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
OA C
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cans
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-4<=x<=5 total 10 options
6<=y<=16 total 11 options
choose point p co-ordinates: 10*11 (10 for x and 11 for y) = 110
now R has the same y as P, option for x = 9 (1 is used by P)
Q has same x as P, option for y = 10 ( 1 is used by P)
total = 110*9*10 = 9900
IMO C
6<=y<=16 total 11 options
choose point p co-ordinates: 10*11 (10 for x and 11 for y) = 110
now R has the same y as P, option for x = 9 (1 is used by P)
Q has same x as P, option for y = 10 ( 1 is used by P)
total = 110*9*10 = 9900
IMO C
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sivaelectric
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Great explanation CANS
If I am wrong correct me 
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Ozlemg
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Hi cans,
Could you make it little bit detailed? Actually I understand the logic but I could not get why multiply 4 numbers but not 6? Because P,R, and Q each has 2 coordinates?
thank you!
Could you make it little bit detailed? Actually I understand the logic but I could not get why multiply 4 numbers but not 6? Because P,R, and Q each has 2 coordinates?
thank you!
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When a question asks for the number of triangles that can be constructed, it's not a geometry question but a combinations question. Why? Because a triangle is a combination of 3 points.irock wrote: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
OA C
We need to determine how many ways we can combine P, Q and R to form a triangle. For each point, we need to choose an x value and a y value.
Point P:
x value: -4≤x≤5, giving us 10 choices.
y value: 6≤y≤16, giving us 11 choices.
Now we have to combine the number of choices for x with the number of choices for y. It's as though we have 10 shirts and 11 ties, and we need to determine how many outfits can be made:
(number of choices for x)*(number of choices for y)=10*11=110 choices for P.
Point Q:
x value: In order to construct a right triangle, Q has to have the same x coordinate as P (so that Q is directly above P and we get a right angle). So we have only 1 choice for x: it must be the same integer that we chose for P's x value.
y value: If P and Q have the same x value, they can't have the same y value, or they will be the same point. We used 1 of our 11 choices for y when we chose P, so we have 11-1=10 choices for Q's y value.
(number of choices for x)*(number of choices for y)=1*10=10 choices for Q.
Point R:
y value: For PR to be parallel to the x axis, P and R have to share the same y value. So the number of choices for y is 1; it must be the same integer that we chose for P's y value.
x value: If P and R have the same y value, they can't have the same x value, or they will be the same point. We used 1 of our 10 choices for x when we chose P, so we have 10-1=9 choices for R's x value.
(number of choices for x)*(number of choices for y)=9*1=9 choices for R.
So we have 110 choices for P, 10 choices for Q, and 9 choices for R. We need to determine how many ways we can combine P, Q and R to make a triangle. It's as though we have 110 shirts, 10 ties, and 9 pairs of pants, and we need to determine the number of outfits that can be made:
(number of choices for P)*(number of choices for Q)*(number of choices for R) = 110*10*9 = 9900.
The correct answer is C.
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cans
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Ozlemg wrote:Hi cans,
Could you make it little bit detailed? Actually I understand the logic but I could not get why multiply 4 numbers but not 6? Because P,R, and Q each has 2 coordinates?
thank you!
This is given in the question. let P be P(x,y). Now as PR is parallel to x-axis, P and R have same value of y (a line parallel to x-axis is of form y=k where k is constant). Thus when we select a y for P, it is automatically selected for R also.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.
Similarly as it is right angled triangle at P, PQ mus be parallel to y-axis. And x co-ordinate of Q is same as that of P. Thus we multiply 4 numbers.
If you have trouble in that, then you can solve it like:
for P : 10*11 (10 for x and 11 for y)
for Q: 1*10 (1 for x (as it is same as P's x) and 10 for y)
for R: 9*1 (9 for x and 1 for y(as it is same as P's y))
Hope it helps
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Cans!!
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vikram4689
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+1 for C,
10 options on x axis for 2 co-ordinates . There total options = 10*9
11 options on y axis for 2 co-ordinates . There total options = 11*10
Total no of triangles with varying co-ordinates = 11*10*10*9 =9900
10 options on x axis for 2 co-ordinates . There total options = 10*9
11 options on y axis for 2 co-ordinates . There total options = 11*10
Total no of triangles with varying co-ordinates = 11*10*10*9 =9900
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I am looking for some comments about the way I solved the problem.
Let the co-ordinate of P be (x1,y1), R(x2,y2) and Q(x3,y3). Since P and R must have the same value for y, R(x2,y1) and since Q should have the same x value as P, Q is (x1,y3). So from P(x1,y1), R(x2,y2) & Q(x3,y3) we are reduced to P(x1,y1),R(x2,y1)& Q is (x1,y3).
We need to work out the possibilities for x1,x2,y1 & y3 now.
x1 can have 10 different values, given in the question.
y1 can have 11 different values, given in the question.
to form a right triangle the possible values of x2 will be (10-1) and the value for y3 would be (11-1), multiplying all of these together results in;
(x1)(y1)(x2)(y3) = 10 X 11 X 9 X 10 = 9900
Is this approach ok? I know it's not a geometry question as such but found it easy to do this way by assigning co-ordinates.
Let the co-ordinate of P be (x1,y1), R(x2,y2) and Q(x3,y3). Since P and R must have the same value for y, R(x2,y1) and since Q should have the same x value as P, Q is (x1,y3). So from P(x1,y1), R(x2,y2) & Q(x3,y3) we are reduced to P(x1,y1),R(x2,y1)& Q is (x1,y3).
We need to work out the possibilities for x1,x2,y1 & y3 now.
x1 can have 10 different values, given in the question.
y1 can have 11 different values, given in the question.
to form a right triangle the possible values of x2 will be (10-1) and the value for y3 would be (11-1), multiplying all of these together results in;
(x1)(y1)(x2)(y3) = 10 X 11 X 9 X 10 = 9900
Is this approach ok? I know it's not a geometry question as such but found it easy to do this way by assigning co-ordinates.
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Brent@GMATPrepNow
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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, 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
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]
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