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# Right triangle ABC is to be drawn in the xy-plane so that

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Right triangle ABC is to be drawn in the xy-plane so that  [#permalink]

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Updated on: 16 Jul 2013, 00:17
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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?

A. 54
B. 432
C. 2,160
D. 2,916
E. 148,824

Originally posted by hrish88 on 09 Jan 2010, 04:50.
Last edited by Bunuel on 16 Jul 2013, 00:17, edited 3 times in total.
Edited the question and added the OA
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Re: tough problem  [#permalink]

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09 Jan 2010, 05:34
24
28
hrish88 wrote:
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?

A.54

B.432

C.2160

D.2916

E.148,824

i ve got it right.but this problem is very time consuming.can anyone suggest shorter method

We have the rectangle with dimensions 9*6 (9 horizontal dots and 6 vertical). AB is parallel to y-axis and AC is parallel to x-axis.

Choose the (x,y) coordinates for vertex A: 9C1*6C1;
Choose the x coordinate for vertex C (as y coordinate is fixed by A): 8C1, (9-1=8 as 1 horizontal dot is already occupied by A);
Choose the y coordinate for vertex B (as x coordinate is fixed by A): 5C1, (6-1=5 as 1 vertical dot is already occupied by A).

9C1*6C*8C1*5C1=2160.

Answer: C.
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Re: Right triangle ABC is to be drawn in the xy-plane so that  [#permalink]

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25 Jan 2013, 13:03
12
7
Slightly different way of thinking:

On the 9x6 grid of possibilities, I can imagine a bunch of rectangles (with sides parallel to x and y axes). Each of these rectangles contains 4 triangles that fit the description of the question stem.

therefore:

Answer = ( # of Rectangles I can make on the grid) x 4

To create the rectangle, I need to pick 2 points on the x direction, and 2 points on the y direction. Therefore:

Answer = C(9,2) * C(6,2) * 4 = 36 * 15 * 4 = 2160 (OPTION C)
##### General Discussion
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Re: tough problem  [#permalink]

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09 Jan 2010, 05:50
Bunuel wrote:
hrish88 wrote:
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?

A.54

B.432

C.2160

D.2916

E.148,824

i ve got it right.but this problem is very time consuming.can anyone suggest shorter method

We have the square with dimensions 9*6(9 horizontal dots and 6 vertical). AB is parallel to y-axis and AC is parallel to x-axis.

Choose the (x,y) coordinates for vertex A: 9C1*6C1;
Choose the x coordinate for vertex C (as y coordinate is fixed by A): 8C1, (9-1=8 as 1 horizontal dot is already occupied by A);
Choose the y coordinate for vertex B (as x coordinate is fixed by A): 5C1, (6-1=5 as 1 vertical dot is already occupied by A).

9C1*6C*8C1*5C1=2160.

Answer: C.

OA is C.very nice explanation.you rock man as always.
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Joined: 07 Aug 2010
Posts: 60
Re: tough problem  [#permalink]

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14 Oct 2010, 22:06
Bunuel wrote:
hrish88 wrote:
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?

A.54

B.432

C.2160

D.2916

E.148,824

i ve got it right.but this problem is very time consuming.can anyone suggest shorter method

We have the rectangle with dimensions 9*6 (9 horizontal dots and 6 vertical). AB is parallel to y-axis and AC is parallel to x-axis.

Choose the (x,y) coordinates for vertex A: 9C1*6C1;
Choose the x coordinate for vertex C (as y coordinate is fixed by A): 8C1, (9-1=8 as 1 horizontal dot is already occupied by A);
Choose the y coordinate for vertex B (as x coordinate is fixed by A): 5C1, (6-1=5 as 1 vertical dot is already occupied by A).

9C1*6C*8C1*5C1=2160.

Answer: C.

Good one. +1 for it. Hope I didn't mess it up.

so what about the triangles that look like the mirror images of the ones above? - think, switching the co-ords of A and C along x axis and switching A and B along y axis....
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Re: tough problem  [#permalink]

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15 Oct 2010, 02:58
BlitzHN wrote:
Bunuel wrote:
hrish88 wrote:
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?

A.54

B.432

C.2160

D.2916

E.148,824

i ve got it right.but this problem is very time consuming.can anyone suggest shorter method

We have the rectangle with dimensions 9*6 (9 horizontal dots and 6 vertical). AB is parallel to y-axis and AC is parallel to x-axis.

Choose the (x,y) coordinates for vertex A: 9C1*6C1;
Choose the x coordinate for vertex C (as y coordinate is fixed by A): 8C1, (9-1=8 as 1 horizontal dot is already occupied by A);
Choose the y coordinate for vertex B (as x coordinate is fixed by A): 5C1, (6-1=5 as 1 vertical dot is already occupied by A).

9C1*6C*8C1*5C1=2160.

