In the xy-plane, both line K and L intersect with axis-y. Is

The answer should be C.

1. x intercept of K is larger than that of L - This means K crosses X axis farther away than L , But as we don’t know the inclination/slope we can’t say how these lines are inclined with respect to X axis . More inclined -> means more bend towards positive x axis - > means shorter y intercept. This is true irrespective of x intercept is +ve or -ve. But this info is not available here , so insufficient.
2. Same slope - not sufficient

1+2 - same slope . K crosses x axis farther away so this will also cross the y axis farther away from origin ,so larger y intercept .
Hence the answer is C .

Thanks,
VCG.

Does your solution hold good for negative slopes as well?

Both statements individually are obviously not sufficient so lets see if C is valid:

Situation 1: Line 1 has points (3,0) and (0,-3), Line 2 has points (1,0) and (0,-1)
It means that Line 1 X and Y intercepts are larger than intercepts of line 2

Situation 2: Line 1 has points (3,0) and (0,-3) and Line 2 has points (-3,0) and (0,3)
Line 1 has larger X intercept but smaller Y intercept

Hence, solution is E as both the statements combined are insufficient. We can draw multiple scenarios to confirm this solution, just make sure to have a same slope each time.

C it is, because if we take negative slope into consideration then K's y intercept would be less than L's.

It is definitely E. Draw a coordinate system and using both conditions at the same time there are two possibilities. One where both the slopes are positive and one where both the slopes are negative. The y-intercept is greater in one case and lesser in the other. Hence insufficient.
There will be no instances where they it is equal since they have distinct x intercepts and same slopes...

E

I cant draw the lines..so let me explain this further ....
X intercept of k is larger than that of L (x intercept can be both +ve or -ve) . Both have same slope . Think two sticks those are touching x axis say at (3,0)-L & (4,0)K OR (-3,0)-L & (-4,0)K respectively . Now if these are in the positive side with postive slope - means the Y intercept will ne -ve . K - will have larger than that of L . With 90 Degree slope they wount touch Y axis - which is not the case . More than 90 degre that is -ve slope they will cut the Y axis at postive side again Kwill have higher Y intercept than that of L . .......... You can replicate this for the sticks if those are crossing the X axis at negative side -(-3,0)-L & (-4,0)K . Hence both 1 & 2 are sufficient and the answer should be C .

Please explain if it is otherwise .
Thanks ,
VCG.

Dear Writer,

I have put together a graphic as a response to the you specific question. Now this is an important question because there is a "HIGH" likelihood of you getting a question like this on the GMAT these days. Me and most of my friends got very similar questions. What is important to realize is that we are looking for the "Best" possible approach to such questions, an approach that will give you the answer in around a minute. I guess the confusion is between two answer choices, C and E. So I am going to satisfy both the statements and you will see that there are still two possibilities. Trust me this can be solved in 30 seconds. You just need to draw a bunch of quick lines and the answer is obvious.
So please, do draw the lines with such questions and you won't have a problem answering such questions.

I hope the above post will put this case to rest !

Guys do not look to solve this question through algebra. Whatever algebraic solution you come up with is bound to take more time and more effort. The trick is to find the simplest, the most definite approach to solving a question.....

Hi Omer - I think we have a disconnect in the way we are interpretting the word 'intercept' . In this context I consider this as an absolute/mod value of the length .
If I re-read the qs , may be this is also asking about the absolute length . ......Is K's intercept with axis-y greater than that of line L?

I see where you are coming from... But rethink a bit. We are just concerned about the value being greater or not. If both lines are parallel (i.e. they have the same slopes) then no matter which way you look at it, the question does not make any sense. Even if we take the Mod of the y-intercept, I could still arrive at . Because we have no idea how far apart they are. The question just heads into limbo that ways and we can never determine the distance between both the intercepts so the answer would still be E, No?

I understand the graphical representation and agree that its the quickest way to approach, but can someone point the error in my approach:

if the two lines, K: y1 - m1x1 + b1 and L: y2 = m2x2 +b2 then

As per the statement 1: -b1/m1>-b2/m2 <Not sufficient>

Statement 2: m1=m2 <not sufficient>

Both: -b1>-b2---> b1<b2...isnt that what we wanted to prove.. can someone please explain the error!

\(m_1=m_2\) and \(-\frac{b_1}{m_1}>-\frac{b_2}{m_2}\) --> \(-\frac{b_1}{m_1}>-\frac{b_2}{m_1}\). But from this you cannot reduce inequality by \(\frac{1}{m_1}\) and write \(-b_1>-b_2\), because you don't know whether \(\frac{1}{m_1}\) is negative or not. If it's negative then when reducing by a negative value you should flip the sign of the inequality and write \(-b_1<-b_2\).

Never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know the sign of it.

So, from \(-\frac{b_1}{m_1}>-\frac{b_2}{m_1}\) we have: \(\frac{b_2}{m_1}-\frac{b_1}{m_1}>0\) --> \(\frac{1}{m_1}(b_2-b_1)>0\) --> if \(\frac{1}{m_1}>0\) then \(b_2>b_1\) but if \(\frac{1}{m_1}<0\) then \(b_2<b_1\).

Hope it's clear.

Prefect .. Thank you!

I thought Intercept is measured as an absolute value..kind of distance of the point of intersection on the axes from the origin...
Always learn smthing new..Nyc 1 bunuel..

Thanks for the explanation. I used to assume that intercept is considered in its absolute value. I'm surprised to know that the sign is also taken into consideration apart from the absolute value of the intercept.

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