--\
|__\__________ X
...... \


another possibility to have product of intercepts +ve is that the line passes thru 2nd,3rd and 4th quadrant, but that need not be considered because of the additional point given in the question.
Last edited by harshavardhanc on Wed Apr 07, 2010 4:41 am, edited 1 time in total.
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by kstv » Sat Apr 03, 2010 6:46 pm
Line D Passes through point(-2,5), and the product of its x-intercept and y-intercept is positive. Which of the following points could be on line D A 5,10 B -3,2 C -1,7 D -1,2 E -3,1

if the eq of a line is y = mx +c , y intercept is when x = 0 and x interpect is when y = 0
c is the y intercept , x intercept is = -c/m the product of the intercepts is + ve
so c x -c/m = c²/m is positive but c X -c/m = -c²/m
c² is always + ve so m the slope has to be negative
only option D gives - ve slope
slope between two points = y1-y/x1-x if x,y ( -2,5) and (x1,y1) are the 5 options A to E
options A will give 5-10/-2-5 = +5/7
B = 5-2/-2+3= + 3/3
C = 5-7/-2+1 = + 2/1
D = 5-2/-2+1 = 3/-1 = -ve 3
E = 5-1/-2 +3 = + 4
Still it is not possible to solve under 2 mins.
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by gmatmachoman » Sat Apr 03, 2010 9:28 pm
kstv wrote:Line D Passes through point(-2,5), and the product of its x-intercept and y-intercept is positive. Which of the following points could be on line D A 5,10 B -3,2 C -1,7 D -1,2 E -3,1

if the eq of a line is y = mx +c , y intercept is when x = 0 and x interpect is when y = 0
c is the y intercept , x intercept is = -c/m the product of the intercepts is + ve
so c x -c/m = c²/m is positive but c X -c/m = -c²/m
c² is always + ve so m the slope has to be negative
only option D gives - ve slope
slope between two points = y1-y/x1-x if x,y ( -2,5) and (x1,y1) are the 5 options A to E
options A will give 5-10/-2-5 = +5/7
B = 5-2/-2+3= + 3/3
C = 5-7/-2+1 = + 2/1
D = 5-2/-2+1 = 3/-1 = -ve 3
E = 5-1/-2 +3 = + 4
Still it is not possible to solve under 2 mins.
@kstv,

the way u approached is very much logical!!
Yeah its not that easy to finish off with in 2 mins r so!!
Agreed!
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by thephoenix » Sat Apr 03, 2010 9:38 pm
gmatmachoman wrote: @kstv,
the way u approached is very much logical!!
Yeah its not that easy to finish off with in 2 mins r so!!
Agreed!
i guess plotting the points graphically will save time
as i did in my earlier post
am i wrong somewhere
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by Stuart@KaplanGMAT » Sun Apr 04, 2010 8:49 am
If you draw it out, this is a 30 second question - no math required.

All that you need to understand is the definition of x- and y-intercept.

For the product of the intercepts to be positive, either they're both positive or both negative. By plotting the point (-2,5) you quickly see that the only way to get two intercepts with the same sign is for the slope to be negative, i.e. our line must go from the top left of (-2,5) to the bottom right of the point.

Looking at the choices:

A) to get from -2,5 to 5,10 we go up and right: eliminate.

B) to get from -2,5 to -3,2 we go down and left: eliminate.

C) to get from -2,5 to -1,7 we go up and right: eliminate.

D) to get from -2,5 to -1,2 we go down and right: winner!

E) to get from -2,5 to -3,1 we go down and left: eliminate.

We also could have solved this playing the "One of these things is not like the others game"; A, B, C and E all give us lines with positive slopes - only D gives us a line with a negative slope. Even if we all that we understand about the question is that slope makes the difference, that's a great reason to select D.
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by kstv » Sun Apr 04, 2010 9:50 am
Bravo!
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by vineetbatra » Wed Apr 07, 2010 4:38 am
Stuart Kovinsky wrote:If you draw it out, this is a 30 second question - no math required.

.
All,

Thanks a lot for the reply, Stuart I was able to identify te slope is negative, but what took me so long is to figure out what to do next.

Usually in difficult problems I just cannot get my head around to figure out the next step. I am able to find the hidden logic, but I am unable to follow it through to the end. Any suggestion on this?

Only if we were to get partial credit :)

Thanks again guys,

Vineet
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by susheelh » Sun May 07, 2017 12:19 am
I too face the exact same problem as mentioned by vineet batra here! Please let me know if someone has an answer for it :-D
vineetbatra wrote:
Stuart Kovinsky wrote:If you draw it out, this is a 30 second question - no math required.

.
All,

Thanks a lot for the reply, Stuart I was able to identify te slope is negative, but what took me so long is to figure out what to do next.

Usually in difficult problems I just cannot get my head around to figure out the next step. I am able to find the hidden logic, but I am unable to follow it through to the end. Any suggestion on this?

Only if we were to get partial credit :)

Thanks again guys,

Vineet
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by [email protected] » Sun May 07, 2017 9:35 am
Hi susheelh,

A number of Quant questions that you'll face on Test Day can be solved with a bit of 'brute force'; you won't need to do any complex math, but you will have to put the pen on the pad and work through whatever basic math is required.

In this question, you should try drawing a graph with the point (-2,5) on it. Then place the 5 answer choices and sketch out the lines (you'll want to do a reasonably accurate drawing of each). Only one answer will make sense.

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