Author 
Message 
TAGS:

Hide Tags

Retired Moderator
Joined: 20 Dec 2010
Posts: 1868

On the xycoordinate plane, a quadrilateral is bounded by
[#permalink]
Show Tags
17 May 2011, 04:07
Question Stats:
50% (01:52) correct 50% (01:39) wrong based on 276 sessions
HideShow timer Statistics
On the xycoordinate plane, a quadrilateral is bounded by the xaxis, yaxis, the line \(y = 6\), and the line \(y = ax + 8\). What is the area of this region? (1) The point (1,6) is on the boundary of the quadrilateral. (2) The point (6,5) is on the boundary of the quadrilateral. Attachment:
screen_shot_2011_03_11_at_3.28.58_pm.png [ 18.66 KiB  Viewed 7844 times ]
©GrockitThere are a few inferences we can start with before we turn to the statements. Since the lines y = 6 and the xaxis are parallel to one another (and the yaxis and the line y = ax + 8 are not parallel), this area is a trapezoid. Furthermore, since the xaxis and yaxis are parallel, this area is a right trapezoid (a trapezoid with two right angles). It is relatively easy to find the area of a right trapezoid; they can be divided into a rectangle and a triangle. All we need is the value of a. Then we will have the equations of all of the lines that form this trapezoid. It is neither advisable nor necessary to actually calculate the area of this trapezoid. It is better just to determine what information we will need to do so. In this case, we just need to find the equation of that fourth line. Now, turning to the statements:
(1) Remember that the line y = 6 is one of the borders of the trapezoid. This line is the horizontal line that passes through (1,6). Therefore, we don't know if the point given is on y = 6 or the vertex intersection with the line y = ax + 8. INSUFFICIENT
(2) The point (6,5) does not lie on the xaxis, the yaxis, or the line y = 6. Thus, it must lie on the line y = ax + 8. Substituting x and y into the equation we have:
5 = a(6) + 8 3 = 6a 0.5 = a
The equation of the line is y = 0.5x + 8
Now that we have this equation, we can solve the questions. SUFFICIENT.
Although we don't want to waste time finding the area, here's how we would do it for review purposes: the line we found crosses the xaxis and the line y = 0.5x + 8. Let's first find the intersection points.
6 = 0.5x + 8 2 = 0.5x x = 4
This is where the line we found meets the line y = 6. Draw an imaginary line from this point to the xaxis. this will split the trapezoid into a rectangle and a triangle. The area of the rectangle is 24 (4 wide x 6 long). Now let's find the other intersection point, the xintercept of y = 0.5x + 8:
0 = 0.5x + 8 8 = 0.5x 16 = x
So we have a right triangle with a base of 12 {from our imaginary line at (4,0) to the xintercept at (16,0)} and a height of 6. Plugging these values into the formula for the area of a triangle, we get:
0.5(12 x 6) = 0.5(72) = 36
Adding up the two areas gives us a final value of 60.
The credited response is (B): Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
Official Answer and Stats are available only to registered users. Register/ Login.



Senior Manager
Joined: 03 Mar 2010
Posts: 394

Re: Area of trapezium using coordinates
[#permalink]
Show Tags
17 May 2011, 05:09
See the attached images. All coordinates mentioned can be found out by substituting correct coordinates. Stmt1: The point (1,6) is on the boundary of the quadrilateral. That means point is on line y=6. But we don't know the value of a. Insufficient. Stmt2:The point (6,5) is on the boundary of the quadrilateral. That means point is on y=ax+8. Substituting (6,5) 5=6a+8 a=1/2 We know the value of a. We can find all the coordinates with a in attached image. (8/a,0)=> 16,0 (2/a,6)=> 4,0 Area of shaded region will be bigger triangle minus smaller triangle. Area of bigger triangle= 1/2 * 16 * 8 =64 Area of smaller triangle=1/2 * 4 * 2 =4 Area of shaded region= 644=60. OA B.
Attachments
soln5.JPG [ 10.3 KiB  Viewed 7795 times ]
_________________
My dad once said to me: Son, nothing succeeds like success.



VP
Status: There is always something new !!
Affiliations: PMI,QAI Global,eXampleCG
Joined: 08 May 2009
Posts: 1097

Re: Area of trapezium using coordinates
[#permalink]
Show Tags
17 May 2011, 09:00
a (1,6) lies on y= ax + 8 or y=6 as 6= ax + 8 means ax = 2. Hence not sufficient. b (6,5) on y = ax+8 gives 5 = 6a + 8 giving a = 1/2 also ax= 2 means x = 4 means y= ax + 8 when y = 0 = 1/2 x + 8 thus x = 16 thus area of trapezium = 1/2 * altitude * (sum of parallel sides ) = 1/2 * 6 * (16 + 4) = 60 Sufficient. Hence B.
_________________
Visit  http://www.sustainablesphere.com/ Promote Green Business,Sustainable Living and Green Earth !!



