Interesting GMATFix Problem-31

The method of solving is pretty simple for this one but i am not sure about the answer.

Soln:-

let eqn of line l be

y=mx+c where m-> slope and c-> y intercept

from given data
eqn is y=(5/4)x + c
since the line passes through (8,0), putting this value in the eqn of line l we get
0=(5/4)*8 +c
c=-10

hence eqn of l => y=(5/4)x-10

now line k meets l at (4,y').
put this value in eqn of l
y'=(5/4)*4 -10
y'=-5
hence point of intersection is (4,-5)

also y intercept of line k is (0,-3)
then using the equation:-
m=(y"-y')/(x"-x')

we get
m=(-5-(-3))/(4-0)
m=-(1/2)

hence eqn of line k is
k=> y=-(1/2)x-3


we have the two equations as
l=> y = (5/4)x - 10
k=> y = -(1/2)x - 3

the shaded region can be taken as all the values of y which are lesser than (5/4)x-10 and greater than -(1/2)x-3.

to verify this one can put a random value for x or y in any one of the equation and check.

eg:- x = 12 in eqn for l
we get y = 5
the shaded region will have all the values of y less than 5.

hence we get the solution as
-(1/2)x-3 < y < (5/4)x-10

we cannot have an equality with the line as there are points on the line which do not come under the shaded region. for eg (0,-3) lies on line k but not in the region and (0,-10) lies on the line l but not in the region.
arora please verify the options for this question.

Ishaan

the shaded region can be taken as all the values of y which are lesser than (5/4)x-10 and greater than -(1/2)x-3.

I was unruffled by this step... somehow found it tough!

arora007 wrote:

the shaded region can be taken as all the values of y which are lesser than (5/4)x-10 and greater than -(1/2)x-3.

I was unruffled by this step... somehow found it tough!

arora i hope u got the solution until the 2 equations.
Further to that:
Consider line l, there are two regions, one above it (where (0,0) lies) and one below it. Similarly for line k there are two regions, above and below.

Now the requirement in our question is to find out the equation for the shaded region, so we can very well establish one fact that the required region is the common part of area below line l and above line k.

now consider any particular point which you know will lie below line l. for instance (8,-1) does not lie on line l and is below it.
put this value in the equation for line l (y=(5/4)x-10).
we will get
L.H.S. =-1
R.H.S. = (5/4)*8-10 = 0
hence L.H.S < R.H.S.
this means that the equation
y<(5/4)x-10 represents the area below line l.

similarly for line k we can find out the equation for the region above line k. Try with point (0,0).

so we get the required solution.
But i still have the confusion if the equality should be there or not, because there are points in line l and in line k which can never come in the shaded region hence no equality.

Ishaan