square

gmatnmein2010 wrote:A certain square is to be drawn on a coordinate plane. One of the vertices must be on the origin, and the square is to have an area of 100. If all coordinates of the vertices must be integers, how many different ways can this square be drawn?


1)4


2)6


3)8


4)10


5)12

https://www.beatthegmat.com/hard-mgmat-7 ... t9944.html

Please do a search - How many times do I remind you. DO you even check the answers after posting these questions??

gmatnmein2010 wrote:A certain square is to be drawn on a coordinate plane. One of the vertices must be on the origin, and the square is to have an area of 100. If all coordinates of the vertices must be integers, how many different ways can this square be drawn?


1)4


2)6


3)8


4)10


5)12

Area of square =100.
Each side =10
Let a side be inclined by x degree with the x axis.
So,we have to find all those angles when 10 cosx and 10 sinx both are integers.

So,the answer should be A.

gmatnmein2010 wrote:A certain square is to be drawn on a coordinate plane. One of the vertices must be on the origin, and the square is to have an area of 100. If all coordinates of the vertices must be integers, how many different ways can this square be drawn?
1)4
2)6
3)8
4)10
5)12

since area of square is = 100
so side of square = sqrt.(100) = 10
now the square can have coordinates of vertices a integers when its two sides lies on x axis and y -axis ...which is possible in [spoiler]4 ways...
Hence Ans 1)[/spoiler]

The answer is 12 and it is well explained in the link I provided earlier

gmatnmein2010 wrote:A certain square is to be drawn on a coordinate plane. One of the vertices must be on the origin, and the square is to have an area of 100. If all coordinates of the vertices must be integers, how many different ways can this square be drawn?


1)4


2)6


3)8


4)10


5)12

though its already solved
but my way is like
question is asking how many combinations of integers x & y that form the relationship (x)^2- (y)^2 = 100
they are (0,10),(6,8) and (8,6)
this shows the 3 options of one vertix in a single quadrant.
Now multiply this by the number of quadrants i.e. 4
You get the number of ways of drawing the square will be 3*4 = 12

thephoenix wrote:

gmatnmein2010 wrote:A certain square is to be drawn on a coordinate plane. One of the vertices must be on the origin, and the square is to have an area of 100. If all coordinates of the vertices must be integers, how many different ways can this square be drawn?
1)4
2)6
3)8
4)10
5)12

though its already solved
but my way is like
question is asking how many combinations of integers x & y that form the relationship (x)^2- (y)^2 = 100
they are (0,10),(6,8) and (8,6)
this shows the 3 options of one vertix in a single quadrant.
Now multiply this by the number of quadrants i.e. 4
You get the number of ways of drawing the square will be 3*4 = 12

Ok thats nice..but I have a querry why u considered (8,6) and (6,8) both ..why don't you consider (10,0) and (0,10) both
then the total solun = 4x4 = 16.

shashank.ism wrote:

thephoenix wrote:

gmatnmein2010 wrote:A certain square is to be drawn on a coordinate plane. One of the vertices must be on the origin, and the square is to have an area of 100. If all coordinates of the vertices must be integers, how many different ways can this square be drawn?
1)4
2)6
3)8
4)10
5)12

though its already solved
but my way is like
question is asking how many combinations of integers x & y that form the relationship (x)^2- (y)^2 = 100
they are (0,10),(6,8) and (8,6)
this shows the 3 options of one vertix in a single quadrant.
Now multiply this by the number of quadrants i.e. 4
You get the number of ways of drawing the square will be 3*4 = 12

Ok thats nice..but I have a querry why u considered (8,6) and (6,8) both ..why don't you consider (10,0) and (0,10) both
then the total solun = 4x4 = 16.

Because he can count only one of them in one quadrant - the answer is still 3*4 = 12 (He did consider all the 4 points (10,0) (-10,0) (0,10) and (0, -10)

ajith wrote:
Because he can count only one of them in one quadrant - the answer is still 3*4 = 12 (He did consider all the 4 points (10,0) (-10,0) (0,10) and (0, -10)

hmm. i got it, we can only get four coordinates out of these (10,0 ) and (0,10)...