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A certain square is to be drawn on a coordinate plane. One

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Re: Tuesday Q1 - # of Squares [#permalink]

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New post 13 May 2011, 03:10
at 90deg to x, at 45 deg to x and at 60 deg to x.

4*3=12 total.
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Re: Tuesday Q1 - # of Squares [#permalink]

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New post 13 May 2011, 13:19
vannu wrote:
IMO, OA is 12

We can draw 12 squares with one vertix at the origin.
The regular 4 squares with all vertices on the x and y coordinates - 4 squares
Tilt the side of a square to 45 degrees - 4 squares
Tilt the side of the square to 60 degrees - 4 squares



Can you post a diagram and explain.
I dont think your answer is right.

With 45 degree the diagonal will lie ong the X aix.

The co-ordinates of diagonal on X aixs will be (0,0) and (10\sqrt{2} , 0)
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Re: Cartesian square [#permalink]

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New post 20 Jul 2011, 21:07
i think E


12 ways....

possibilities are of type -
6,8
8,6
10,0

4 quadrants... so 4*3 =12 ways
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Re: Cartesian square [#permalink]

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New post 21 Jul 2011, 05:02
You're correct. I wouldn't come up with that in 90 seconds.

agdimple333 wrote:
i think E


12 ways....

possibilities are of type -
6,8
8,6
10,0

4 quadrants... so 4*3 =12 ways
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Re: MGMAT How many squares? [#permalink]

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New post 31 Jul 2011, 07:57
Can anybody explain how can we apply combination concepts in this question?
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Re: how many squares ? [#permalink]

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New post 19 Aug 2011, 04:14
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Let the other vertex by (x,y)

Then sqrt (x^2 + y^2) = 10
=> x^2 + y^2 = 100
This has 12 solutions as the pairs can be (0,+/- 10); (+/- 10,0); (+/- 6, +/- 8); (+/- 8, +/- 6)

Option (E)
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Re: how many squares ? [#permalink]

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New post 19 Aug 2011, 15:04
I am totally confused, shouldn't x and y be the same as it is a square and there are only 4 possibilities.
Isn't it (+/- 10, 10) and (10, +/- 10)? Help Please!
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Re: how many squares ? [#permalink]

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New post 23 Aug 2011, 20:31
Hi Gyanone .. Can you please describe your solution in detail ?

I really want to know hw you reached the solution ...

Thanks
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Re: how many squares ? [#permalink]

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New post 24 Aug 2011, 06:32
For the square to have an area of 100, the length of any one side of the square must be 10.

For this to be true, if one vertex of the square is given to be the origin (0,0) then we can take one vertex adjacent to this vertex as (x,y)

For the length of this vertex to be 10, using the distance formula,
sqrt [(x-0)^2 + (y-0)^2] = 10
=> x^2 + y^2 = 100

Therefore the possible coordinates of the adjacent vertex are the values of x and y which satisfy this equation. There are 12 such values (as detailed above), so there are 12 possible squares.

Note that (10,10) is not a solution for a vertex adjacent to (0,0) as its distance from (0,0) is not 10.
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Re: how many squares ? [#permalink]

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New post 24 Aug 2011, 16:51
Thank you very much Gyanone. Yes, (6,8) and (0,0) don't make the diagonal but one of the sides of the square. I got it now. Thank you again for the explanation.
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Re: Looking for modern floor lamps........ [#permalink]

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Re: how many squares ? [#permalink]

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New post 12 Sep 2011, 02:20
GyanOne wrote:
Let the other vertex by (x,y)

Then sqrt (x^2 + y^2) = 10
=> x^2 + y^2 = 100
This has 12 solutions as the pairs can be (0,+/- 10); (+/- 10,0); (+/- 6, +/- 8); (+/- 8, +/- 6)

Option (E)


Thanks Gyanone.. MGMAT CAT solution is much more complicated than yours..

Is there a possibility of picking decimals in this question(if the question did not say that they have to be integers)
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Re: Zumit PS 023 [#permalink]

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New post 22 Sep 2011, 15:04
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scthakur wrote:
vr4indian wrote:
Hey dude I am not getting how you come with (6,8) and (8,6) as valid vertices point.

As per my knowledge... if we have one vertices to be (0,0) then one of vertices will lie on X axis (10,0) and other one must lie on Y axis (0,10) ..

So we are left with only one choice for fourth vertices... (10,10)

This is for 1 quadrants , as we have total 4 quadrants answer should be 4 .


Hope i have not done anything wrong here and i had understand question.

Please correct me if i am wrong

Thanks
Vishal Shah



Distance between (0,0) and (6,8) or (8,6) will be 10.



And this distance will make the side of square. So do I think that the answer is 12 (3 square in 4 areas)
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Re: Zumit PS 023 [#permalink]

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New post 23 Sep 2011, 06:11
shahideh wrote:
scthakur wrote:
vr4indian wrote:
Hey dude I am not getting how you come with (6,8) and (8,6) as valid vertices point.

As per my knowledge... if we have one vertices to be (0,0) then one of vertices will lie on X axis (10,0) and other one must lie on Y axis (0,10) ..

So we are left with only one choice for fourth vertices... (10,10)

This is for 1 quadrants , as we have total 4 quadrants answer should be 4 .


Hope i have not done anything wrong here and i had understand question.

Please correct me if i am wrong

Thanks
Vishal Shah


Distance between (0,0) and (6,8) or (8,6) will be 10.



And this distance will make the side of square. So do I think that the answer is 12 (3 square in 4 areas)


I still don't understand.
The distance between (0,0) and (6,8) equals to 10 - you're right, but area of the square is not 100, but 48.
Or where am I mistaken?
4 is correct IMO.
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Re: Zumit PS 023 [#permalink]

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New post 23 Sep 2011, 06:14
I think i got it:)
It isn't said that square cannot cross any axis :)
Probably yes, 12 - is correct
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Re: MGMAT PS- How many ways can the square be drawn [#permalink]

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New post 25 Sep 2011, 10:04
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given that origin is one of the vertex of the square and also area of the square is 100.

=> In other words its saying length of side of square has to be 10.

let (x,y) be the point that is at a distance of 10 from origin => x^2+y^2 = 100

36,64
0,100 are such square combinations that are possible here.

Number of possible points in first quadrant = (6,8),(8,6)
Number of possible points in all four quadrants = 2*4 = 8

there are also 4 points on the co-ordinate axis = (10,0) (-10,0)(0,10)(0,-10) = 4

=> total possible points = 12

Answer is E.
Re: MGMAT PS- How many ways can the square be drawn   [#permalink] 25 Sep 2011, 10:04

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