Right triangle PQR is to be constructed in the xy plane so that the right angle is at P and PR is parallel to X axis.The x and y coordinates of P,Q and R are to be integers that satisfy the following inequalities
i. x greater than equal to -4 & less than equal to 5
ii. y is greater than equal to 6 & less than equal to 16.
How many different triangles can be constructed with these properties?
Ans-9,900
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Since x and y should be integers,
x: 5 - (-4) + 1 = 10
y: 16 - 6 + 1 = 11
we have 10 options for x and 11 options for y. The total number of ways that we can put p on xy plane according to the given ranges for x and y is:
10 × 11 = 110
Since p and r have the same y, once we put p on xy plane, we are no longer concerned about the y of r, we know that yp = yr. So we should only choose x for the point r. Since we selected one x for point p before, now we have one less options for x. In other words, 10 - 1 = 9.
And point q should have the same x as p. So we have 11 - 1 = 10 options.
=> the total triangles = 110 × 9 × 10 = 9900
x: 5 - (-4) + 1 = 10
y: 16 - 6 + 1 = 11
we have 10 options for x and 11 options for y. The total number of ways that we can put p on xy plane according to the given ranges for x and y is:
10 × 11 = 110
Since p and r have the same y, once we put p on xy plane, we are no longer concerned about the y of r, we know that yp = yr. So we should only choose x for the point r. Since we selected one x for point p before, now we have one less options for x. In other words, 10 - 1 = 9.
And point q should have the same x as p. So we have 11 - 1 = 10 options.
=> the total triangles = 110 × 9 × 10 = 9900
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The question should read:
Stage 1: Select any point where the right angle will be (point P).
The point can be selected from a 10x11 grid. So, there 110 points to choose from.
This means that stage 1 can be completed in 110 ways.
Stage 2: Select a point that is on the same horizontal line as the first point. This point will be point R.
The 2 legs of the right triangle are parallel to the x- and y-axes.
The first point we select (in stage 1) dictates the y-coordinate of point R.
In how many ways can we select the x-coordinate of point R?
Well, we can choose any of the 10 coordinates from -4 to 5 inclusive EXCEPT for the x-coordinate we chose for point P (in stage 1).
So, there are 9 coordinates to choose from.
This means that stage 2 can be completed in 9 ways.
Stage 3: Select a point that is on the same vertical line as the first point. This point will be point Q.
The 2 legs of the right triangle are parallel to the x- and y-axes.
The first point we select (in stage 1) dictates the x-coordinate of point Q.
In how many ways can we select the y-coordinate of point Q?
Well, we can choose any of the 11 coordinates from 6 to 16 inclusive EXCEPT for the y-coordinate we chose for point P (in stage 1).
So, there are 10 coordinates to choose from.
This means that stage 3 can be completed in 10 ways.
So, by the Fundamental Counting Principle (FCP), the total number of triangles = (110)(9)(10) = [spoiler]9900 = C[/spoiler]
------------------------------------------------
Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmat-counting?id=775
Then you can try solving the following questions:
EASY
- https://www.beatthegmat.com/what-should- ... 67256.html
- https://www.beatthegmat.com/counting-pro ... 44302.html
- https://www.beatthegmat.com/picking-a-5- ... 73110.html
- https://www.beatthegmat.com/permutation- ... 57412.html
- https://www.beatthegmat.com/simple-one-t270061.html
- https://www.beatthegmat.com/mouse-pellets-t274303.html
MEDIUM
- https://www.beatthegmat.com/combinatoric ... 73194.html
- https://www.beatthegmat.com/arabian-hors ... 50703.html
- https://www.beatthegmat.com/sub-sets-pro ... 73337.html
- https://www.beatthegmat.com/combinatoric ... 73180.html
- https://www.beatthegmat.com/digits-numbers-t270127.html
- https://www.beatthegmat.com/doubt-on-sep ... 71047.html
- https://www.beatthegmat.com/combinatoric ... 67079.html
DIFFICULT
- https://www.beatthegmat.com/wonderful-p- ... 71001.html
- https://www.beatthegmat.com/ps-counting-t273659.html
- https://www.beatthegmat.com/permutation- ... 73915.html
- https://www.beatthegmat.com/please-solve ... 71499.html
- https://www.beatthegmat.com/no-two-ladie ... 75661.html
- https://www.beatthegmat.com/laniera-s-co ... 15764.html
Cheers,
Brent
Take the task of building triangles and break it into stages.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 can be constructed?
(A) 110
(B) 1100
(C) 9900
(D) 10000
(E) 12100
C
Stage 1: Select any point where the right angle will be (point P).
The point can be selected from a 10x11 grid. So, there 110 points to choose from.
This means that stage 1 can be completed in 110 ways.
Stage 2: Select a point that is on the same horizontal line as the first point. This point will be point R.
The 2 legs of the right triangle are parallel to the x- and y-axes.
The first point we select (in stage 1) dictates the y-coordinate of point R.
In how many ways can we select the x-coordinate of point R?
Well, we can choose any of the 10 coordinates from -4 to 5 inclusive EXCEPT for the x-coordinate we chose for point P (in stage 1).
So, there are 9 coordinates to choose from.
This means that stage 2 can be completed in 9 ways.
Stage 3: Select a point that is on the same vertical line as the first point. This point will be point Q.
The 2 legs of the right triangle are parallel to the x- and y-axes.
The first point we select (in stage 1) dictates the x-coordinate of point Q.
In how many ways can we select the y-coordinate of point Q?
Well, we can choose any of the 11 coordinates from 6 to 16 inclusive EXCEPT for the y-coordinate we chose for point P (in stage 1).
So, there are 10 coordinates to choose from.
This means that stage 3 can be completed in 10 ways.
So, by the Fundamental Counting Principle (FCP), the total number of triangles = (110)(9)(10) = [spoiler]9900 = C[/spoiler]
------------------------------------------------
Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmat-counting?id=775
Then you can try solving the following questions:
EASY
- https://www.beatthegmat.com/what-should- ... 67256.html
- https://www.beatthegmat.com/counting-pro ... 44302.html
- https://www.beatthegmat.com/picking-a-5- ... 73110.html
- https://www.beatthegmat.com/permutation- ... 57412.html
- https://www.beatthegmat.com/simple-one-t270061.html
- https://www.beatthegmat.com/mouse-pellets-t274303.html
MEDIUM
- https://www.beatthegmat.com/combinatoric ... 73194.html
- https://www.beatthegmat.com/arabian-hors ... 50703.html
- https://www.beatthegmat.com/sub-sets-pro ... 73337.html
- https://www.beatthegmat.com/combinatoric ... 73180.html
- https://www.beatthegmat.com/digits-numbers-t270127.html
- https://www.beatthegmat.com/doubt-on-sep ... 71047.html
- https://www.beatthegmat.com/combinatoric ... 67079.html
DIFFICULT
- https://www.beatthegmat.com/wonderful-p- ... 71001.html
- https://www.beatthegmat.com/ps-counting-t273659.html
- https://www.beatthegmat.com/permutation- ... 73915.html
- https://www.beatthegmat.com/please-solve ... 71499.html
- https://www.beatthegmat.com/no-two-ladie ... 75661.html
- https://www.beatthegmat.com/laniera-s-co ... 15764.html
Cheers,
Brent
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simply awesome question - seriously.
P = 11*10 ways
R = 9 ways of selecting x
Q = 10 ways of selecting y
11*10*9*10
9900
P = 11*10 ways
R = 9 ways of selecting x
Q = 10 ways of selecting y
11*10*9*10
9900