Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

In the xy-plane, a triangle has vertexes (0,0), (4,0) and [#permalink]
16 Nov 2009, 06:27

1

This post received KUDOS

4

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

55% (hard)

Question Stats:

58% (02:22) correct
42% (01:40) wrong based on 145 sessions

In the xy-plane, a triangle has vertexes (0,0), (4,0) and (4,5). If a point (x,y) is selected at random from the triangular region, What is the probability that x-y>0 ?

In the xy-plane, a triangle has vertexes (0,0), (4,0) and [#permalink]
16 Nov 2009, 06:50

5

This post received KUDOS

Expert's post

apoorvasrivastva wrote:

In the xy- plane, a triangle has vertexes (0,0), (4,0) and (4,5). If a point (x,y) is selected at random from the triangular region, What is the probabilty that x-y>0 ?

A. 1/5 B. 1/3 c. 1/2 D. 2/3 E. 4/5

I am getting 3/5 as the answer..please explain the reasoning..

We have right triangle with the area=4*5/2=10. Consider the line y<x. All the points which satisfy this equation (are below the line y=x) and lie in the triangular region obviously will have x more than y, which is exactly what we want (as x>y --> x-y>0).

The probability that the point will be from this region is: (Area of this region)/(Area of the triangle).

Favorable region is also right triangle with vertexes at (0,0) (0,4) and (4,4). As y=x intersects the side of our original triangle at the point (4,4). You''ll see it easily if you draw it. So favorable are=4*4/2=8.

Re: In the xy-plane, a triangle has vertexes (0,0), (4,0) and [#permalink]
16 Nov 2009, 08:24

Bunuel wrote:

We have right triangle with the area=4*5/2=10. Consider the line y<x. All the points which satisfy this equation (are below the line y=x) and lie in the triangular region obviously will have x more than y, which is exactly what we want (as x>y --> x-y>0).

The probability that the point will be from this region is: Area of this region/Area of the triangle.

Favorable region is also right triangle with vertexes at (0,0) (0,4) and (4,4). As y=x intersects the side of our original triangle at the point (4,4). You''ll see it easily if you draw it. So favorable are=4*4/2=8.

P=8/10=4/5

Answer: E.

well my reasoning goes like this..... P(e) = F(e) / T(e)

T(e) = 4C1 ( ways of selecting any x co-ordinate in the region ) * 5C1 ( ways of selecting any y co ordinate in the region) = 4*5= 20

for F(e) we are required to find x>y .... so for this X co-ordinate can be chosen in 4 ways and the y co-ordinate can be chosen in 3 ways to satisfy the x>y condition

so T(e) 4*3=12

so p(e) = 12/20 = 3/5

i am considering only integers is this the problem??? kindly guide

Re: In the xy-plane, a triangle has vertexes (0,0), (4,0) and [#permalink]
16 Nov 2009, 08:28

apoorvasrivastva wrote:

Bunuel wrote:

We have right triangle with the area=4*5/2=10. Consider the line y<x. All the points which satisfy this equation (are below the line y=x) and lie in the triangular region obviously will have x more than y, which is exactly what we want (as x>y --> x-y>0).

The probability that the point will be from this region is: Area of this region/Area of the triangle.

Favorable region is also right triangle with vertexes at (0,0) (0,4) and (4,4). As y=x intersects the side of our original triangle at the point (4,4). You''ll see it easily if you draw it. So favorable are=4*4/2=8.

P=8/10=4/5

Answer: E.

well my reasoning goes like this..... P(e) = F(e) / T(e)

T(e) = 4C1 ( ways of selecting any x co-ordinate in the region ) * 5C1 ( ways of selecting any y co ordinate in the region) = 4*5= 20

for F(e) we are required to find x>y .... so for this X co-ordinate can be chosen in 4 ways and the y co-ordinate can be chosen in 3 ways to satisfy the x>y condition

so T(e) 4*3=12

so p(e) = 12/20 = 3/5

i am considering only integers is this the problem??? kindly guide

T(e) = 4C1 ( ways of selecting any x co-ordinate in the region ) * 5C1 ( ways of selecting any y co ordinate in the region) = 4*5= 20

this would make the region a rectangle (4 x 5) but you need to multiply by 1/2 since it's a right triangle

Re: In the xy-plane, a triangle has vertexes (0,0), (4,0) and [#permalink]
16 Nov 2009, 08:31

Bunuel wrote:

We have right triangle with the area=4*5/2=10. Consider the line y<x. All the points which satisfy this equation (are below the line y=x) and lie in the triangular region obviously will have x more than y, which is exactly what we want (as x>y --> x-y>0).