Manager
Status: GMAT in 4 weeks
Joined: 28 Mar 2010
Posts: 164
GPA: 3.89

Re: Area of trapezium using coordinates
[#permalink]
Show Tags
17 May 2011, 10:51
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
_________________
If you liked my post, please consider a Kudos for me. Thanks!



Retired Moderator
Joined: 16 Nov 2010
Posts: 1452
Location: United States (IN)
Concentration: Strategy, Technology

Re: Area of trapezium using coordinates
[#permalink]
Show Tags
17 May 2011, 21:15
We need (Parallel Side 1 + Parallel Side2)/2 * Distance Distance = 6 y = ax + 8 intersects xaxis 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
_________________
Formula of Life > Achievement/Potential = k * Happiness (where k is a constant)
GMAT Club Premium Membership  big benefits and savings



Manager
Joined: 19 Apr 2011
Posts: 88

Re: Area of trapezium using coordinates
[#permalink]
Show Tags
09 Jun 2011, 07:31
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.



Intern
Status: Don't worry about finding inspiration. It eventually come >>>>>>>
Joined: 31 May 2011
Posts: 23
Location: Î Ñ D Ï Â

Re: Area of trapezium using coordinates
[#permalink]
Show Tags
09 Jun 2011, 08:30
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.



Current Student
Joined: 26 May 2005
Posts: 525

Re: Area of trapezium using coordinates
[#permalink]
Show Tags
09 Jun 2011, 08:33
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



Retired Moderator
Joined: 20 Dec 2010
Posts: 1868

Re: Area of trapezium using coordinates
[#permalink]
Show Tags
09 Jun 2011, 08:50
sudhir18n wrote: 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.
_________________
~fluke
GMAT Club Premium Membership  big benefits and savings



Intern
Joined: 27 Apr 2011
Posts: 36
Location: India
Concentration: Marketing, Social Entrepreneurship
GMAT 1: 660 Q47 V33 GMAT 2: 730 Q50 V40
GPA: 3.37
WE: Programming (Computer Software)

Re: Area of trapezium using coordinates
[#permalink]
Show Tags
09 Jun 2011, 21:45
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 xaxis, yaxis and the line y = 6 so the line y = ax + 8, y=6 and xaxis 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 yaxis 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.
Attachments
trapeziumsoln.JPG [ 27.4 KiB  Viewed 7374 times ]



Intern
Joined: 27 Apr 2011
Posts: 36
Location: India
Concentration: Marketing, Social Entrepreneurship
GMAT 1: 660 Q47 V33 GMAT 2: 730 Q50 V40
GPA: 3.37
WE: Programming (Computer Software)

Re: Area of trapezium using coordinates
[#permalink]
Show Tags
09 Jun 2011, 21:54
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.



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8281
Location: Pune, India

Re: Area of trapezium using coordinates
[#permalink]
Show Tags
10 Jun 2011, 11:55
phanideepak wrote: 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 coordinate will be greater than 8 in the first quadrant.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!



Intern
Joined: 27 Apr 2011
Posts: 36
Location: India
Concentration: Marketing, Social Entrepreneurship
GMAT 1: 660 Q47 V33 GMAT 2: 730 Q50 V40
GPA: 3.37
WE: Programming (Computer Software)

Re: Area of trapezium using coordinates
[#permalink]
Show Tags
10 Jun 2011, 21:16
VeritasPrepKarishma wrote: phanideepak wrote: 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 coordinate 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



Intern
Joined: 20 Apr 2013
Posts: 20
Concentration: Finance, Finance
GMAT Date: 06032013
GPA: 3.3
WE: Accounting (Accounting)

Re: On the xycoordinate plane, a quadrilateral is bounded by
[#permalink]
Show Tags
10 May 2013, 11:02
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.



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8281
Location: Pune, India

Re: On the xycoordinate plane, a quadrilateral is bounded by
[#permalink]
Show Tags
06 Oct 2014, 03:12
VeritasPrepKarishma wrote: phanideepak wrote: 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 coordinate 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).
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!



NonHuman User
Joined: 09 Sep 2013
Posts: 8095

Re: On the xycoordinate plane, a quadrilateral is bounded by
[#permalink]
Show Tags
09 May 2017, 19:54
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources




Re: On the xycoordinate plane, a quadrilateral is bounded by &nbs
[#permalink]
09 May 2017, 19:54