Answer: C.

so what about the triangles that look like the mirror images of the ones above? - think, switching the co-ords of A and C along x axis and switching A and B along y axis....

Above solution counts all position:

AC and CA;

A
B
and
B
A;

For example point C with 8C1 can be placed to the right as well to the left of A and point B with 5C1 can be placed below as well as above of A. So all cases are covered.

More here: arithmetic-og-question-88380.html?hilit=dimensions

Hope it helps.
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Re: tough problem  [#permalink]

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17 Oct 2010, 20:24
3
C.

I am not sure if this approach is correct. I used Elimination. There can be only 5 possible values of C if we fix A. So the number of triangles possible has to be multiple of 5. The only answer that satisfies the criterion is C.
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Re: Right triangle ABC is to be drawn in the xy-plane so that  [#permalink]

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25 Jan 2013, 07:08
4
2
hrish88 wrote:
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?

A. 54
B. 432
C. 2,160
D. 2,916
E. 148,824

First, get the integer points available for x-axis: 2 - (-6) + 1 = 9
Second, get the interger points available for y-axis: 9-4+1 = 6

How many ways to select the location of line AB in the x-axis? 9
How many ways to select the location of point C in the x-axis? 8 (Note: we cannot select the location of line AB)
How many ways to select the location of the base? 2 (Is it BC or AB?)
How many ways to position line AB parallel to y axis? 6!/2!4! = 15

Multiple all that:$$9*8*2*15 = 2,160$$

Answer: C
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Re: tough problem  [#permalink]

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18 May 2013, 10:49
Bunuel wrote:
hrish88 wrote:
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?

A.54

B.432

C.2160

D.2916

E.148,824

i ve got it right.but this problem is very time consuming.can anyone suggest shorter method

We have the rectangle with dimensions 9*6 (9 horizontal dots and 6 vertical). AB is parallel to y-axis and AC is parallel to x-axis.

Choose the (x,y) coordinates for vertex A: 9C1*6C1;
Choose the x coordinate for vertex C (as y coordinate is fixed by A): 8C1, (9-1=8 as 1 horizontal dot is already occupied by A);
Choose the y coordinate for vertex B (as x coordinate is fixed by A): 5C1, (6-1=5 as 1 vertical dot is already occupied by A).

9C1*6C*8C1*5C1=2160.

Answer: C.

Hi Bunuel ,

That was a fantastic solution , but i have a small doubt . How do we ensure that by selecting points in this way the properties of a triangle are satisfied always . Could there be some points through which we cannot even form a triangle leave alone right angled triangle. I hope i am clear in my question .
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Posts: 55272
Re: tough problem  [#permalink]

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19 May 2013, 04:09
venkat18290 wrote:
Bunuel wrote:
hrish88 wrote:
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?

A.54

B.432

C.2160

D.2916

E.148,824

i ve got it right.but this problem is very time consuming.can anyone suggest shorter method

We have the rectangle with dimensions 9*6 (9 horizontal dots and 6 vertical). AB is parallel to y-axis and AC is parallel to x-axis.

Choose the (x,y) coordinates for vertex A: 9C1*6C1;
Choose the x coordinate for vertex C (as y coordinate is fixed by A): 8C1, (9-1=8 as 1 horizontal dot is already occupied by A);
Choose the y coordinate for vertex B (as x coordinate is fixed by A): 5C1, (6-1=5 as 1 vertical dot is already occupied by A).

9C1*6C*8C1*5C1=2160.

Answer: C.

Hi Bunuel ,

That was a fantastic solution , but i have a small doubt . How do we ensure that by selecting points in this way the properties of a triangle are satisfied always . Could there be some points through which we cannot even form a triangle leave alone right angled triangle. I hope i am clear in my question .

ANY 3 non-collinear points on a plane form a triangle.
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Re: Right triangle ABC is to be drawn in the xy-plane so that  [#permalink]

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30 May 2013, 21:14
Bunuel... you're a freaking genius. Get a job with NASA already.
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Re: Right triangle ABC is to be drawn in the xy-plane so that  [#permalink]

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13 Nov 2013, 08:38
Another way of looking at the problem.
According to the given constraints, the co-ordinates have to be chosen this way :-
A(a,b) B(a,c) C(d,b) where a,b,c and d are arbitrary integers. If you check this satisfies the constraint that AB must be parallel to the Y-axis.
Drawing the triangle and rotating it will give you a rectangle whose sides will measure length= |b-c| and breadth= |a-d|.
This rectangle's area will be = |b-c| X |a-d|
Now after having realized this, you just have to choose values from the given ranges such that the area is always non-zero,
and this can be done in the following way,
!.
selecting a and d from the range [-6,2] which has 9 elements, derived as --> 2 - (-6) +1 = 9.

9C2 X 2 (2 because both a>d and d>a are permissible).

2. selecting b and c similarly
6C2 X 2.

