-
vineeshp
- Legendary Member
- Posts: 965
- Joined: Thu Jan 28, 2010 12:52 am
- Thanked: 156 times
- Followed by:34 members
- GMAT Score:720
Source: GROCKIT.
Hi,
I need help with the question Q1 below.
Q2) Right triangle ABC is to be drawn in the xy-plane so that the right angle is at A and AB is parallel to the y-axis. If the x- and y-coordinates of A, B, and C are to be integers that are consistent with the inequalities -6 ≤ x ≤ 2 and 4 ≤ y ≤ 9 , then how many different triangles can be drawn that will meet these conditions?
To solve this, I went with the following.
that A can be chosen in 9 * 6 = 54 ways.
B will have the same Y coordinate only x varies and should not be identical to A. So B can be chosen in 9-1 = 8 ways.
C can similarly be chosen in 6 -1 = 5 ways.
So total no. of ways of choosing the triangle is 54 * 8 * 5 = 2160 ways. This came out correct!
Let us go to Q1:
Q1) Rectangle ABCD is constructed in the coordinate plane parallel to the x- and y-axes. If the x- and y-coordinates of each of the points are integers which satisfy 3 ≤ x ≤ 11 and -5 ≤ y ≤ 5, how many possible ways are there to construct rectangle ABCD?
(Note that two rectangles that have the same four vertices that are labeled differently are considered to be the same rectangle.)
Answer is 1980.
Now in Q1 above, I must be able to employ the same technique I used for the triangle in Q1 because a rectangle's 4th coordinate must use the x and y coordinates of the adjacent vertices. When I apply that method I get to 7920. But the answer seems to be 1980.
Where am I going wrong?
Hi,
I need help with the question Q1 below.
Q2) Right triangle ABC is to be drawn in the xy-plane so that the right angle is at A and AB is parallel to the y-axis. If the x- and y-coordinates of A, B, and C are to be integers that are consistent with the inequalities -6 ≤ x ≤ 2 and 4 ≤ y ≤ 9 , then how many different triangles can be drawn that will meet these conditions?
To solve this, I went with the following.
that A can be chosen in 9 * 6 = 54 ways.
B will have the same Y coordinate only x varies and should not be identical to A. So B can be chosen in 9-1 = 8 ways.
C can similarly be chosen in 6 -1 = 5 ways.
So total no. of ways of choosing the triangle is 54 * 8 * 5 = 2160 ways. This came out correct!
Let us go to Q1:
Q1) Rectangle ABCD is constructed in the coordinate plane parallel to the x- and y-axes. If the x- and y-coordinates of each of the points are integers which satisfy 3 ≤ x ≤ 11 and -5 ≤ y ≤ 5, how many possible ways are there to construct rectangle ABCD?
(Note that two rectangles that have the same four vertices that are labeled differently are considered to be the same rectangle.)
Answer is 1980.
Now in Q1 above, I must be able to employ the same technique I used for the triangle in Q1 because a rectangle's 4th coordinate must use the x and y coordinates of the adjacent vertices. When I apply that method I get to 7920. But the answer seems to be 1980.
Where am I going wrong?
Vineesh,
Just telling you what I know and think. I am not the expert.
Just telling you what I know and think. I am not the expert.
