For most people -- a bit of adrenaline will kick in during the test and that slight bit of nervousness (that funny feeling in the stomach) will actually help people focus better.

It's much harder to get "psyched" up for a practice test at home -- so naturally there is a tendency to let the mind wander and drift.

finance wrote:Thank you sss2534. My GMat Prep is 710,,Princeton 730...but when I am distracted I get even 630..I think it depends a lot on my mood:(...

The problem can be best done by visualizing the co-ordinate plane
Since one point must be origin, 4 squares can be drawn with sides on the axes. 4 more can be drawn with sides on the lines x=y and x = -y.
Also since 6,8 and 10 are pythagorean triplets, there 4 more squares which can be drawn with vertices on (+/-6/,+/-8) or (+/-8,+/-6)
Hence 12 squares.

GMATGuruNY....
How to approach this problem, without spending more time on this?

GMATGuruNY wrote:Here is a drawing of the 12 squares that can be formed:



The squares comprise:

-- horizontal and vertical line segments, each with a length of 10
-- the hypotenuses of 6-8-10 triangles, each hypotenuse with a length of 10

GMATGuruNY wrote:Here is a drawing of the 12 squares that can be formed:



The squares comprise:

-- horizontal and vertical line segments, each with a length of 10
-- the hypotenuses of 6-8-10 triangles, each hypotenuse with a length of 10

Hi GMATGuru,

Please share your thought process to approach this problem. How do we come up with these 12 squares within 2 mins

Regards,
Vishal

vishal.pathak wrote:

GMATGuruNY wrote:Here is a drawing of the 12 squares that can be formed:



The squares comprise:

-- horizontal and vertical line segments, each with a length of 10
-- the hypotenuses of 6-8-10 triangles, each hypotenuse with a length of 10

Hi GMATGuru,

Please share your thought process to approach this problem. How do we come up with these 12 squares within 2 mins

Regards,
Vishal

Can anyone tell me the level of this question. I believe I will have to skip all such questions and the ones harder than this one on the actual exam. Any thoughts on the level of this??

Regards,
Vishal

vishal.pathak wrote:

vishal.pathak wrote:

GMATGuruNY wrote:Here is a drawing of the 12 squares that can be formed:



The squares comprise:

-- horizontal and vertical line segments, each with a length of 10
-- the hypotenuses of 6-8-10 triangles, each hypotenuse with a length of 10

Hi GMATGuru,

Please share your thought process to approach this problem. How do we come up with these 12 squares within 2 mins

Regards,
Vishal

Can anyone tell me the level of this question. I believe I will have to skip all such questions and the ones harder than this one on the actual exam. Any thoughts on the level of this??

Regards,
Vishal

Vishal, this was ranked as 700-800 question in the Manhattan test i took

An easy way to solve this more quickly is to realize that any square we can find for one quadrant will have a a corresponding square for each other quadrant. If we focus only on squares that lie predominantly in quadrant 1, we can find 3:

1) (0,0) (10,0) (10,10) (0,10)
2) (0,0) (8,6) (2,14) (-6,8)
3) (0,0) (8,-6) (14,2) (6,8)

For each of these, we can mirror across the y-axis to find the corresponding squares for quadrant 2. We can mirror these new squares across the x-axis to find the squares for quadrant 3. Finally, we can mirror those squares across the y-axis to find the squares for quadrant 4. This gives us a total of 12, as we can see in GMATGuruNY's helpful diagram.

Bill@VeritasPrep wrote:An easy way to solve this more quickly is to realize that any square we can find for one quadrant will have a a corresponding square for each other quadrant. If we focus only on squares that lie predominantly in quadrant 1, we can find 3:

1) (0,0) (10,0) (10,10) (0,10)
2) (0,0) (8,6) (2,14) (-6,8)
3) (0,0) (8,-6) (14,2) (6,8)

For each of these, we can mirror across the y-axis to find the corresponding squares for quadrant 2. We can mirror these new squares across the x-axis to find the squares for quadrant 3. Finally, we can mirror those squares across the y-axis to find the squares for quadrant 4. This gives us a total of 12, as we can see in GMATGuruNY's helpful diagram.

Hi Bill,

How do we figure out that (0,0) (8,6) (2,14) (-6,8) could be a possible square.
It is easy to see that the distance between (0,0) and (8,6) or (-6,8) is 10 but we will have to do sufficient maths to find that the 4th point of this square is (2,14). (2,14) will be the solution of 2 equations formed using the initial 3 points.

Regards,
Vishal