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In the xy-plane, a triangle has vertexes (0,0), (4,0) and

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16 Nov 2009, 07:27
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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 ?

A. 1/5
B. 1/3
C. 1/2
D. 2/3
E. 4/5
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In the xy-plane, a triangle has vertexes (0,0), (4,0) and  [#permalink]

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24 Dec 2009, 08:35
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In the xy- plane, a triangle has vertexes (0,0), (4,0) and  [#permalink]

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24 Dec 2009, 08:09
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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 ?

A. 1/5
B. 1/3
C. 1/2
D. 2/3
E. 4/5
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In the xy-plane, a triangle has vertexes (0,0), (4,0) and  [#permalink]

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16 Nov 2009, 07:50
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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 ?

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

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 area = 4*4/2 = 8.

$$P=\frac{8}{10}=\frac{4}{5}$$.

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In the xy-plane, a triangle has vertexes (0,0), (4,0) and  [#permalink]

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16 Nov 2009, 09:31
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

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...
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Re: In the xy-plane, a triangle has vertexes (0,0), (4,0) and  [#permalink]

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16 Nov 2009, 12:26
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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.
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Re: In the xy-plane, a triangle has vertexes (0,0), (4,0) and  [#permalink]

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16 Nov 2009, 15:39
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

OA is 4/5

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.
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Re: In the xy-plane, a triangle has vertexes (0,0), (4,0) and  [#permalink]

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16 Nov 2009, 15:48
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.
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Re: In the xy- plane,  [#permalink]

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03 Jan 2013, 01:31
1
sagarsabnis 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 probability that x-y>0 ?

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

My approach is to use the x=y line which is the boundary that divides x>y and x<y coordinate system. Below the x=y line is the region of points with x > y. Above the x=y line is the region of points with x < y.

Get the area of the given triangle: (4)(5) / 2 = 10
Get the area of the smaller triangle drawn below the line x=y which has (0,0), (4,0) and (4,4) for its coordinates: (4)(4)/2 = 8

P = 8/10 = 4/5

Detailed Solution: Line x=y as a Boundary
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Re: In the xy- plane, a triangle has vertexes (0,0), (4,0) and  [#permalink]

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15 Mar 2013, 14:55
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3
See attached drawing.

The probability will be the area of the dark triangle divided by the area of the original triangle:

Area dark triangle (where X>Y is true): $$\frac{1}{2}*4*4$$

Area original triangle: $$\frac{1}{2}*4*5$$

$$Probability=\frac{(0.5*4*4)}{(0.5*4*5)}=\frac{4}{5}$$

Solution E
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File comment: Drawing resolution

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

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26 Mar 2013, 08:55
1
johnwesley wrote:
See attached drawing.

The probability will be the area of the dark triangle divided by the area of the original triangle:

Area dark triangle (where X>Y is true): $$\frac{1}{2}*4*4$$

Area original triangle: $$\frac{1}{2}*4*5$$

$$Probability=\frac{(0.5*4*4)}{(0.5*4*5)}=\frac{4}{5}$$

Solution E

What about (4,4)? that is in the shaded region, but is not greater than 0 (or algebraically x=4 is not greater than y=4)?
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Re: In the xy- plane, a triangle has vertexes (0,0), (4,0) and  [#permalink]

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27 Mar 2013, 08:47
perilm41 wrote:
johnwesley wrote:
See attached drawing.

The probability will be the area of the dark triangle divided by the area of the original triangle:

Area dark triangle (where X>Y is true): $$\frac{1}{2}*4*4$$

Area original triangle: $$\frac{1}{2}*4*5$$

$$Probability=\frac{(0.5*4*4)}{(0.5*4*5)}=\frac{4}{5}$$

Solution E

What about (4,4)? that is in the shaded region, but is not greater than 0 (or algebraically x=4 is not greater than y=4)?

A point by definition has no length, area or any other dimension, thus this won't affect the answer.

Check this discussion for more: a-5-meter-long-wire-is-cut-into-two-pieces-if-the-longer-106448.html
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Re: In the xy-plane, a triangle has vertexes (0,0), (4,0) and  [#permalink]

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25 Jul 2013, 00:10
Bunuel wrote:
Favorable region is also right triangle with vertexes at (0,0) (0,4) and (4,4).

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).
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Re: In the xy-plane, a triangle has vertexes (0,0), (4,0) and  [#permalink]

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02 Sep 2013, 14:15
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).

Yes, The points fall below the line y = x will give us x > y and hence it is the favorable region.

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 probability that x-y>0 ?

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

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In the xy-plane, a triangle has vertexes (0,0), (4,0) and  [#permalink]

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20 Jul 2017, 17:08
Taking the third vertex as (4,4) means, to me, x=y not x>y. am I wrong?
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In the xy-plane, a triangle has vertexes (0,0), (4,0) and  [#permalink]

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20 Jul 2017, 20:51
Yoohanna wrote:
Taking the third vertex as (4,4) means, to me, x=y not x>y. am I wrong?

A point has no dimension, so including (4, 4) won't affect the probability.
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Re: In the xy- plane, a triangle has vertexes (0,0), (4,0) and  [#permalink]

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19 Aug 2017, 10:54
Bunuel wrote:
sagarsabnis 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 probability that x-y>0 ?

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

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) (4,0) 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

Just out of curiosity!

In the above case, what would be the probability that x-y<0 and that x=y?

I think P(x-y<0)=0.2, but don't know exactly the P(x=y). It may be 0, but don't know why.

P(x-y>0)=0.8
P(x-y<0)=0.2
P(x=y)=0
Total 1.0

Would you help me?
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Re: In the xy- plane, a triangle has vertexes (0,0), (4,0) and  [#permalink]

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19 Aug 2017, 11:03
1
Mahmud6 wrote:
Bunuel wrote:
sagarsabnis 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 probability that x-y>0 ?

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

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) (4,0) 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

Just out of curiosity!

In the above case, what would be the probability that x-y<0 and that x=y?

I think P(x-y<0)=0.2, but don't know exactly the P(x=y). It may be 0, but don't know why.

P(x-y>0)=0.8
P(x-y<0)=0.2
P(x=y)=0
Total 1.0

Would you help me?

A line segment has only one dimension, length, no area. So, P(x=y) = 0.
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Re: In the xy- plane, a triangle has vertexes (0,0), (4,0) and  [#permalink]

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19 Aug 2017, 11:13
Ans is shown in picture as part of my calculation
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File comment: Clearly ans is E

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

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18 Apr 2018, 09:11
+1 for option E. A great question combining the concepts of probability and coordinate geometry. Draw the graph and the required probability will be the ratio of two areas. 8/10 i.e, 4/5 or option E.
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Re: In the xy-plane, a triangle has vertexes (0,0), (4,0) and   [#permalink] 18 Apr 2018, 09:11

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