# Beat The GMAT Challenge Question – August 30, 2010:

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Triangular region

Tlies on thexy-coordinate plane and has vertices at (0,4), (10,0) and (0,0). If a point in regionTis randomly selected, what is the probability that the point’sx-coordinate will be greater than itsy-coordinate?

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## 21 comments

Varun on August 30th, 2010 at 7:20 am

Is it 1/5?

Praveen on August 30th, 2010 at 7:45 am

Let the point (0,4) is A and (10,0) is B, origin is O

eqtn of the line passing thru (0,4) and (10,0) is

x/10 = 1-y/4

The line x=y will intersect this line at (20/7,20/7)

Let the intersection point be C

Now, Any point in the triangle COB will have the x coordinate greater than y coordinate.

hence probability = area of triangle COB / area of trianlge AOB

area of triangle COB = 1/2 * 20/7 * 10 = 100/7

Area of triangle AOB = 1/2 * 10 * 4 = 20

Probability = (100/7)/20 = 5/7

ZHANG-CHN on August 30th, 2010 at 8:04 am

5/7

equation of the line is y=-2/5x+4

x>y,so x>-2/5x+4,x>20/7

x ranges from 0 to 10

the proportion of the length where 20/7<x<10 to 0<x<10 equals to 5/7,which is the answer to the question.

the key is to draw a chart, and pay attention to the meaning in it.

Ankur on August 30th, 2010 at 10:01 am

Is it 4/5.

plg_cp on August 30th, 2010 at 11:17 am

5/7. The key is knowing to find the intersection of the hypotenuse and the line x=y.

I am only starting my GMAT study. I understand how to get the answer to this question but it took me a lot longer than 2 minutes. Should this question be solved in under 2 minutes by someone who is ready to write the test?

rybalgri on August 30th, 2010 at 11:43 am

The answer is 4/5.

The area where y>x is a half of the rectangle area (0;0) (0;4) (4;4) (4;0). To check this dray the function x=y. It will go trough the points (0;0) (4;4). Any point above the function will meet the inequation y>x. It's exactly what we are NOT looking for. The area of this area is 4*4/2 = 8. The whole area is 10*4 = 40. Thus, the area which will meet the requirement of the tast is 40 - 8 = 32. The probobility is required area/whole are, 32/40 = 4/5

Gurpreet singh on August 30th, 2010 at 2:34 pm

http://img411.imageshack.us/img411/4862/47768147.jpg

Check the image above. We need to find the ratio of yellow area and the given triangle.

The yellow area is basically the area formed by the three lines : Hypotenuse of the given triangle, y=0 and x=y. Area on the right hand side of x-y =0 is basically for x-y>0. This can be verified by putting 10,0 in x-y = 10-0 >0.

Now we need to find out the point of intersection of x=y and the hypotenuse of the triangle.

Hypotenuse equation is : (y-0)/(x-10) = (4-0)/(0-10)

=> 5y+2x = 20

For the point of intersection put x=y, which gives x=y=20/7.

Required probability = (Area of yellow region)/ (Area of given triangle)

= (1/2 * 20/7 * 10 )/ (1/2* 4* 10) = 5/7

Ashok on August 31st, 2010 at 12:33 am

Nice job Gurpreet!!

MBA Dreamer on August 31st, 2010 at 9:55 am

Sardar ji, tussi great ho!~

Eric Bahn on August 31st, 2010 at 6:40 pm

Congrats Gurpeet, you've been selected as this week's Challenge Question winner! We'll be reaching out to you in the next few minutes with instructions to get free premium access to Beat The GMAT Practice Questions.

Nice work!

Gurpreet singh on August 31st, 2010 at 7:00 pm

Thanks Ashok and MBADreamer for your comments.

Thanks Eric for considering me. This will certainly help me a lot in my preparation.

hemanth on August 30th, 2010 at 6:40 pm

Same as Praveen's - 5/7.