3. Multiplying the two terms :-
9C2 X 6C2 X 2 X 2 = 2160.
Kudos if you liked it.
Do have a look at this approach Bunuel
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Re: tough problem  [#permalink]

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13 Nov 2013, 09:38
Bunuel wrote:
hrish88 wrote:
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?

A.54

B.432

C.2160

D.2916

E.148,824

i ve got it right.but this problem is very time consuming.can anyone suggest shorter method

We have the rectangle with dimensions 9*6 (9 horizontal dots and 6 vertical). AB is parallel to y-axis and AC is parallel to x-axis.

Choose the (x,y) coordinates for vertex A: 9C1*6C1;
Choose the x coordinate for vertex C (as y coordinate is fixed by A): 8C1, (9-1=8 as 1 horizontal dot is already occupied by A);
Choose the y coordinate for vertex B (as x coordinate is fixed by A): 5C1, (6-1=5 as 1 vertical dot is already occupied by A).

9C1*6C*8C1*5C1=2160.

Answer: C.

Kudos Bunuel. Nice explanation.
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Re: Right triangle ABC is to be drawn in the xy-plane so that  [#permalink]

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26 Jun 2018, 01:40
Making a rough diagram for the axis helps.
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Re: Right triangle ABC is to be drawn in the xy-plane so that  [#permalink]

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19 Sep 2018, 20:58
1
hrish88 wrote:
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?

A. 54
B. 432
C. 2,160
D. 2,916
E. 148,824

Please check solution as attached.

Answer:Option C
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Screen Shot 2018-09-20 at 9.26.09 AM.png [ 659.51 KiB | Viewed 3002 times ]

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Right triangle ABC is to be drawn in the xy-plane so that  [#permalink]

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28 Sep 2018, 14:32
Bunuel wrote:
hrish88 wrote:
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?

A.54

B.432

C.2160

D.2916

E.148,824

i ve got it right.but this problem is very time consuming.can anyone suggest shorter method

We have the rectangle with dimensions 9*6 (9 horizontal dots and 6 vertical). AB is parallel to y-axis and AC is parallel to x-axis.

Choose the (x,y) coordinates for vertex A: 9C1*6C1;
Choose the x coordinate for vertex C (as y coordinate is fixed by A): 8C1, (9-1=8 as 1 horizontal dot is already occupied by A);
Choose the y coordinate for vertex B (as x coordinate is fixed by A): 5C1, (6-1=5 as 1 vertical dot is already occupied by A).

9C1*6C*8C1*5C1=2160.

Answer: C.

Hi @bunnel ,
I have a doubt.
If AB is parallel to Y-axis, how can we count X=0 as a possibility for vertex A?
If X=0, when vertex A lies of the Y-axis and therefore, AB can't be parallel to Y-axis..
With this in mind, I got 8*6 possibilities for vertex A.

please help.
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Re: Right triangle ABC is to be drawn in the xy-plane so that  [#permalink]

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28 Sep 2018, 20:49
ganeshvenugopal wrote:
Bunuel wrote:
hrish88 wrote:
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?

A.54

B.432

C.2160

D.2916

E.148,824

i ve got it right.but this problem is very time consuming.can anyone suggest shorter method

We have the rectangle with dimensions 9*6 (9 horizontal dots and 6 vertical). AB is parallel to y-axis and AC is parallel to x-axis.

Choose the (x,y) coordinates for vertex A: 9C1*6C1;
Choose the x coordinate for vertex C (as y coordinate is fixed by A): 8C1, (9-1=8 as 1 horizontal dot is already occupied by A);
Choose the y coordinate for vertex B (as x coordinate is fixed by A): 5C1, (6-1=5 as 1 vertical dot is already occupied by A).

9C1*6C*8C1*5C1=2160.

Answer: C.

Hi @bunnel ,
I have a doubt.
If AB is parallel to Y-axis, how can we count X=0 as a possibility for vertex A?
If X=0, when vertex A lies of the Y-axis and therefore, AB can't be parallel to Y-axis..
With this in mind, I got 8*6 possibilities for vertex A.

please help.

Here comes some cotradiction.

One definition says that parallel lines are lines that never intersect but they lack mentioning the fact that the lines should lie in one plane for it to happen.

Here the line is parallel to Y axis and Y-Axis is a direction while origin is just a point of reference from where the Y direction may be referenced so I believe that X=0 may be taken for the line which is parallel to Y-Axis.
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Re: Right triangle ABC is to be drawn in the xy-plane so that  [#permalink]

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18 Mar 2019, 12:00
Bunuel

Thank you for your solution. I have a question: How can we ensure that the sum of any 2 sides of the triangle is greater than the third side and that the length is greater than the difference between the lengths of the other 2 sides? Thank you!
Re: Right triangle ABC is to be drawn in the xy-plane so that   [#permalink] 18 Mar 2019, 12:00
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# Right triangle ABC is to be drawn in the xy-plane so that

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