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chandni012: Newbie | Next Rank: 10 Posts; Posts: 1; Joined: Tue May 12, 2009 12:40 am; Location: India

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by chandni012 » Wed Mar 10, 2010 10:24 am

A room measures 10 feet by 10 feet by 10 feet.
Each basketball measures 6 inches in radius.

The question is how many basketballs can u fit inside this room?

Last edited by chandni012 on Wed Mar 10, 2010 11:21 am, edited 1 time in total.

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Source: Beat The GMAT — Problem Solving | Wed Mar 10, 2010 10:24 am

Osirus@VeritasPrep: GMAT Instructor; Posts: 1578; Joined: Thu May 28, 2009 8:02 am; Thanked: 128 times; Followed by:34 members; GMAT Score:760

by Osirus@VeritasPrep » Wed Mar 10, 2010 11:12 am

the volume of the room is 1000 feet. You have to convert this to inches in order to solve this problem (because the radius of the basketball is in inches).

1000 feet = 1000 * 12 inches = 12000 inches. The volume of the basketball is 4/3 pi r^3. This is roughly 904.32. Divide this number by 12000 and you have your answer (Roughly 13 basketballs)

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happyzzim: Newbie | Next Rank: 10 Posts; Posts: 4; Joined: Fri Mar 05, 2010 4:09 pm; Thanked: 1 times

by happyzzim » Wed Mar 10, 2010 12:58 pm

Isn't it 1000?

The diameter of the basketball is 12 inch(= 1 feet).

In the space, 10 feet * 10 feet * 10 feet, you can put 1000 basketball side by side.

Picture it.

OOOOOOOOOO
OOOOOOOOOO
OOOOOOOOOO
OOOOOOOOOO
OOOOOOOOOO
OOOOOOOOOO
OOOOOOOOOO
OOOOOOOOOO
OOOOOOOOOO
OOOOOOOOOO

width = 10 feet, length = 10 feet, and then height = 10 feet, so the rectangle like that is 10.

Please let me know the answer.

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Shawshank: Master | Next Rank: 500 Posts; Posts: 305; Joined: Mon Jul 27, 2009 5:38 am; Thanked: 10 times

by Shawshank » Wed Mar 10, 2010 10:58 pm

happyzzim wrote:Isn't it 1000?

The diameter of the basketball is 12 inch(= 1 feet).

In the space, 10 feet * 10 feet * 10 feet, you can put 1000 basketball side by side.

Picture it.

OOOOOOOOOO
OOOOOOOOOO
OOOOOOOOOO
OOOOOOOOOO
OOOOOOOOOO
OOOOOOOOOO
OOOOOOOOOO
OOOOOOOOOO
OOOOOOOOOO
OOOOOOOOOO

width = 10 feet, length = 10 feet, and then height = 10 feet, so the rectangle like that is 10.

Please let me know the answer.

The calculation shown here holds good for a cube with side as 1 feet .. So volume would be 1 x 1 x 1 = 1
But for the circular object .. it has to be 4/3 pie (1/2)^3 .
hence, as suggested by Phoenix,, the answer is somewhr arnd 13.

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kstv: Legendary Member; Posts: 610; Joined: Fri Jan 15, 2010 12:33 am; Thanked: 47 times; Followed by:2 members

by kstv » Thu Mar 11, 2010 2:49 am

If the balls are placed on the floor. They will sit side by side. Length = Breath = Height
Lengthwise we can place = 10 * 12 / 6*2 = 10 balls (1 feet = 12 inches, diameter is 12 inches = 1 feet)
Similarly Beathwise = 10feet/ 1 feet = 10
So on the floor we have 10*10 = 100 balls.
Stacking them one on top of another we can fit 10 balls i.e 9 on top of each ball on the floor .
So total number is 100*10 = 1000

Assume u do't have the options cos' without them you have to work the complete problem and that may not be the best technique to use.