Data Sufficiency

Problem from GMATPrep

Let squares side be a and rectangular's side 2x and 3x (because their ratio is 2:3). Their perimeters are equal 4a=10x => a^2=(25x^2)/4

The area of the rectangular 2x*3x=6x^2
The area of the circle a*a=a^2

Ratio =(6x^2)/a^2. Substitute instead a^2=(25x^2)/4. Then 24:25

bomond wrote:Let squares side be a and rectangular's side 2x and 3x (because their ratio is 2:3). Their perimeters are equal 4a=10x => a^2=(25x^2)/4

The area of the rectangular 2x*3x=6x^2
The area of the circle a*a=a^2

Ratio =(6x^2)/a^2. Substitute instead a^2=(25x^2)/4. Then 24:25

WOW - You don't have any idea how long I mulled over this during and after the test all while thinking S is a circle. you should have seen some of the formulae I came up with. It was a wild pi party

Anyway - thank you. Your approach makes perfect sense.

By the way - it's actually quite funny. You have a typo talking about circles in your solution as well. There must be something about this question

LSB wrote: By the way - it's actually quite funny. You have a typo talking about circles in your solution as well. There must be something about this question

I'm sorry. Yes you are right. By the area of the circle I mean square.

a/b = 2/3, 4s = 2(a+b)

if you solve these two eq you get

a= 4s/5, b= 6s/5

area of r/area of s = ab/s^2 = 24/26

so b