BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

tricky geometry

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Data Sufficiency |
User avatar
Legendary Member
Posts: 2134
Joined: Mon Oct 20, 2008 11:26 pm
Thanked: 237 times
Followed by:25 members
GMAT Score:730

Re: tricky geometry

by logitech » Mon Nov 03, 2008 5:02 pm
Fab wrote:Please be clear with the explanations.

THANKS!
Please post them seperately with the source. This way, people can solve one question at a time, and search it in the future. Thanks.
LGTCH
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"
Top
User avatar
Senior | Next Rank: 100 Posts
Posts: 52
Joined: Thu Jul 31, 2008 12:30 pm
Thanked: 5 times

by dhanda.arun » Wed Nov 05, 2008 1:09 pm
Q1
consider statement 1
if AB= 3 and BC =2
area of smaller = pi.Ab^2
and area of larger = pi.(AB + BC)^2)
so we can subtract and get the desired area.
Statement b does not help because we can't calculate area of smaller with this.
Top
Senior | Next Rank: 100 Posts
Posts: 87
Joined: Tue Jul 15, 2008 10:23 am
Location: Lima
Thanked: 4 times
Followed by:1 members

by Fab » Wed Nov 05, 2008 1:22 pm
I agree with you, however the OA for question 1 is D (apparently).
I don't know what's the source, a fried of mine passed me the exams.

THANKS.
Top
Legendary Member
Posts: 891
Joined: Sat Aug 16, 2008 4:21 am
Thanked: 27 times
Followed by:1 members
GMAT Score:660(

by 4meonly » Thu Nov 06, 2008 1:25 am
Q30:
Answer B? B. (sqrt2)/4

Area of BCE = Area of BEO - Area of BOC

(1+(sqrt2)/2)*((sqrt2)/2)*1/2 - (sqrt2)/2)*(sqrt2)/2)*1/2 = (sqrt2)/4
Attachments
123.GIF
Top
Senior | Next Rank: 100 Posts
Posts: 95
Joined: Fri Sep 28, 2007 10:16 am
Thanked: 6 times
GMAT Score:710

by muzali » Mon Nov 24, 2008 11:01 pm
1. D

We need the difference in area, i.e. the radii of both circles
First statement: Gives us the radii of both circles: AB (smaller circle), and AB+BC (larger circle) Sufficient
Second statement: CD+DE = radius of larger circle; 2 (CD+DE)-DE = AD = diameter of smaller circle. So we have the radii of both circles again. Sufficient

2. Max height of the semicircular tunnel = 20/2 = 10ft
Clearance required = 1/2 ft
Therefore, max height allowed = 10-(1/2) = 9(1/2) ft

3. See the previous post w/ diagram
Top
Master | Next Rank: 500 Posts
Posts: 247
Joined: Sun Jul 27, 2008 1:39 am
Thanked: 2 times
GMAT Score:660

by orel » Mon Nov 24, 2008 11:41 pm
for the second question:

I got the max height as 7,5.

i think, the max height without clearance can't be 10, because the width of the vehicle is 12, the width remains the same for the top part of the vehicle as well. well, here we have a rectangilar inscribed in a semicircle. since the radius of the semicircle is 10, and the width if the rectangle 12, its length will be sqrt(10^2-6^2)=8
8-1/2=7 1/2

am i missing smth?
Top
Senior | Next Rank: 100 Posts
Posts: 87
Joined: Tue Jul 15, 2008 10:23 am
Location: Lima
Thanked: 4 times
Followed by:1 members

by Fab » Tue Nov 25, 2008 6:13 am
Second statement: CD+DE = radius of larger circle; 2 (CD+DE)-DE = AD = diameter of smaller circle. So we have the radii of both circles again. Sufficient
Are you sure??

Isn't it CD + DE - DE = CD?
Top
Senior | Next Rank: 100 Posts
Posts: 95
Joined: Fri Sep 28, 2007 10:16 am
Thanked: 6 times
GMAT Score:710

by muzali » Tue Nov 25, 2008 10:15 am
Fab wrote:
Second statement: CD+DE = radius of larger circle; 2 (CD+DE)-DE = AD = diameter of smaller circle. So we have the radii of both circles again. Sufficient
Are you sure??

Isn't it CD + DE - DE = CD?
See carefully,
(CD+DE) = radius of larger circle
2*(CD+DE) = diameter of the larger circle
2*(CD+DE)-DE= diameter of smaller circle...enough info to get the area of both circles.
Top
Master | Next Rank: 500 Posts
Posts: 168
Joined: Thu Nov 13, 2008 4:34 am
Location: Pittsburgh
Thanked: 9 times

by Tryingmybest » Fri Nov 28, 2008 10:35 am
Muzali-- Excellent approach to solve these problems...Thanks!!
Top
Post new topic Post Reply
• Page 1 of 1