5 x 5 array of lattice points.

Question

A point at the intersection of two or more grid lines is called a lattice point. S is a 5 x 5 array of lattice points. How many squares have their vertices in S?

(1) 30 (2) 36 (3) 44 (4) 50 (5) none of these

Let me know if you think this kind of questions are beyond GMAT difficulty level.
So that even I dont waste time studying them.

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amit2k9 · Answer

its same as counting number of squares in a chess board.

the technique is :

squares with 
 
1X1 = 5 horizontal * 5 vertical positions = 25
2X2 = 4 * 4 = 16

3X3 = 3 * 3 = 9

4X 4 = 2* 2 = 4

5X5 = 1* 1 = 1

hence total = 25 + 16 + 9 + 4 + 1 = 55

hussi9 · Answer

Squares can be formed by diagonal lines also ....

amit2k9 · Answer

I am not sure what do you mean by diagonal lines.
As far as I understand this is a 5X5 matrix.

Diagonal lines will essentially require additional constructions,not mentioned here otherwise.

hussi9 · Answer

Solution:

For 3X3 array we have in all 5 squares (when joined vertically or horizontally) such that we have 2^2 squares of side 1X1, and 1 square of size 2X2. 
But when mid-points of the 2X2 square is joined, we have another square whose all vertices are on the perimeter of the array of 3X3. 
Please note that both the 2X2 square have all the vertices on the perimeter of 3X3 array. Thus, in all we have 4 + 2 = 6 such squares for 3X3.

For 4X4 array we have 9 squares of size 1X1, 3 squares (whose vertices lie on the perimeter of the array) of size 3X3 and 4*2 squares of size 2x2   => In all 9 + 8 + 3 squares.

Now for our problem in hand 5X5:
We have 16 ((n-1)^2, when array is nXn) squares of size 1X1, 4 (n-1, when array is nXn) squares (whose vertices lie on the perimeter of the array) of size 3X3, 18 squares (2*(n-2)^2 of size 2X2) and 12 squares (3*(n-3)^2 of size 3X3)  
=> In all 4 + 16 + 18 + 12 = 50 squares.

=> Choice (4) is the right answer

hussi9 · Answer

5X5 matrix is used to define the latices points and not the orientation of the squares .

amit2k9 · Answer

strange where did Gurpreet's post vanish.

1/2 ( 5c2 * 5c2) is the solution i had thought too. But any two points would have given rectangle too.
Thats what I thought.

some thing like 2 lines are there with m and n points on each. calculate how many triangles can be made ?

so, m (nC2) + n(mC2). logic used is the same there too.

Nice solution Gurpreet,lots of learning from you mate.

gurpreetsingh · Answer

My friends called me for the movie, so instead of editing it, I deleted the post to write again later. Now.. If you select 2 points on the one side, then the other two points are fixed. In case you don;t fix the other two, the result would be a rectangle. but at the same time when you would select the later 2 points then the former points would be fixed. So total squares considered by us = 5c2*5c2. But, we have counted the same square twice. so the answer must be 1/2 of it.

diebeatsthegmat · Answer

this mathematic doesnt sound like an gmat... which source is it from?

5 x 5 array of lattice points.

This kind of questions are beyond GMAT difficulty level?

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