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:
Re: slope l1 greater than slope l2 [#permalink]
30 Jan 2012, 03:17
Expert's post
1
This post was BOOKMARKED
Topic moved to DS subforum.
kenguva wrote:
If the slopes of the line l1 and l2 are of the same sign, is the slope of the line l1 greater than that of line l2?
(1) Lines l1 and l2 intersect at point(a,b) where a and b are positive (2) The X-intercept of line l1 is greater than the X-intercept of line l2
If the slopes of the line l1 and l2 are of the same sign, is the slope of the line l1 greater than that of line l2?
(1) Lines l1 and l2 intersect at point (a,b) where a and b are positive --> clearly insufficient: we can rotate the lines on this point so that we get an YES, as well as NO answers. Not sufficient.
(2) The X-intercept of line l1 is greater than the X-intercept of line l2 --> also insufficient: we can draw numerous lines with the same sign slopes so that we get an YES, as well as NO answers. Not sufficient.
(1)+(2) We know that: the slopes of the lines are of the same sign, that they intersect in I quadrant and the X-intercept of line l1 is greater than the X-intercept of line l2. There can be two cases:
Attachment:
graph 5.png [ 19.87 KiB | Viewed 6297 times ]
You can see that in both cases the slope of L1 is greater than the slope of L1 as a steeper incline indicates a higher absolute value of the slope.
In red case: both lines have positive slope, L1 is steeper thus its slope is greater than the slope of L2; In blue case: both lines have negative slope, L2 is steeper thus its slope is more negative (the absolute value of its slope is greater) than the slope of L1, which also means that the slope of L1 is greater than the slope of L2.
Re: If the slopes of the line l1 and l2 are of the same sign, is [#permalink]
31 Jul 2013, 04:01
Hi can anyone help me here, in the figure above for the blue case it looks to me that L2 is steeper than L1, lets take a very large value and a very small value for the x intercept.Lets suppose L1 passes through 100 and L2 passes through 5,so x intercept of L1 is more than L2,since L2 is closer to the y axis on account of lesser x intercept it is more vertical hence its steeper and hence it looks to me that for Blue case( when both the lines have negative slope ) slope of L2 is greater than L1.
Also Visually in the Graph above for the Blue case it looks as though L2 is steeper than L1, what am I missing here? _________________
Re: If the slopes of the line l1 and l2 are of the same sign, is [#permalink]
01 Aug 2013, 21:08
anirudh777 wrote:
ALTERNATE SOLUTION:
Equation 1 : l1 : y=(m1)X+c1 , where m1 is slope and c1 is constant Equation 2 : l2 : y=(m2)X+c2 , where m2 is slope and c2 is constant
Also given that slope are of same sign. Means--either both m1, m2 are positive or both are negative.
Statement 1 : intersect at point (a,b) where in both a and b are positive.
(1)==> y=b=(m1)a+c1=(m2)a+c2 -simply putting coordinates in both the equations of lines.
But from here we cannot conclude whether m1 > or < m2
Hence Insufficient.
Statement 2 : The X-intercept of line l1 is greater than the X-intercept of line l2
==> As value of x, when y=0, is the X intercept. Thus, X intercept for l1 and l2 are: l1: -c1/(m1) l2: -c2/(m2)
(2)==> -c1/(m1) > -c2/(m2), Since we dont know the sogn of c1 and c2 and we dont know there values we cannot conclude anything from this.
Hence Insufficient.
Combining both the statements : and solving (1) and (2) we get,
a-b/(m1) >a-b/(m2) ---> m1<m2 Hence SUFFICIENT. Thus answer is C.
Hi
I have been able to figure out the graphical way by myself, but I cannot figure out how you combined the two equations to get \(\frac{(a-b)}{m1} >\frac{(a-b)}{m2}\) in your algebraic method, can anyone help with the algebraic method here.
Re: If the slopes of the line l1 and l2 are of the same sign, is [#permalink]
01 Aug 2013, 23:57
1
This post received KUDOS
stne wrote:
anirudh777 wrote:
ALTERNATE SOLUTION:
Equation 1 : l1 : y=(m1)X+c1 , where m1 is slope and c1 is constant Equation 2 : l2 : y=(m2)X+c2 , where m2 is slope and c2 is constant
Also given that slope are of same sign. Means--either both m1, m2 are positive or both are negative.
Statement 1 : intersect at point (a,b) where in both a and b are positive.
(1)==> y=b=(m1)a+c1=(m2)a+c2 -simply putting coordinates in both the equations of lines.
But from here we cannot conclude whether m1 > or < m2
Hence Insufficient.
Statement 2 : The X-intercept of line l1 is greater than the X-intercept of line l2
==> As value of x, when y=0, is the X intercept. Thus, X intercept for l1 and l2 are: l1: -c1/(m1) l2: -c2/(m2)
(2)==> -c1/(m1) > -c2/(m2), Since we dont know the sogn of c1 and c2 and we dont know there values we cannot conclude anything from this.
Hence Insufficient.
Combining both the statements : and solving (1) and (2) we get,
a-b/(m1) >a-b/(m2) ---> m1<m2 Hence SUFFICIENT. Thus answer is C.
