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From among all the triangles that could be drawn in the coordinate
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26 Aug 2018, 20:24
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From among all the triangles that could be drawn in the coordinate plane with vertices with integer coordinates (x,y) satisfying 0 ≤ x ≤ 3 and 0 ≤ y ≤ 3, one triangle is to be chosen at random. If one of the edges of this triangle lies on the x-axis, what is the probability the area of the triangle is an integer ? a) 1/3 B)1/2 C)5/9 D)2/3 E)3/4
From among all the triangles that could be drawn in the coordinate
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27 Aug 2018, 00:14
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Raksat wrote:
From among all the triangles that could be drawn in the coordinate plane with vertices with integer coordinates (x,y) satisfying 0 ≤ x ≤ 3 and 0 ≤ y ≤ 3, one triangle is to be chosen at random. If one of the edges of this triangle lies on the x-axis, what is the probability the area of the triangle is an integer ? a) 1/3 B)1/2 C)5/9 D)2/3 E)3/4
since edge is on x axis, two vertices will be on x axis that is \(( x_1,0)\) and \((x_2,0)\) and x can take any of the 4 values - 0,1,2,3
I. total triangles possible = choose any two of 0,1,2,3 so 4C2=6 and for this the vertex can be any of the 3*4=12, 3 for y= 1,2, and 3 and 4 for x as 0,1,2,3 total 12*6=72
II. when area = integer for this one of the side is EVEN as area = \(\frac{1}{2} * x*y\) when the edge on x axis is odd ... (0,0) and (1,0); (1,0) and (2,0); (2,0) and (3,0); (0,0) and (3,0); so 4 cases y has to be even so 2 so (0,2), (1,2),(2,2),(3,2) so 4 cases total 4*4=16 when the edge on x axis is even ... (0,0) and (2,0); (1,0) and (3,0); so 2 cases y can be anything so 2 so 1,2 and 3 and x 0,1,2,3 so 3*4 = 12 cases total 2*12=24 Overall = 16+24=40
Re: From among all the triangles that could be drawn in the coordinate
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27 Aug 2018, 04:00
chetan2u wrote:
Raksat wrote:
From among all the triangles that could be drawn in the coordinate plane with vertices with integer coordinates (x,y) satisfying 0 ≤ x ≤ 3 and 0 ≤ y ≤ 3, one triangle is to be chosen at random. If one of the edges of this triangle lies on the x-axis, what is the probability the area of the triangle is an integer ? a) 1/3 B)1/2 C)5/9 D)2/3 E)3/4
since edge is on x axis, two vertices will be on x axis that is \(( x_1,0)\) and \((x_2,0)\) and x can take any of the 4 values - 0,1,2,3
I. total triangles possible = choose any two of 0,1,2,3 so 4C2=6 and for this the vertex can be any of the 3*4=12, 3 for y= 1,2, and 3 and 4 for x as 0,1,2,3 total 12*6=72
II. when area = integer for this one of the side is EVEN as area = \(\frac{1}{2} * x*y\) when the edge on x axis is odd ... (0,0) and (1,0); (1,0) and (2,0); (2,0) and (3,0); (0,0) and (3,0); so 4 cases y has to be even so 2 so (0,2), (1,2),(2,2),(3,2) so 4 cases total 4*4=16 when the edge on x axis is even ... (0,0) and (2,0); (1,0) and (3,0); so 2 cases y can be anything so 2 so 1,2 and 3 and x 0,1,2,3 so 3*4 = 12 cases total 2*12=24 Overall = 16+24=40
prob = \(\frac{40}{72}=\frac{5}{9}\)
hi chetan2u, nice explanation though i do have questions if you dont mind
can you please help me to understand your solution ... i definately dont understand one thing how can we create a triangle with the coordinates in which
x or y is 0 for instance you wrote these coordinates, see in red (0,0) and (1,0); (1,0) and (2,0); (2,0) and (3,0); (0,0) and (3,0); so 4 cases y has to be even so 2 so (0,2), (1,2),(2,2),(3,2) so 4 cases
see attached diagram this is how i understood the problem as you see for one of the edges of triangles to be on X axis i got folowing options (3:3), (2:3) (2;2) , (2:1), (1:1)
Also i didnt get this "and for this the vertex can be any of the 3*4=12, 3 for y= 1,2, and 3 and 4 for x as 0,1,2,3 total 12*6=72" can you please somehow rephrase the explanation the only thing i got is that triangle has 3 vertices ....
