GMAT prep DS question
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- cubicle_bound_misfit
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Is slope of line n < slope of line p?
(1) the point of intersection tells us nothing about the respective slopes: insufficient.
(2) the y intercepts tell us nothing about the respective slopes: insufficient.
Together:
from (1), we know that the lines intersect in quadrant 1 (+,+).
From (2), we know that line n crosses the y-axis higher than does line p.
For a line that has points in quadrant 1, the LOWER the y-intercept, the HIGHER the slope will be. Let's examine the 3 possible cases:
(1) y-intercept = 1
This line will be parallel to the x-axis and have a slope of 0.
(2) y-intercept > 1
This line will slope from upper right to bottom left and have a negative slope. The higher we start, the MORE negative the slope will be.
(3) y-intercept < 1
This line will slope from lower left to upper right and have a positive slope. The lower we start, the MORE positive the slope will be.
So, if n has a bigger y-int than does p, line n will definitely have a smaller slope: sufficient, choose (C).
(1) the point of intersection tells us nothing about the respective slopes: insufficient.
(2) the y intercepts tell us nothing about the respective slopes: insufficient.
Together:
from (1), we know that the lines intersect in quadrant 1 (+,+).
From (2), we know that line n crosses the y-axis higher than does line p.
For a line that has points in quadrant 1, the LOWER the y-intercept, the HIGHER the slope will be. Let's examine the 3 possible cases:
(1) y-intercept = 1
This line will be parallel to the x-axis and have a slope of 0.
(2) y-intercept > 1
This line will slope from upper right to bottom left and have a negative slope. The higher we start, the MORE negative the slope will be.
(3) y-intercept < 1
This line will slope from lower left to upper right and have a positive slope. The lower we start, the MORE positive the slope will be.
So, if n has a bigger y-int than does p, line n will definitely have a smaller slope: sufficient, choose (C).
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
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- cubicle_bound_misfit
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hi stuart,
please let me know if this can be used
general equation of st line
ax+by+c =0
x intercept = -c/a => a = -c/x-intercept
y intercept = -c/b ==> b = -c/y intercept
slope = -a/b ==> k/x-intercept/k/y-intercept
so as the lines have point in quad 1. the value of y intercept is inversely proportional to slope.
Hence C.
Is this approach right?
Please help.
if
now to compare slopes of two lines
please let me know if this can be used
general equation of st line
ax+by+c =0
x intercept = -c/a => a = -c/x-intercept
y intercept = -c/b ==> b = -c/y intercept
slope = -a/b ==> k/x-intercept/k/y-intercept
so as the lines have point in quad 1. the value of y intercept is inversely proportional to slope.
Hence C.
Is this approach right?
Please help.
if
now to compare slopes of two lines
Cubicle Bound Misfit
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IMO C
Equation : y=mx+c
From 1 ,
N-----y1= m1x1+c1
Therefore, 1 = 5m1+c1
Similarly, L---- 1=5m2+c2
From 2, c1>c2
subtracting equation in 1
5(m1-m2) + c1-c2) = 0
5(m1-m2) = - (c1-c2) = c2-c1
But c1>c2
Therefore, 5(m1-m2) <0
i.e m1<m2
ANs C
Equation : y=mx+c
From 1 ,
N-----y1= m1x1+c1
Therefore, 1 = 5m1+c1
Similarly, L---- 1=5m2+c2
From 2, c1>c2
subtracting equation in 1
5(m1-m2) + c1-c2) = 0
5(m1-m2) = - (c1-c2) = c2-c1
But c1>c2
Therefore, 5(m1-m2) <0
i.e m1<m2
ANs C