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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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47% (01:38) correct 53% (01:59) wrong based on 66 sessions
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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 xaxis, 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
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From among all the triangles that could be drawn in the coordinate
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27 Aug 2018, 00:14
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 xaxis, 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=16when 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=24Overall = 16+24=40 prob = \(\frac{40}{72}=\frac{5}{9}\)
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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 xaxis, 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=16when 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=24Overall = 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!
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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 xaxis, 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=16when 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=24Overall = 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 ?



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Re: From among all the triangles that could be drawn in the coordinate
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27 Aug 2018, 08:07
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 xaxis, 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=16when 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=24Overall = 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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1) Absolute modulus : http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effectsofarithmeticoperationsonfractions269413.html
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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"
just need to understand it



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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 xaxis, 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=16when 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=24Overall = 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. Thank you in advance.



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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
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 xaxis, 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=16when 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=24Overall = 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 &nbs
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