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Case 1: 2Ï€r = 30 and h = 40
r = 15/Ï€
The volume is πr^2*h = π(15/π)^2 * 40 = ( 225 * 40 ) / π = 9000 / π.

Case 2: 2Ï€r = 40 and h = 30
r = 20/Ï€
The volume is πr^2*h = π(20/π)^2 * 30 = ( 400 * 30 ) / π = 12000 / π.

Thus, the difference in the two volumes is 12000 / π - 9000 / π = 3000 / π.


Therefore, C is the answer.
Answer: C

If the cylinder is rolled along 30...
Circumference of circle so formed = 2Ï€r=30..... r=15/Ï€
Height =40
$$V=\pi\cdot\frac{15^2}{\pi}\cdot40$$
For the other volume 30 and 40 will interchange
So $$V=\pi\cdot\frac{20^2}{\pi}\cdot30$$
Difference = (20^2*30-15^2*40)/π = (12000-9000)/ π = 3000/π.

Regards!

Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

A piece of cardboard measures 30 cm by 40 cm. We can make two different right circular cylinders by rolling the cardboard along its length and by rolling the cardboard along its width. What is the difference in their volumes?

A. 1000 / π
B. 2000 / π
C. 3000 / π
D. 4000 / π
E. 5000 / π

If we roll the cardboard along its length, we will have the circumference of the circular base of the cylinder = 40 cm and the height of the cylinder = 30 cm. In this case, we have radius = 40/2π = 20/π and the volume = π r^2 h = π x (20/π)^2 x 30 = π x 400/π^2 x 30 = 12,000/π.

Similarly, if we roll the cardboard along its width, we will have the circumference of the circular base of the cylinder = 30 cm and the height of the cylinder = 40 cm. In this case, we have radius = 30/2π = 15/π and the volume = π r^2 h = π x (15/π)^2 x 40 = π x 225/π^2 x 40 = 9,000/π.

Thus, the difference between the two volumes is 12,000/Ï€ - 9,000/Ï€ = 3,000/Ï€.

Answer: C