Find the area of lower triangle when you draw the line (with equation x=y). All the points here are where you have x>y.

siddharth on August 30th, 2010 at 6:41 pm

The problem specifies a specific area of the X-Y plane & then asks- for what% of this area is x>y. This clearly tells us that we need to find the area of the shape bounded by x>y (i.e. x-y>0) & the traingle specified. The hypotenuse of the triangle specified by the given intercepts is- X/10 + Y/4=1 (incidentallly, this is a s standard and useful form of line equation if X & Y intercepts are given). The point of intersection of X=Y & this line is (20/7,20/7). So the area of X>Y intercepted by the triangle given is- 1/2*10*20/7 = 100/7. Total area of traingle is - 1/2*10*4= 20. So the required probability is- 100/7/20=5/7.

shibbirahamed on August 30th, 2010 at 10:47 pm

5/7

the intersection point of the hypotenuse and the line x=y is (20/7,20/7). let the point be 'P' and the origin be 'O' and the vertic (10,0) be 'Q'. thus the probability is

(Area of OPQ)/Area of whole=5/7=(.5*10*20/7)/(.5*10*4).

isn't it?

Pardeep on August 30th, 2010 at 11:58 pm

5/7

Ashok on August 31st, 2010 at 12:34 am

The answer is indeed 5/7.

Udit on August 31st, 2010 at 1:41 am

Anwer is 5/7. Look at the intersection of line y=x and the line formed by the given cordinates (0,4) and (10,0). the intersection is (20/7,20/7). So the points which lie right to the line y=x will have x>y. So, the probability is area of Triangle formed by (0,0), (10,0) and (20/7,20/7) over area of triangle formed by (0,0), (10,0) and (0,4) which comes out to be 5/7.

yulia on August 31st, 2010 at 4:01 am

My answer is 5/7.

1. find the equation : y = mx+b --> y = -2/5x+4. Set x=y (x intersects with y at the point 20/7).

2. Probability = (1/2*10*20/7) / (1/2*10*4) = (100/7) / 20 = 5/7.

niksworth on August 31st, 2010 at 5:55 am

The region T is defined as area bounded by points O(0,0), A(10,0) and B(0,4).

Let the straight line y=x intersect AB at P. Then the area in region T where x>y will be triangle OPA.

Since we know A(10,0) and B(0,4), equation of straight line AB is :

y = -0.4x + 4.

Solving with y=x, we get the coordinates of P as (20/7,20/7).

The required region where x>y is the region bounded by OPB.

Thus required probability = Area of OPB/Area of OAB.

Now, Area of OPB = 1/2 * Base * Height = 1/2*10*20/7 = 100/7

Area of OAB = 1/2 * Base * Height = 1/2*10*4 = 20.

Thus required probability = Area of OPB/Area of OAB = 5/7.

niksworth on August 31st, 2010 at 5:58 am

Oops, Error in naming points.

Area of OPA = 1/2 * Base * Height = 1/2*10*20/7 = 100/7

Area of OAB = 1/2 * Base * Height = 1/2*10*4 = 20.

Required probability = Area of OPA/Area of OAB = 5/7.

Swarna on August 31st, 2010 at 2:38 pm

Yes.. The answer is 5/7 for this.

The coordinates given form a right angle triangle , with one vertex at origin, one at x axis and one at y axis.

the area of this triangle = 1/2 * 10 * 4 = 20 units.

now, we now that if we draw a line which passes through points (0,0) , (1,1), (2,2) .. then all points below this line will have x co-ordinate greater than y co-ordinate. so if compute the area of the triangle formed by the above line , x axis , and the hypotenuse of the triangle can give us the area in which x co-ordinate will always be greater than y co-ordinate. OK.

so to compute - we know the base = 10

height = ?

to compute the height, we need to compute the intersection point of line y =x and line joining the points (10,0) & (0,4). the intersection point will be (20/7, 20/7).

so here comes the height = 20/7

area of the traingle = 1/2 * 10 * 20 / 7 = 100 /7

probability = chances in favour / total chances = (100/7) / 20

= 5/7

so the answer 5/7 as given my many others also, seem correct to me.