In the coordinate plane, line L passes above the points (50, 70) and (

Question

In the coordinate plane, line L passes above the points (50, 70) and (100, 89) but below the point (80, 84). Which of the following could be the slope of line L?

(A) 0
(B) 1/2
(C) 1/4
(D) 2/5
(E) 6/7

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This is one of a set of 15 challenging GMAT Quant problems. For the whole collection, as well as the OE for this particular problem, see:
Challenging GMAT Math Practice Questions

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Official Answer

D

khrystal · Accepted Answer

The slope of the line L (lets call it m ) has to greater than the slope of the line which is formed by ((50, 70), (100, 89) -> lets call it Line A ) and has to be less than the slope of the line which is formed by ((50, 70), (80,84) -> Line B) The slope of Line A = (y2-y1)/(x2-x1) => 19/50 =>which is slightly less than 0.4 (since 20/50 = 0.4) The slope of Line B = 14/30 => slightly less than 0.5 (since 15/30 = 0.5) so =>19/50(slightly less than 0.4) < m < 14/30(slightly less than 0.5) Lets check with answers now A, B, E, C => Doesn't fit in above mentioned criteria Only 2/5 =0.4 fits the required condition So the answer is D Thanks

grantcke · Answer

Please post a solution to this problem.

Bunuel · Answer

Check Q10 in  Challenging GMAT Math Practice Questions .

aragonn · Answer

Official Magoosh Explanation:

One point is (50, 70) and one is (100, 89): the line has to pass above both of those. Well, round the second up to (100, 90)—if the line goes above (100, 90), then it definitely goes above (100, 89)!

What is the slope from (50, 70) to (100, 90)? Well, the rise is 90 – 70 = 20, and the run is 100 – 50 = 50, so the slope is rise/run = 20/50 = 2/5. A line with slope of 2/5 could pass just above these points.

Now, what about the third point? For the sake of argument, let’s say that the line has a slope of 2/5 and goes through the point (50, 71), so it will pass above both of the first two points. Now, move over 5, up 2: it would go through (55, 73), then (60, 75), then (65, 77), then (70, 79), then (75, 81), then (80, 83). This means, it would pass under the third point, (80, 84). A slope of 2/5 works for all three points.

We don’t have to do all the calculations, but none of the other slope values works.

Answer = (D)

EncounterGMAT · Answer

Not able to grasp such questions. Can someone explain it in a much simpler way?

Thanks a lot in advance!

JiddiHuMain · Answer

The slope of the line L (lets call it m) has to greater than the slope of the line which is formed by ((50, 70), (100, 89) -> lets call it Line A ) and has to be less than the slope of the line which is formed by ((50, 70), (80,84) -> Line B)

The slope of Line A = (y2-y1)/(x2-x1) => 19/50 =>which is slightly less than 0.4 (since 20/50 = 0.4)
The slope of Line B = 14/30 => slightly less than 0.5 (since 15/30 = 0.5)
so =>19/50(slightly less than 0.4) < m < 14/30(slightly less than 0.5)

Lets check with answers now
A, B, E, C => Doesn't fit in above mentioned criteria

Only 2/5 =0.4 fits the required condition

So the answer is D

Thanks

Kinshook · Answer

Slope of line is greater than (89-70)/(100-50) = 19/50 < .4

Slope of the line is less than (84-70)/(80-50) = 14/30 < .5

IMO D

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Bambi2021 · Answer

I used the "below-point" (80, 84) as anchor here.

The slope of a line from the lowest point (50,70) to (80,84) must be more than what we are looking for, and the slope from (80,84) to the higher point (100, 89) must be less than what we are looking for. If slope = k, then:

(89-84)/(100-80) < k < (84-70)/(80-50)

15/60 < k < 28/60

Only 2/5 lies in this range. Answer D.

For an easier to grasp example, consider a line that goes above (1,1) and above (11,2), but below (6,6).

amoljain · Answer

Alternate Solution: Let the slope of the line to be found be t and a general point on the line be (a,b) Equation of the line to be found comes out to be y-b=t(x-a) Rearranging equation: y-x*t+a*t-b=0 We know that if any point (a,b) lies above a curve f(x,y)=0 , then f(a,b)>0 and if point (a,b) lies below the curve, then f(a,b)<0 Using the above property, we satisfy the given points in the derived equation, we get 84-80t+at-b>0 (80,84) lies above the line]... eq(1) 70-50t+at-b<0 (50,70) lies below the line]... eq(2) 89-100t+at-b<0 100,89 ) lies below the line]... eq(3) As we know that we can only add inequalities, multiplying -1 to eqn(1) , we get -84+80t-(at-b)<0 ... eq(4) Adding eqn(2) to eqn(4) -14+30t<0 t<7/15 Adding eqn(3) to eqn(4) 5-20t<0 t>1/4 As value of t will be the intersection of both the condition, we get t belongs to (1/4,7/15) Ans D

Crytiocanalyst · Answer

There is 2 method to solve either resort to inequality or by picking up the right example 
with the latter first we can take :
105,90; 60, 65 ; gives us 2/5
other method being 
:

(89-84)/(100-80) < m < (84-70)/(80-50)

15/60 < m < 28/60

And 2/5 lies in this range.
Hence IMO D.

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In the coordinate plane, line L passes above the points (50, 70) and (

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In the coordinate plane, line L passes above the points (50, 70) and (

Prep Toolkit

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