The probability that the point will be from this region is: Area of this region/Area of the triangle.

Favorable region is also right triangle with vertexes at (0,0) (0,4) and (4,4). As y=x intersects the side of our original triangle at the point (4,4). You''ll see it easily if you draw it. So favorable are=4*4/2=8.

P=8/10=4/5

Answer: E.

Also if i go by the way u say.... shouldn't the probability be <4/5?

probability = area of region where x-y>0 / total are of region = x/0.5*4*5 = x/10

point (4,4) is where x-y=0. point(4, <4) is where x-y>0.

x = 0.5 * 4*(<4) = <8 hence x/10 = <8/10 = <4/5

the point here is that to get 4/5 you would have to take the region where x and y = 4 in which case x-y is not >0. any value of y below 4 would give x-y>0 hence the probability should be less than 4/5...I dont think you can place an exact value on the probability as x and y can be decimals...

Re: In the xy-plane, a triangle has vertexes (0,0), (4,0) and [#permalink]
16 Nov 2009, 08:32

1

This post received KUDOS

for F(e) we are required to find x>y .... so for this X co-ordinate can be chosen in 4 ways and the y co-ordinate can be chosen in 3 ways to satisfy the x>y condition

the y can be chosen in 4 ways as well (4 x 4) * 1/2 (for a triangle) = 8

think of the y = x coordinate

0,0 1,1 2,2 3,3 4,4

anything below this line satisfies the equation x>y

Re: In the xy-plane, a triangle has vertexes (0,0), (4,0) and [#permalink]
16 Nov 2009, 08:53

lagomez wrote:

for F(e) we are required to find x>y .... so for this X co-ordinate can be chosen in 4 ways and the y co-ordinate can be chosen in 3 ways to satisfy the x>y condition

the y can be chosen in 4 ways as well (4 x 4) * 1/2 (for a triangle) = 8

think of the y = x coordinate

0,0 1,1 2,2 3,3 4,4

anything below this line satisfies the equation x>y

ok now i get it!!!!

The probability for x>y is nothing but = 1 - P(y>x) = 1- 1/5 = 4/5

reasoning as follows

probability = area of region where y>0 / total are of region = x/0.5*4*5 = x/10

Re: In the xy-plane, a triangle has vertexes (0,0), (4,0) and [#permalink]
16 Nov 2009, 11:26

Expert's post

apoorvasrivastva wrote:

Also if i go by the way u say.... shouldn't the probability be <4/5?

probability = area of region where x-y>0 / total are of region = x/0.5*4*5 = x/10

point (4,4) is where x-y=0. point(4, <4) is where x-y>0.

x = 0.5 * 4*(<4) = <8 hence x/10 = <8/10 = <4/5

the point here is that to get 4/5 you would have to take the region where x and y = 4 in which case x-y is not >0. any value of y below 4 would give x-y>0 hence the probability should be less than 4/5...I dont think you can place an exact value on the probability as x and y can be decimals...

I responded to this same post on BTG, but I'll paste that here:

To respond specifically to the concern above: the answer is still exactly 4/5, regardless of whether the question says x-y > 0 or x-y > 0. Note that the region where x-y=0 is just the line x=y; its area is zero. So you can subtract its area in your calculation if you like, but it won't change the answer. _________________

GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

Re: In the xy-plane, a triangle has vertexes (0,0), (4,0) and [#permalink]
16 Nov 2009, 14:39

apoorvasrivastva wrote:

In the xy- plane, a triangle has vertexes (0,0), (4,0) and (4,5). If a point (x,y) is selected at random from the triangular region, What is the probabilty that x-y>0 ?

A. 1/5 B. 1/3 c. 1/2 D. 2/3 E. 4/5

I am getting 3/5 as the answer..please explain the reasoning..

x - y > 0 or x > y is possible below the line x = y in the given triangle / co-ordinate system. i.e. the area below x = y line in the triangle is the prob that x>y. So draw a line from origin to the point (4, 4). This makes a triangle with vertexes (0,0), (4,0) and (4,4).

Given triangle with vertexes (0,0), (4,0) and (4,5) has an area of 10. The area of the triangle with vertexes (0,0), (4,0) and (4,4) is 8.

