Tricky - A solid cylinder with radius 3 inches sits in a

[Brent@GMATPrepNow]

A solid cylinder with radius 3 inches sits in a cylindrical container containing water. The cylindrical container has radius 4 inches, and the water is 6/Ï€ inches deep. If the solid cylinder is removed from the container, what will be the depth of the water (in inches)?

A) 2/(3pi)

B) 9/(4pi)

C) 7/(3pi)

D) 21/(8pi)

E) 3/pi

Source: GMAT Prep Now
Difficulty level: 700+

Answer: D

EDIT: I changed the answer choices after I was alerted to a mistake I made.

[regor60]

Must be tricky. OK, I'll bite.

Volume of cylinder and water together = pi(r^2)h = pi(16)(6/pi) = 96 cubic inches

Volume of cylinder = pi(3^2)(6/pi) = 54 cubic inches

Therefore, volume of water = 42 cubic inches.

H after cylinder removed = 42/(pi(4^2) = [spoiler]21/pi(8)[/spoiler]

[Brent@GMATPrepNow]

Argh!!
I solved that question TWICE before posting it and, each time I subtracted 96 - 54, I got 32 (instead of 42)

I have edited the answer choices accordingly.

Cheers,
Brent

[regor60]

Another, more direct, approach.

What is the volume of water directly ? Area of surface x height

Area of surface = length x width > width is difference in radii > =1"

Length is trickier: It's the average of the length of the inner radius and the outer radius

Length = 2(pi)[(3+4)/2)] = 7(pi)

Surface Area = Length x width = 7(pi) x 1 = 7(pi) square inches

Volume = Surface area x height = 7(pi) x (6/pi) = 42 cubic inches

New height of water = Volume/area = 42/16(pi)=

D