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3 sides are fixed with 3 lines
1. X axis
2. Y axis
3. y=6

Forth side need to be fixed. We have a equation of 4th line with one variable.
Anything that giving 4th equation will be sufficent to ans the q.

statement 2 give a point on that line and we can find the 4th line equation as well.
Hence B
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We need (Parallel Side 1 + Parallel Side2)/2 * Distance

Distance = 6

y = ax + 8 intersects x-axis at x = -8/a ( y = 0) and the line y = 6 at (-2/a,6)

So area = -10/a * 1/2 * 6




(1)

(1,6) can lies on y = 6, but it may lie on y = ax + 8 in a nonverlapping manner,i.e, it may be common point to both or may not be so.

InSufficient

(2)

Point (6,5) can lie only on y = ax + 8

=> 5 = 6x + 8

=> a = -3/6 = -1/2

Sufficient

Answer - B
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for 1)
point (1,6) lies on y= ax + 8 or 6= a.1 + 8 means a = -2.

Hence sufficient.

2) is already sufficient as proved earlier.

So my ans is D.
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I also agree with toughmat.
At Coordinate (1,6)=> x=1 and y=6
Now we also know that y=mx+c
According to question y=ax+8, by putting x=1 and y= 6 wecan get the value of a = -2, thus slope is -2.
Now we can calculate area of quadrilateral which comes to be 18.
Same in the case for second option. so, my answer is D.
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Hi Fluke/Subhash,
Can you explain me statement 1 in a better way .
if points (1,6) boundaries.. then why can't it lie on y =ax+8 ; 6 =a*1+8 a=-2?
or is it that we are given another y=6 ( which we know has to be a horizontal line parallel to x axis) and hence (1,6) can't lie on y = ax+8
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Hi Fluke/Subhash,
Can you explain me statement 1 in a better way .
if points (1,6) boundaries.. then why can't it lie on y =ax+8 ; 6 =a*1+8 a=-2?
or is it that we are given another y=6 ( which we know has to be a horizontal line parallel to x axis) and hence (1,6) can't lie on y = ax+8

Please check the OE and see whether it helps.
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I don't understand why the OA is B

as in a trapezium formed by 4 lines so each line has to intersect at 2 points

we know that (0,0) and (6,0) are intersected by the x-axis, y-axis and the line y = 6 so the line y = ax + 8, y=6 and x-axis intersect at 2 other points.

so if the 3rd vertex is given as (1,6) then the line has to definitely pass through the intersection of y=6 and y = ax+8 how can we assume that the line intersects y-axis in a non contributing way? Can someone explain that with the help of a diagram if thats the case?

My reasoning is that each pairs of lines have unique intersection points and each point represents a vertex of the quadrilateral. this can be confirmed by the diagrams given below.

Sorry in the diagram below the slopes have to be -ve and +ve respectively.
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trapeziumsoln.JPG
trapeziumsoln.JPG [ 27.4 KiB | Viewed 12587 times ]

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Are you by any chance talking about the trapezium in which the 4 points lie within the bounded region??

and if that were the case the answer should be E because we can only find the bounds of the bounded region and not the specific points in that region.
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phanideepak
Are you by any chance talking about the trapezium in which the 4 points lie within the bounded region??

and if that were the case the answer should be E because we can only find the bounds of the bounded region and not the specific points in that region.

The question and solution are fine.

You have four lines: x axis, y axis, y = 6 and y = ax + 8
You need the value of a to get the area bounded by these lines.

Stmnt1: This tells you that (1, 6) lies on the boundary (one of the lines) of the quadrilateral (not that it is a vertex necessarily). We know that y = 6 passes through this point anyway. All this says is that y = 6 intersects with y = ax + 8 at a point (m, 6) such that m is greater than 1. We still do not know the value of a and hence this is not sufficient.

Stmnt2: None of x axis, y axis and y = 6 can pass through (6, 5). If (6, 5) lies on the boundary of the quadrilateral, y = ax + 8 has to pass through it. So we know a point on y = ax + 8 which will give us the value of 'a' and hence the equation of the fourth line. This will give us the area of the quadrilateral. Sufficient

Also, in your figure, the second diagram is not possible because that line cannot pass through (6, 5). Its y co-ordinate will be greater than 8 in the first quadrant.
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phanideepak
Are you by any chance talking about the trapezium in which the 4 points lie within the bounded region??

and if that were the case the answer should be E because we can only find the bounds of the bounded region and not the specific points in that region.

The question and solution are fine.

You have four lines: x axis, y axis, y = 6 and y = ax + 8
You need the value of a to get the area bounded by these lines.

Stmnt1: This tells you that (1, 6) lies on the boundary (one of the lines) of the quadrilateral (not that it is a vertex necessarily). We know that y = 6 passes through this point anyway. All this says is that y = 6 intersects with y = ax + 8 at a point (m, 6) such that m is greater than 1. We still do not know the value of a and hence this is not sufficient.

Stmnt2: None of x axis, y axis and y = 6 can pass through (6, 5). If (6, 5) lies on the boundary of the quadrilateral, y = ax + 8 has to pass through it. So we know a point on y = ax + 8 which will give us the value of 'a' and hence the equation of the fourth line. This will give us the area of the quadrilateral. Sufficient

Also, in your figure, the second diagram is not possible because that line cannot pass through (6, 5). Its y co-ordinate will be greater than 8 in the first quadrant.

Ah! Now i understand... I was assuming it be the vertex. Thanks a lot for your solution :) I need to improve my reading skills :(
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The mistake I did was substitute (1,6) in the equation y=ax+b. I got Ans: D. But each statement gives different slopes and intercepts, so I am hesitant to keep answer D. I think in GMAT if the option is D, the answer will be same.

The point (1,6) may be on line y=6.

I think this is the trap. Otherwise this question is easy.
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phanideepak
Are you by any chance talking about the trapezium in which the 4 points lie within the bounded region??

and if that were the case the answer should be E because we can only find the bounds of the bounded region and not the specific points in that region.

The question and solution are fine.

You have four lines: x axis, y axis, y = 6 and y = ax + 8
You need the value of a to get the area bounded by these lines.

Stmnt1: This tells you that (1, 6) lies on the boundary (one of the lines) of the quadrilateral (not that it is a vertex necessarily). We know that y = 6 passes through this point anyway. All this says is that y = 6 intersects with y = ax + 8 at a point (m, 6) such that m is greater than 1. We still do not know the value of a and hence this is not sufficient.

Stmnt2: None of x axis, y axis and y = 6 can pass through (6, 5). If (6, 5) lies on the boundary of the quadrilateral, y = ax + 8 has to pass through it. So we know a point on y = ax + 8 which will give us the value of 'a' and hence the equation of the fourth line. This will give us the area of the quadrilateral. Sufficient

Also, in your figure, the second diagram is not possible because that line cannot pass through (6, 5). Its y co-ordinate will be greater than 8 in the first quadrant.

Quote:
Quick query regarding your answer below for a DS questions. From the first statement, are we wrong to assume that the point (1,6) is a vertex? Or is it because it could be a vertex or just on the line y=6, that the statement is insufficient?

We cannot assume that the point (1, 6) is a vertex. It could be a vertex but it is not necessary. Boundary of the quadrilateral is defined by the 4 line segments that make it. (1, 6) could be anywhere on those lines (vertex or otherwise).
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