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 Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

I suggest to draw quickly an XY plan to figure the problem.

for n: Y = a(n)*X + b(n)
for p: Y = a(p)*X + b(p)

1) intersecting on 1 points <=> a(n) <> a(p)
INSUFF

2) b(n) > b(p)
INSUFF

(1) and (2), it exists 3 cases:

case 1 : b(n) > 1 and b(p) >= 1 To pass by the point (5,1):
> The line n has a negative slop, a(n) < 0
> The line p has a negative slop, a(p) =< 0
as b(n) > b(p), a(p) > a(n)

case 2 : b(n) >= 1 and b(p) < 1 To pass by the point (5,1):
> The line n has a negative slop, a(n) =< 0
> The line p has a positive slop, a(p) > 0
thus a(p) > a(n)

case 3 : b(n) < 1 and b(p) < 1 To pass by the point (5,1):
> The line n has a positive slop, a(n) > 0
> The line p has a positive slop, a(p) > 0
as b(n) > b(p), a(p) > a(n)

[quote="haas_mba07"]I am getting C and here's why.
1) Lines n and p intersect at the point (5,1)
2) The y-intercept of line n is greater than the y-intercept of line p

Stmt1: Intersection point doesn't tell us anything. Not sufficient.

Stmt2: Yn > Yp .. not sufficient either.

Together:

Let intersection point (x2, y2) = (5, 1)
Let the other intersection points be on the y-axis so that x = 0. Given Yn > Yp

Line n and p lie in the xy-plane.Is the slope of the line less than the slope of line p?
1) Lines n and p intersect at the point (5,1)
2) The y-intercept of line n is greater than the y-intercept of line p

For line n: y = m1x + b1
For line p: y = m2x + b2

at (5,1), for line n: 5 = m1+b1
at (5,1), for line p: 5 = m2+b2

Conclusion from statement 1: m1+b1 = m2+b2
When using statement 2 in addition to statement 1,
b1 > b2 ---> m1 < m2
[ where m1 is the slope of line n and m2 is the slope of line p ]

Stmnt 1: Obviously insufficient. Line N and P could virtually go anyway you want them to so it could be yes and no = insuf.

Stmnt 2: Insuffc. Just make some points where N has a great y intercept (where x is 0) n: has point (0,2) p: (0,1) If u give the points (1,2) also for N. and (1,1) for p. then u can figure out the slopes. (the lines will be parallel) ul get a slope 0/1 for both. I dunno if u can have a slope of 0/1, but thats what it came out to. This means that p=n. You could also just pick some other points or just see that the lines could be made almost anyway. N could have a positive or negative slope and same with P. = insuf.

Stmnts Together.

Here N is forced to have a negative slope, while P is forced to have a positive slope. This is because the lines cant be perpendicular b/c they both have to have a y-intercept. They can't be parallel or be the same line since they have to have different y-intercepts. N has a greater y-intercept than P, which forces N to go from top left to bottom right since it has to meet at (5,1). Likewise it forces P to go from bottom left to top right.

Thus P's slope>N's slope.

Hope this wasn't too confusing. I wrote it like this to so I could better understand it myself.

The point of intersection is known. Draw a rough sketch and the answer is before you

I'm adding the diagram now...

<phew> time for a beer...

paddyboy:

i agree that the answer is C... however i disagree with you that you have the only config. possible. i sketched two general situations to get my answer based one with negative slopes the other positve. I guess there should also be a third graph where slope of n=0 and slope of p is positive.

why is that the only configuration possible? you do not take
negative slopes into consideration, no? also, if the y intercept of n>y intercept of p don't you have your labels wrong (i think you labelebd your lines based on x intercepts)

Has to be E. It is being seen that in the explanation provided by paddyboy that slope(n) > slope(p), Fig proves that slope(p) > slope(n) and mishari again proves slope(p) > slope(n) so why is it C? _________________

My will shall shape the future. Whether I fail or succeed shall be no man's doing but my own.

Has to be E. It is being seen that in the explanation provided by paddyboy that slope(n) > slope(p), Fig proves that slope(p) > slope(n) and mishari again proves slope(p) > slope(n) so why is it C?

Graphic approach:

Lines n and p lie in the xy plane. Is the slope of line n less than the slope of line p?

(1) Lines n and p intersect at (5,1) (2) The y-intercept of line n is greater than y-intercept of line p

The two statements individually are not sufficient.

(1)+(2) Note that a higher absolute value of a slope indicates a steeper incline.

Now, if both lines have positive slopes then as the y-intercept of line n (blue) is greater than y-intercept of line p (red) then the line p is steeper hence its slope is greater than the slope of the line n:

Attachment:

1.PNG [ 14.29 KiB | Viewed 18037 times ]

If both lines have negative slopes then again as the y-intercept of line n (blue) is greater than y-intercept of line p (red) then the line n is steeper hence the absolute value of its slope is greater than the absolute value of the slope of the line p, so the slope of n is more negative than the slope of p, which means that the slope of p is greater than the slope of n:

Attachment:

2.PNG [ 13.66 KiB | Viewed 18042 times ]

So in both cases the slope of p is greater than the slope of n. Sufficient.