Area of new triangle/Area of the original triangle = 8/10 = 4/5. This is the prob that x>y. _________________

Re: In the xy-plane, a triangle has vertexes (0,0), (4,0) and [#permalink]
16 Nov 2009, 14:48

Expert's post

GMAT TIGER wrote:

x - y > 0 or x > y is possible below the line x = y in the given triangle / co-ordinate system. i.e. the area below x = y line in the given triangle is the prob that x>y. So draw a line from origin to the point (4, 4). This line makes a triangle with vertexes (0,0), (4,0) and (4,4).

Given triangle with vertexes (0,0), (4,0) and (4,5) has an area of 20. The area of the triangle with vertexes (0,0), (4,0) and (4,4) is 16.

Area of new triangle/Area of the original triangle = 16/20 = 4/5. This is the prob that x>y.

Areas are 10 and 8 accordingly. You forgot to divide by 2 in both cases. Though it didn't affect the final answer. _________________

Re: In the xy-plane, a triangle has vertexes (0,0), (4,0) and [#permalink]
17 Nov 2009, 13:30

Bunuel wrote:

GMAT TIGER wrote:

x - y > 0 or x > y is possible below the line x = y in the given triangle / co-ordinate system. i.e. the area below x = y line in the given triangle is the prob that x>y. So draw a line from origin to the point (4, 4). This line makes a triangle with vertexes (0,0), (4,0) and (4,4).

Given triangle with vertexes (0,0), (4,0) and (4,5) has an area of 20. The area of the triangle with vertexes (0,0), (4,0) and (4,4) is 16.

Area of new triangle/Area of the original triangle = 16/20 = 4/5. This is the prob that x>y.

Areas are 10 and 8 accordingly. You forgot to divide by 2 in both cases. Though it didn't affect the final answer.

Thanks Bunuel. You are amezing..

Updated ...taking advantage of being a moderator. _________________

Re: In the xy-plane, a triangle has vertexes (0,0), (4,0) and [#permalink]
24 Jul 2013, 23:10

Bunuel wrote:

Favorable region is also right triangle with vertexes at (0,0) (0,4) and (4,4). Answer: E.

Hi Bunuel,

Since we have to select a point within the given triangular region formed by (0,0(, (4,0) and (4,5), isn't the favorable region formed by (0,0), (4,4) and (4,5)? If not, kindly clarify how the vertexes for favorable region would be (0,0) (0,4) and (4,4).

Re: In the xy-plane, a triangle has vertexes (0,0), (4,0) and [#permalink]
02 Sep 2013, 12:21

Here is how I thought of the problem. Considering only integers, the X values can be 1,2,3 or 4. For either of these four values, in order to get X-Y > 0, Y values have to be one of (0,1,2 or 3) Probability of both of these events is (4/5)*(4/4). Hence I got the value 4/5

idk if this worked out only because of a fluke or is this wrong?

Re: In the xy-plane, a triangle has vertexes (0,0), (4,0) and [#permalink]
02 Sep 2013, 13:15

Expert's post

Jaisri wrote:

Hi Bunuel, Since we have to select a point within the given triangular region formed by (0,0(, (4,0) and (4,5), isn't the favorable region formed by (0,0), (4,4) and (4,5)? If not, kindly clarify how the vertexes for favorable region would be (0,0) (0,4) and (4,4).

In the xy-plane, a triangle has vertexes (0,0), (4,0) and (4,5). If a point (x,y) is selected at random from the triangular region, What is the probability that x-y>0 ?

Re: In the xy-plane, a triangle has vertexes (0,0), (4,0) and [#permalink]
20 Jul 2014, 05:28

Expert's post

Bunuel wrote:

apoorvasrivastva wrote:

In the xy- plane, a triangle has vertexes (0,0), (4,0) and (4,5). If a point (x,y) is selected at random from the triangular region, What is the probabilty that x-y>0 ?

A. 1/5 B. 1/3 c. 1/2 D. 2/3 E. 4/5

I am getting 3/5 as the answer..please explain the reasoning..

We have right triangle with the area=4*5/2=10. Consider the line y<x. All the points which satisfy this equation (are below the line y=x) and lie in the triangular region obviously will have x more than y, which is exactly what we want (as x>y --> x-y>0).

The probability that the point will be from this region is: (Area of this region)/(Area of the triangle).

Favorable region is also right triangle with vertexes at (0,0) (0,4) and (4,4). As y=x intersects the side of our original triangle at the point (4,4). You''ll see it easily if you draw it. So favorable are=4*4/2=8.

Harvard asks you to write a post interview reflection (PIR) within 24 hours of your interview. Many have said that there is little you can do in this...