If the slopes of the line l1 and l2 are of the same sign, is

Question

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

Bunuel · Answer

Topic moved to DS subforum. 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: 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: math-coordinate-geometry-87652.html P.S. Please post PS questions in the PS subforum: gmat-problem-solving-ps-140/ Please post DS questions in the DS subforum: gmat-data-sufficiency-ds-141/ No posting of PS/DS questions is allowed in the main Math forum.

pavanpuneet · Answer

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??

Bunuel · Answer

Yes, that is correct.

Bunuel · Answer

Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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

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stne · Answer

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?

Also Visually in the Graph above for the Blue case it looks as though L2 is steeper than L1, what am I missing here?

anirudh777 · Answer

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.

stne · Answer

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

anirudh777 · Answer

______________________ 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) ==> - /m1 > - /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 :lol:

stne · Answer

many thanks +1 however slope of m1>m2 and not m1 \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.

jlgdr · Answer

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

Gmatboat · Answer

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!

CEdward · Answer

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?

WiziusCareers1 · Answer

To determine if the slope of line l1 (m1) is greater than the slope of line l2 (m2), we need to analyze the relationship between their intercepts and their intersection point.

We are given that m1 and m2 have the same sign (both positive or both negative).

Considering Statement (1) 
Lines l1 and l2 intersect at point (a,b) where a, b > 0.
Knowing only the intersection point in the first quadrant doesn't tell us about the steepness of the lines. For any point (a,b), we could draw two lines with positive slopes where m1 >m2, or two lines where m2 >m1. The same applies to negative slopes.
 Insufficient.

Considering Statement (2) 
The x-intercept of l1 is greater than the x-intercept of l2.
Let x1 be the x-intercept of l1 and x2 be the x-intercept of l2. We know x1 > x2. However, without knowing a second point (like the intersection) or whether the slopes are positive or negative, we cannot determine which line is steeper.
 Insufficient.

Combining (1) and (2) 
When we combine the statements, we have a fixed intersection point (a, b) in the first quadrant and a specific relationship between their x-intercepts (x1 > x2).

The formula for the slope m given an intersection (a, b) and an x-intercept (xi, 0) is:
m = (b - 0)/(a - xi) = b/(a - xi)

Since b is positive (from statement 1), the sign of the slope depends entirely on the denominator (a−xi). Because the prompt states both slopes have the same sign, both x-intercepts must be on the same side of a (either both are less than a, or both are greater than a).

Case 1: Positive Slopes (x1, x2 < a) 
If the slopes are positive, the x-intercepts must be to the left of a. Given x1 > x2:

- The denominator (a − x1) is smaller than (a − x2).
- Since the denominator is smaller, the fraction m1 = b/(a−x1) is larger.
- Therefore, m1 > m2.

Case 2: Negatiive Slopes (x1, x2 > a) 
- (a - x1) is more negative than (a - x2) (For example a - x1 = -10 and a - x2 = -5)
- m1 = -b/10 and ma = -b/5
- m1 > m2

So, the answer is (C)

If the slopes of the line l1 and l2 are of the same sign, is

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If the slopes of the line l1 and l2 are of the same sign, is

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