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superrrom: Senior | Next Rank: 100 Posts; Posts: 39; Joined: Mon Sep 01, 2008 6:27 am; Thanked: 2 times

by superrrom » Mon Sep 01, 2008 7:16 am

Thanks

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Source: Beat The GMAT — Problem Solving | Mon Sep 01, 2008 7:16 am

pepeprepa: Legendary Member; Posts: 661; Joined: Tue Jul 08, 2008 12:58 pm; Location: France; Thanked: 48 times

by pepeprepa » Mon Sep 01, 2008 7:29 am

This one is faster.

Let's say diameter of S is x.
The sides of the square are x.
The diagonal of the square is x*sqrt(2).
And the diameter of T is x*sqrt(2).

So the area of T is pi*(x*sqrt(2)/2)²=pi*x²/2
The area of S is pi*(x/2)²=pi*x²/4

And T/S = 2

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4meonly: Legendary Member; Posts: 891; Joined: Sat Aug 16, 2008 4:21 amFollowed by: 1 member; Thanked: 27 times; GMAT Score:660(

by 4meonly » Mon Sep 01, 2008 9:12 am

Good rule to use here:

the ratio of squares is the same as the ratio of the sides in the power of 2

the ratio of volumes is the same as the ratio of the sides in the power of 3

Here you are to find the second power of the ratios of the diameters (or the sides of right isolescence triagle formed by the side and the diagonal of the square)

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