Geometry Toughies

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vini vidi vici: Newbie | Next Rank: 10 Posts; Posts: 9; Joined: Sun Oct 05, 2008 8:30 am

by vini vidi vici » Mon Nov 17, 2008 2:30 am

Check these geometry ones out. do share your solutions as im not getting the right answers for either of the two

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GMAT geometry toughies.doc: Cant sort out the first one the second one appears ok though; (24.5 KiB) Downloaded 194 times

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Source: Beat The GMAT — Problem Solving | Mon Nov 17, 2008 2:30 am

rtfact: Newbie | Next Rank: 10 Posts; Posts: 9; Joined: Tue Sep 16, 2008 9:44 am

by rtfact » Mon Nov 17, 2008 3:46 am

1. C

from (1) we can't determine the circumference.
from (2) we don't know where the cord is located in the circle - the chord can be the diameter or smaller than the diameter.

from (1+2): we can determine that the chord is on a central angle of 60 degrees 360/6. The hight from the center of the circle (mark as: point k) to the chord divides the chord in to two straight triangles with a base of 3 (half of the chord's length). The angles of each triangle are 30:60:90, hence the hypotenuse equals 2*3=6=Radius.

2. B
There are three options to create a cylinder with a maximum volume, because there are 3 bases possible.

diameter=8, R=4, H=12, Volume=pi*R^2*H=pi*192
diameter=10, R=5, H=8, Volume=pi*R^2*H=pi*200
diameter=8, R=4, H=10, Volume=pi*R^2*H=pi*160

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tomato1: Junior | Next Rank: 30 Posts; Posts: 26; Joined: Wed Apr 02, 2008 1:53 am; Location: Noida; Thanked: 4 times

by tomato1 » Mon Nov 17, 2008 4:03 am

1. C

Check th attachment for explanation

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download.doc: (28 KiB) Downloaded 138 times

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gmat009: Master | Next Rank: 500 Posts; Posts: 458; Joined: Sun Aug 31, 2008 10:44 amFollowed by: 1 member; Thanked: 3 times

by gmat009 » Wed Nov 19, 2008 12:11 pm

rtfact wrote: 2. B
There are three options to create a cylinder with a maximum volume, because there are 3 bases possible.

diameter=8, R=4, H=12, Volume=pi*R^2*H=pi*192
diameter=10, R=5, H=8, Volume=pi*R^2*H=pi*200
diameter=8, R=4, H=10, Volume=pi*R^2*H=pi*160

Can someone explain me why we cann't take Diameter=12 and H=10
Vol = pi*(6*6)*10=360pi which is maximum volume

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mals24: Legendary Member; Posts: 683; Joined: Tue Jul 22, 2008 1:58 pmFollowed by: 2 members; Location: Dubai; Thanked: 73 times

by mals24 » Wed Nov 19, 2008 12:34 pm

@gmat009

We cannot have the diameter as 12 and height as 10 because in this case the cylinder will not fit in the box.
The cylinder would have fit if the dimensions were diameter = 12 and height also 12.

However if the length is 12 and the width 10 then the cylinder will easily fit. So the maximum radius should be 5 and the maximum height will be 8.

Hope its clear

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