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14 May 2012, 15:32
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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59% (02:18) correct
41%
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wrong
based on 628
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Triangle ABC will be constructed in a xy-plane according to the following conditions: Angle ABC is 90 degrees, and AB is parallel to the y-axis; for each of points A, B and C, both the x-coordinate and the y-coordinate must be integers; the range of possible x-coordinates is 0=<x=<5 and the range of possible y-coordinates is -4=<y=<6. Any two triangles with non-identical vertices are considered different. Given these constraints, how many different triangles could be constructed?
A. 50
B. 66
C. 2500
D. 3300
E. 4356
A. 50
B. 66
C. 2500
D. 3300
E. 4356
14 May 2012, 23:47
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alexpavlos
Triangle ABC will be constructed in a xy-plane according to the following conditions: Angle ABC is 90 degrees, and AB is parallel to the y-axis; for each of points A, B and C, both the x-coordinate and the y-coordinate must be integers; the range of possible x-coordinates is 0=<x=<5 and the range of possible y-coordinates is -4=<y=<6. Any two triangles with non-identical vertices are considered different. Given these constraints, how many different triangles could be constructed?
A. 50
B. 66
C. 2500
D. 3300
E. 4356
A. 50
B. 66
C. 2500
D. 3300
E. 4356
Since give that 0=<x=<5 and -4=<y=<6 then we have a rectangle with dimensions 6*11 (6 horizontal and 11 vertical dots). AB is parallel to y-axis, BC is parallel to x-axis and the right angle is at B.
Choose the (x,y) coordinates for vertex B: 6C1*11C1;
Choose the x coordinate for vertex C (as y coordinate is fixed by B): 5C1, (6-1=5 as 1 horizontal dot is already occupied by B);
Choose the y coordinate for vertex A (as x coordinate is fixed by B): 10C1, (11-1=10 as 1 vertical dot is already occupied by B).
6C1*11C1*5C1*10C1=3,300.
Answer: D.
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right-triangle-pqr-is-to-be-constructed-in-the-xy-plane-so-71597.html
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Hope it helps.
12 Feb 2017, 21:07
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Bookmarks
Hi All,
While this prompt is a bit wordy (and even looks a bit 'complex), it can be broken down into small 'steps' rather easily.
We're given a number of 'restrictions' in terms of how a right triangle can drawn into an XY-plane:
1) The vertices must be comprised of integer values for the X and Y co-ordinates.
2) Segment AB is parallel to the Y-axis.
3) Angle B is the 90 degree angle.
4) The range of possible X co-ordinates is 0 <= X <= 5 and the range of possible Y-coordinates is -4 <= Y <= 6
We're asked to find the number of possible right triangles.
To start, let's place the first point ('Point A'). Considering the limits on the X and Y co-ordinates, there are 6 possible values for X and 11 possible values for Y. This means that there are (6)(11) = 66 possible places that we can place Point A.
Once we place Point A, we now have limitations on where we can place Point B. Since segment AB must be PARALLEL to the Y-axis, it has to have the SAME X-coordinate as Point A. Since we've placed Point A already, the Y-coordinate of Point B must be DIFFERENT from the one that Point A uses, so there are only 10 possibilities for Point B.
Once we place Point B, we now have limitations on where we can place Point C. Since angle B is a 90 degree angle, segment BC must be PARALLEL to the X-axis and Point C has to have the SAME Y-coordinate as Point B. Since we've placed Point B already, the X-coordinate of Point C must be DIFFERENT from the one that Point B uses, so there are only 5 possibilities for Point C.
In total, that gives us (66)(10)(5) = 3300 possible triangles.
Final Answer:
GMAT assassins aren't born, they're made,
Rich
While this prompt is a bit wordy (and even looks a bit 'complex), it can be broken down into small 'steps' rather easily.
We're given a number of 'restrictions' in terms of how a right triangle can drawn into an XY-plane:
1) The vertices must be comprised of integer values for the X and Y co-ordinates.
2) Segment AB is parallel to the Y-axis.
3) Angle B is the 90 degree angle.
4) The range of possible X co-ordinates is 0 <= X <= 5 and the range of possible Y-coordinates is -4 <= Y <= 6
We're asked to find the number of possible right triangles.
To start, let's place the first point ('Point A'). Considering the limits on the X and Y co-ordinates, there are 6 possible values for X and 11 possible values for Y. This means that there are (6)(11) = 66 possible places that we can place Point A.
Once we place Point A, we now have limitations on where we can place Point B. Since segment AB must be PARALLEL to the Y-axis, it has to have the SAME X-coordinate as Point A. Since we've placed Point A already, the Y-coordinate of Point B must be DIFFERENT from the one that Point A uses, so there are only 10 possibilities for Point B.
Once we place Point B, we now have limitations on where we can place Point C. Since angle B is a 90 degree angle, segment BC must be PARALLEL to the X-axis and Point C has to have the SAME Y-coordinate as Point B. Since we've placed Point B already, the X-coordinate of Point C must be DIFFERENT from the one that Point B uses, so there are only 5 possibilities for Point C.
In total, that gives us (66)(10)(5) = 3300 possible triangles.
Final Answer:
GMAT assassins aren't born, they're made,
Rich