And one more question how did you manage to solve it without visualizing the problem ?
i would appreciate your taking time to explain thank you!
From among all the triangles that could be drawn in the coordinate
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27 Aug 2018, 07:59
chetan2u wrote:
Raksat wrote:
From among all the triangles that could be drawn in the coordinate plane with vertices with integer coordinates (x,y) satisfying 0 ≤ x ≤ 3 and 0 ≤ y ≤ 3, one triangle is to be chosen at random. If one of the edges of this triangle lies on the x-axis, what is the probability the area of the triangle is an integer ? a) 1/3 B)1/2 C)5/9 D)2/3 E)3/4
since edge is on x axis, two vertices will be on x axis that is \(( x_1,0)\) and \((x_2,0)\) and x can take any of the 4 values - 0,1,2,3
I. total triangles possible = choose any two of 0,1,2,3 so 4C2=6 and for this the vertex can be any of the 3*4=12, 3 for y= 1,2, and 3 and 4 for x as 0,1,2,3 total 12*6=72
II. when area = integer for this one of the side is EVEN as area = \(\frac{1}{2} * x*y\) when the edge on x axis is odd ... (0,0) and (1,0); (1,0) and (2,0); (2,0) and (3,0); (0,0) and (3,0); so 4 cases y has to be even so 2 so (0,2), (1,2),(2,2),(3,2) so 4 cases total 4*4=16 when the edge on x axis is even ... (0,0) and (2,0); (1,0) and (3,0); so 2 cases y can be anything so 2 so 1,2 and 3 and x 0,1,2,3 so 3*4 = 12 cases total 2*12=24 Overall = 16+24=40
prob = \(\frac{40}{72}=\frac{5}{9}\)
how is it possible to draw a triangle with these coordinates (0,0) and (1,0) ? can someone draw such triangle ?
Re: From among all the triangles that could be drawn in the coordinate
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27 Aug 2018, 08:07
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dave13 wrote:
chetan2u wrote:
Raksat wrote:
From among all the triangles that could be drawn in the coordinate plane with vertices with integer coordinates (x,y) satisfying 0 ≤ x ≤ 3 and 0 ≤ y ≤ 3, one triangle is to be chosen at random. If one of the edges of this triangle lies on the x-axis, what is the probability the area of the triangle is an integer ? a) 1/3 B)1/2 C)5/9 D)2/3 E)3/4
since edge is on x axis, two vertices will be on x axis that is \(( x_1,0)\) and \((x_2,0)\) and x can take any of the 4 values - 0,1,2,3
I. total triangles possible = choose any two of 0,1,2,3 so 4C2=6 and for this the vertex can be any of the 3*4=12, 3 for y= 1,2, and 3 and 4 for x as 0,1,2,3 total 12*6=72
II. when area = integer for this one of the side is EVEN as area = \(\frac{1}{2} * x*y\) when the edge on x axis is odd ... (0,0) and (1,0); (1,0) and (2,0); (2,0) and (3,0); (0,0) and (3,0); so 4 cases y has to be even so 2 so (0,2), (1,2),(2,2),(3,2) so 4 cases total 4*4=16 when the edge on x axis is even ... (0,0) and (2,0); (1,0) and (3,0); so 2 cases y can be anything so 2 so 1,2 and 3 and x 0,1,2,3 so 3*4 = 12 cases total 2*12=24 Overall = 16+24=40
prob = \(\frac{40}{72}=\frac{5}{9}\)
how is it possible to draw a triangle with these coordinates (0,0) and (1,0) ? can someone draw such triangle ?
