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p&c in triangles

Expert replies
by hey_thr67 » Fri Jun 08, 2012 3:39 am
Right triangle PQR is to be constructed in the xy-plane so that the right angle is at P and PR is parallel to the x-axis. The x and y coordinates of P, Q and R are to be integers that satisfy the inequalities -4 <= x <= 5 and 6 <= y <= 16. How many different triangles with these properties could be constructed?
(A) 110
(B) 1,100
(C) 9,900
(D) 10,000
(E) 12,100

[spoiler]OA is C. I am getting 990 as the answer. Where am I getting wrong ?[/spoiler]
Last edited by hey_thr67 on Fri Jun 08, 2012 3:47 am, edited 1 time in total.
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Source: — Problem Solving |

by Anurag@Gurome » Fri Jun 08, 2012 3:45 am
hey_thr67 wrote:..Where am I getting wrong ?
It's difficult to comment on that without knowing your approach.

Anyway, refer to this post >> https://www.beatthegmat.com/tough-coordi ... tml#327412
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by hey_thr67 » Fri Jun 08, 2012 3:53 am
There are 10 points on x axis. So, to select P we have 10 points. After choosing P we are left with 9 points for R. To select Q we have 11 points.

So, number of ways are : 10 X 9 X 11 = 990
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by Anurag@Gurome » Fri Jun 08, 2012 4:10 am
hey_thr67 wrote:There are 10 points on x axis. So, to select P we have 10 points...
Actually that would be 10*11 points.
There are 10 possible values for the x-coordinate and 11 possible values for y-coordinates of P.

Hope that helps.
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