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# A perfectly spherical satellite with a radius of 4 feet is

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## Global Stats

Difficult

A perfectly spherical satellite with a radius of 4 feet is being packed for shipment to its launch site. If the inside dimensions of the rectangular crates available for shipment, when measured in feet, are consecutive even integers, then what is the volume of the smallest available crate that can be used?
(Note: the volume of a sphere is given by the equation v=(4/3)pie r^3 .)

(A) 48
(B) 192
(C) 480
(D) 960
(E) 1,680

OA D

Source: Princeton Review

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BTGmoderatorDC wrote:
A perfectly spherical satellite with a radius of 4 feet is being packed for shipment to its launch site. If the inside dimensions of the rectangular crates available for shipment, when measured in feet, are consecutive even integers, then what is the volume of the smallest available crate that can be used?
(Note: the volume of a sphere is given by the equation v=(4/3)pie r^3 .)

(A) 48
(B) 192
(C) 480
(D) 960
(E) 1,680

OA D

Source: Princeton Review
Given: Radius of the spherical satellite = 4ft, so diameter = 8 ft

The minimum dimension of the rectangular box (crate) must be 8*8*8 in order for the spherical satellite to perfectly fit into the crate.

Since the dimensions of the crate are consecutive even integers, the minimum values of the dimensions of the crates must be 8, 8 + 2 = 10 and 10 + 2 = 12

=> Volume of the rectangular box (Crate) = 8*10*12 = 960 ft^3

Hope this helps!

-Jay
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Since the radius is 4 feet, the smallest side should be equal to the diameter. Hence the shortest side will be 8 feet and therefore the rest of the sides will be 10 and 12 feet.

Volume will be 8 x 10 x 12 = 960 ft^3. Regards!

### GMAT/MBA Expert

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BTGmoderatorDC wrote:
A perfectly spherical satellite with a radius of 4 feet is being packed for shipment to its launch site. If the inside dimensions of the rectangular crates available for shipment, when measured in feet, are consecutive even integers, then what is the volume of the smallest available crate that can be used?
(Note: the volume of a sphere is given by the equation v=(4/3)pie r^3 .)

(A) 48
(B) 192
(C) 480
(D) 960
(E) 1,680
The shortest dimension of the rectangular crate has to be at least the diameter of the spherical satellite; therefore, that dimension has to be at least 8. Since the dimensions of the crate are consecutive even integers, the smallest possible volume of the crate is:

8 x 10 x 12 = 960

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