Hi.. Now coordinate(0,0) & (1,0) will give two vertices say A and B now C can be any of the 12 points above. So C can be (0,1) or (1,1) or (2,1), or (3,1) or (0,2) or (1,2) and so on and all these C coordinates will give different triangles Similarly for (0,0) and (0,2) , again C can be any of the coordinates as shown above
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Re: From among all the triangles that could be drawn in the coordinate
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27 Aug 2018, 08:15
thank you got it ! could you please rephrase the following sentence if you don't mind "and for this the vertex can be any of the 3*4=12, 3 for y= 1,2, and 3 and 4 for x as 0,1,2,3 total 12*6=72"
Re: From among all the triangles that could be drawn in the coordinate
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13 Oct 2018, 07:13
chetan2u wrote:
Raksat wrote:
From among all the triangles that could be drawn in the coordinate plane with vertices with integer coordinates (x,y) satisfying 0 ≤ x ≤ 3 and 0 ≤ y ≤ 3, one triangle is to be chosen at random. If one of the edges of this triangle lies on the x-axis, what is the probability the area of the triangle is an integer ? a) 1/3 B)1/2 C)5/9 D)2/3 E)3/4
since edge is on x axis, two vertices will be on x axis that is \(( x_1,0)\) and \((x_2,0)\) and x can take any of the 4 values - 0,1,2,3
I. total triangles possible = choose any two of 0,1,2,3 so 4C2=6 and for this the vertex can be any of the 3*4=12, 3 for y= 1,2, and 3 and 4 for x as 0,1,2,3 total 12*6=72
II. when area = integer for this one of the side is EVEN as area = \(\frac{1}{2} * x*y\) when the edge on x axis is odd ... (0,0) and (1,0); (1,0) and (2,0); (2,0) and (3,0); (0,0) and (3,0); so 4 cases y has to be even so 2 so (0,2), (1,2),(2,2),(3,2) so 4 cases total 4*4=16 when the edge on x axis is even ... (0,0) and (2,0); (1,0) and (3,0); so 2 cases y can be anything so 2 so 1,2 and 3 and x 0,1,2,3 so 3*4 = 12 cases total 2*12=24 Overall = 16+24=40
prob = \(\frac{40}{72}=\frac{5}{9}\)
Hi chetan2u
If you don't mind could you pls explain a bit more how you are calculating the total number of triangles.
Re: From among all the triangles that could be drawn in the coordinate
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13 Oct 2018, 21:33
chetan2u wrote:
Raksat wrote:
From among all the triangles that could be drawn in the coordinate plane with vertices with integer coordinates (x,y) satisfying 0 ≤ x ≤ 3 and 0 ≤ y ≤ 3, one triangle is to be chosen at random. If one of the edges of this triangle lies on the x-axis, what is the probability the area of the triangle is an integer ? a) 1/3 B)1/2 C)5/9 D)2/3 E)3/4
since edge is on x axis, two vertices will be on x axis that is \(( x_1,0)\) and \((x_2,0)\) and x can take any of the 4 values - 0,1,2,3
I. total triangles possible = choose any two of 0,1,2,3 so 4C2=6 and for this the vertex can be any of the 3*4=12, 3 for y= 1,2, and 3 and 4 for x as 0,1,2,3 total 12*6=72
II. when area = integer for this one of the side is EVEN as area = \(\frac{1}{2} * x*y\) when the edge on x axis is odd ... (0,0) and (1,0); (1,0) and (2,0); (2,0) and (3,0); (0,0) and (3,0); so 4 cases y has to be even so 2 so (0,2), (1,2),(2,2),(3,2) so 4 cases total 4*4=16 when the edge on x axis is even ... (0,0) and (2,0); (1,0) and (3,0); so 2 cases y can be anything so 2 so 1,2 and 3 and x 0,1,2,3 so 3*4 = 12 cases total 2*12=24 Overall = 16+24=40
prob = \(\frac{40}{72}=\frac{5}{9}\)
Hi Chetan, Can you please elaborate on the line " for this the vertex can be any of the 3*4=12, 3 for y= 1,2, and 3 and 4 for x as 0,1,2,3 total 12*6=72".
I am nt very clear how you got 12 combinations for those 6 ways
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Re: From among all the triangles that could be drawn in the coordinate
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13 Oct 2018, 21:33
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50% (02:30) correct 
for instance you wrote these coordinates, see in red 
thank you! 