Answer: C.

Algebraic approach:

Lines n and p lie in the xy-plane. Is the slope of line n less than the slope of line p?

We have two lines: \(y_n=m_1x+b_1\) and \(y_p=m_2x+b_2\). Q: \(m_1<m_2\) true?

(1) Lines n and p intersect at the point (5,1) --> \(1=5m_1+b_1=5m_2+b_2\) --> \(5(m_1-m_2)=b_2-b_1\). Not sufficient. (2) The y-intercept of line \(n\) is greater than the y-intercept of line \(p\) --> y-intercept is value of \(y\) for \(x=0\), so it's the value of \(b\) --> \(b_1>b_2\) or \(b_2-b_1<0\). Not sufficient.

(1)+(2) \(5(m_1-m_2)=b_2-b_1\), as from (2) \(b_2-b_1<0\) (RHS), then LHS (left hand side) also is less than zero \(5(m_1-m_2)<0\) --> \(m_1-m_2<0\) --> \(m_1<m_2\). Sufficient.

Q: If \(m_n\) and \(m_m\) are slopes for n and m respectively,

Is \(m_n<m_m\)

Generic formula for a line passing through a single point \((x_1,y_1)\)is: \(y-y_1=m(x-x_1)\) where m is the slope the line

1. (x,y)=(5,1) Point (5,1) statisfies equations for both lines. Plugging in the values, we get the equation for both the lines as:

Equation for the line n: \(y-1=m_n(x-5)\) \(y=m_n*x-m_n*5+1\) \(y=m_n*x+(1-m_n*5)\)

Likewise for the line m: \(y-1=m_m(x-5)\) \(y=m_m*x-m_m*5+1\) \(y=m_m*x+(1-m_m*5)\)

\((1-m_n*5)\) and \((1-m_m*5)\) are Y-intercepts for n and m respectively

But we don't know any comparable information about the slopes or intercepts to deduce anything concrete; NOT SUFFICIENT.

2. Y-intercept for n is more than Y-intercept for m.

If equation for line n is \(y=m_n*x+C_n\)

If equation for line m is \(y=m_m*x+C_m\)

where \(C_n\) and \(C_m\) are Y-intercepts for n and m respectively We just know that \(C_n>C_m\). We know nothing about the slopes. NOT SUFFICIENT.

However, using both statements;

Y Intercept for n \(C_n\) is \((1-m_n*5)\) Y Intercept for m \(C_m\) is \((1-m_m*5)\)

And

\(C_n > C_m\)

\((1-m_n*5) > (1-m_m*5)\) \(-m_n*5 > -m_m*5\) ###subtracting 1 from both sides### \(m_n*5 < m_m*5\) ###multiplying both sides by -ve sign### \(m_n < m_m\) ###Dividing both sides by 5###

Re: Line n and p lie in the xy-plane.Is the slope of the line [#permalink]

Show Tags

24 Apr 2012, 05:20

what condition in the question leads to conclude that either both slopes will be positive or both negative. i think because difference of y intercept of n & y intercept of p is positive,so slopes of n & p will either be both positive or both negative. Algebraically its fine

But how to satisfy this condition in graphs. Can't it be one positive and one negative slope.

Thanks _________________

The proof of understanding is the ability to explain it.

Re: Line n and p lie in the xy-plane.Is the slope of the line [#permalink]

Show Tags

24 Apr 2012, 12:39

Expert's post

GMATD11 wrote:

what condition in the question leads to conclude that either both slopes will be positive or both negative. i think because difference of y intercept of n & y intercept of p is positive,so slopes of n & p will either be both positive or both negative. Algebraically its fine

But how to satisfy this condition in graphs. Can't it be one positive and one negative slope.

Thanks

It's not so that the slopes of the lines are either both positive to both negative. For example it's possible line n (blue) to have negative slope and line p (red) to have positive slope (naturally in this case slope of n will be less than the slope of p). We are just considering the cases which are not that obvious to see that in ALL cases slope of n is less than the slope of p.

Re: Lines n and p lie in the xy plane. Is the slope of the line [#permalink]

Show Tags

19 Jun 2012, 21:04

Hi Bunuel, Request you to explain the following part. I'm unable to comprehend it.. though I am able to solve the questions by assuming the coordinates. Thanks H

Bunuel wrote:

If both lines have negative slopes then again as the y-intercept of line n (blue) is greater than y-intercept of line p (red) then the line n is steeper hence the absolute value of its slope is greater than the absolute value of the slope of the line p, so the slope of n is more negative than the slope of p, which means that the slope of p is greater than the slope of n:

Excellent posts dLo saw your blog too..!! Man .. you have got some writing skills. And Just to make an argument = You had such an amazing resume ; i am glad...

So Much $$$ Business school costs a lot. This is obvious, whether you are a full-ride scholarship student or are paying fully out-of-pocket. Aside from the (constantly rising)...

They say you get better at doing something by doing it. then doing it again ... and again ... and again, and you keep doing it until one day you look...