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If the slopes of the line l1 and l2 are of the same sign, is Updated on: 30 Jan 2012, 04:18
Originally posted by kenguva on 30 Jan 2012, 00:00.
Last edited by Bunuel on 30 Jan 2012, 04:18, edited 1 time in total.
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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
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If the slopes of the line l1 and l2 are of the same sign, is 30 Jan 2012, 04:17
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Topic moved to DS subforum.

kenguva
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
Show more

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
graph 5.png [ 19.87 KiB | Viewed 30077 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.

Answer: C.

For more on this topic check Coordinate Geometry chapter of Math Book: math-coordinate-geometry-87652.html

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If the slopes of the line l1 and l2 are of the same sign, is 18 Apr 2012, 10:11
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L2 is steeper thus its slope is more negative (the absolute value of its slope is greater) than the slope of L2 ... Does it mean than L1??
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If the slopes of the line l1 and l2 are of the same sign, is 18 Apr 2012, 11:06
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L2 is steeper thus its slope is more negative (the absolute value of its slope is greater) than the slope of L2 ... Does it mean than L1??
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Yes, that is correct.
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If the slopes of the line l1 and l2 are of the same sign, is 27 Jun 2013, 22:30
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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

Theory on Coordinate Geometry: math-coordinate-geometry-87652.html

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If the slopes of the line l1 and l2 are of the same sign, is 31 Jul 2013, 05:01
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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?
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If the slopes of the line l1 and l2 are of the same sign, is 01 Aug 2013, 11:44
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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.
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If the slopes of the line l1 and l2 are of the same sign, is 01 Aug 2013, 22:08
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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.
Show more

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
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If the slopes of the line l1 and l2 are of the same sign, is 02 Aug 2013, 00:57
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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
Show more
______________________


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. :lol:
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If the slopes of the line l1 and l2 are of the same sign, is 02 Aug 2013, 03:32
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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. :lol:
Show more


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

Hope I have done everything correctly


Thank you for showing the way.
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If the slopes of the line l1 and l2 are of the same sign, is 21 Jan 2014, 06:43
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My approach

y = mx + f (Line 1)

y = nx + g (Line 2)

Is m>n or m-n>0?

(1) b=am + f

b = an + g

m-n = g-f/a where 'a' is a positive integer

So m>n only when g>f Insuff

(2) n=-f/x
m=-g/x

-f>-g
g>f

Insuff

(1) and (2)

We know that g>f therefore m>n

C
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If the slopes of the line l1 and l2 are of the same sign, is 24 Aug 2020, 08:38
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Bunuel
Topic moved to DS subforum.

kenguva
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
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.

Show more

Hi Bunuel, could you kindly enlighten on 2 questions please?
1 - How do we know the slopes of the lines are of the same sign as per the highlighted quote above?
2 - Why is that information on X-intercept can be used to solve this question, but not on another question when the intersection point of 2 lines is (-2,4).
This "other" question is referenced below, posted by dreambeliever

Other question: Lines m and n lie in the xy-plane and intersect at the point (-2; 4). Is the slope of line m less than the slope of line n?

(1) The x-intercept of line m is greater than the x-intercept of line n.
(2) The y-intercept of line n is greater than the y-intercept of line m.

PS: Apologies for any errors in posting, this is my first post.

Thank you!
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If the slopes of the line l1 and l2 are of the same sign, is 28 Apr 2021, 15:49
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Topic moved to DS subforum.

kenguva
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
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.

Answer: C.

For more on this topic check Coordinate Geometry chapter of Math Book: https://gmatclub.com/forum/math-coordina ... 87652.html

P.S.
Please post PS questions in the PS subforum: https://gmatclub.com/forum/gmat-problem-solving-ps-140/
Please post DS questions in the DS subforum: https://gmatclub.com/forum/gmat-data-sufficiency-ds-141/

No posting of PS/DS questions is allowed in the main Math forum.
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Hi Bunuel, this distinction between 'greater/lesser' and 'steeper/less steep' still eludes me.

So basically for the red lines with POSITIVE slope, we say that the slope of line 1 is GREATER because it is STEEPER.
But, for the blue lines with NEGATIVE slope, we say that the slope of line 1 is SMALLER because it is LESS STEEP

How would you compare the slope of red line1 with blue line2? Which one has a 'greater' slope?
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If the slopes of the line l1 and l2 are of the same sign, is 29 Nov 2024, 08:21
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