Hi
I have been able to figure out the graphical way by myself, but I cannot figure out how you combined the two equations to get \(\frac{(a-b)}{m1} >\frac{(a-b)}{m2}\) in your algebraic method, can anyone help with the algebraic method here.
Thanks
______________________
solving Equations (1) and (2) means, substituting the value of c1 and c2 from equation (1) in equation (2)-
From (1) : c1= b-a(m1) c2= b-1(m2)
putting these values in eq2
-c1/(m1) > -c2/(m2)
==> -[b-a(m1)]/m1 > -[b-a(m2)]/m2
--> -b/m1 + a > -b/m2 + a
--> since we know a and b are positive constant values, we can cancel them out from the equality and we are then left with
Re: If the slopes of the line l1 and l2 are of the same sign, is [#permalink]
02 Aug 2013, 02:32
anirudh777 wrote:
stne wrote:
anirudh777 wrote:
ALTERNATE SOLUTION:
Equation 1 : l1 : y=(m1)X+c1 , where m1 is slope and c1 is constant Equation 2 : l2 : y=(m2)X+c2 , where m2 is slope and c2 is constant
Also given that slope are of same sign. Means--either both m1, m2 are positive or both are negative.
Statement 1 : intersect at point (a,b) where in both a and b are positive.
(1)==> y=b=(m1)a+c1=(m2)a+c2 -simply putting coordinates in both the equations of lines.
But from here we cannot conclude whether m1 > or < m2
Hence Insufficient.
Statement 2 : The X-intercept of line l1 is greater than the X-intercept of line l2
==> As value of x, when y=0, is the X intercept. Thus, X intercept for l1 and l2 are: l1: -c1/(m1) l2: -c2/(m2)
(2)==> -c1/(m1) > -c2/(m2), Since we dont know the sogn of c1 and c2 and we dont know there values we cannot conclude anything from this.
Hence Insufficient.
Combining both the statements : and solving (1) and (2) we get,
a-b/(m1) >a-b/(m2) ---> m1<m2 Hence SUFFICIENT. Thus answer is C.
Hi
I have been able to figure out the graphical way by myself, but I cannot figure out how you combined the two equations to get \(\frac{(a-b)}{m1} >\frac{(a-b)}{m2}\) in your algebraic method, can anyone help with the algebraic method here.
Thanks
______________________
solving Equations (1) and (2) means, substituting the value of c1 and c2 from equation (1) in equation (2)-
From (1) : c1= b-a(m1) c2= b-1(m2)
putting these values in eq2
-c1/(m1) > -c2/(m2)
==> -[b-a(m1)]/m1 > -[b-a(m2)]/m2
--> -b/m1 + a > -b/m2 + a
--> since we know a and b are positive constant values, we can cancel them out from the equality and we are then left with
==> m1<m2.
Hence Sufficient.
many thanks +1
however slope of m1>m2 and not m1<m2 as you have arrived at, guess small oversight in your calculation.
Here we go:
\(b=m_1.a+c_1\), and \(b=m_2.a+c_2\) these are the two equations of the line from statement 1
\(c_1= b-m_1a\) and \(c_2=b-m_2a\).. ( 1)
x intercept's \(\frac{-c_1}{m_1}\) and \(\frac{-c_2}{m_2}\)...(2)
given \(\frac{-c_1}{m_1} > \frac{-c_2}{m_2}\). from statement 2
now substituting the values of \(c_1\) and \(c_2\) from ..(1) in ...(2)
\(\frac{-(b-m_1a)}{m_1} > \frac{-(b-m_2a)}{m_2}\)
\(\frac{-b+m_1a}{m_1} > \frac{-b+m_2a}{m_2}\) ,multiplying the negative sign in the numerator
\(\frac{-b}{m_1}+a >\frac{-b}{m_2}+a\) , dividing LHS by \(m_1\) and RHS by \(m_2\) \(\frac{-b}{m_1}>\frac{-b}{m_2}\), cancelling out a from both sides
\(\frac{b}{m_1}<\frac{b}{m_2}\), Multiplying both sides by negative and reversing the > sign to < , since in inequality we reverse the operator when we multiply( or divide)by a negative sign. ( Guess you forgot this part )
so finally we are left with \(\frac{b}{m_1}<\frac{b}{m_2}\)
\(\frac{1}{m_1}<\frac{1}{m_2}\) , cancelling b from both sides
\(\frac{1}{m_1}<\frac{1}{m_2}\)
Now we are still not sure of the signs of slope , both could be + or both could be -
if both are positive then \(m_1>m_2\)
if both are negative
then we have \(\frac{1}{- m_1}<\frac{1}{- m_2}\)
here also we have \(m_1>m_2\)
hence \(m_1>m_2\) for all cases and hence sufficient
Re: If the slopes of the line l1 and l2 are of the same sign, is [#permalink]
08 Mar 2015, 04:48
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________
As I’m halfway through my second year now, graduation is now rapidly approaching. I’ve neglected this blog in the last year, mainly because I felt I didn’...
Perhaps known best for its men’s basketball team – winners of five national championships, including last year’s – Duke University is also home to an elite full-time MBA...
Hilary Term has only started and we can feel the heat already. The two weeks have been packed with activities and submissions, giving a peek into what will follow...