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

Redeem

MGMAT Prep Test 2

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Newbie | Next Rank: 10 Posts
Posts: 5
Joined: Tue Sep 16, 2008 10:40 pm

MGMAT Prep Test 2

by domi84 » Tue Nov 18, 2008 8:01 am
Hi guys,


my question is how to solve a question like the one below within two minutes. Are there special techniques I am not aware of? I'm aiming for 700+ and am quite content with my verbal scores (around 90% quantile according to MGMAT), but quant is driving me crazy.

Don't get me wrong here, I fully understand every single step as proposed in the solution, but I don't have any idea how to solve these kind of problems in a reasonable amount of time:

"A cylindrical tank has a base with a circumference of meters and an equilateral triangle painted on the interior side of the base. A grain of sand is dropped into the tank, and has an equal probability of landing on any particular point on the base. If the probability of the grain of sand landing on the portion of the base outside the triangle is 3/4, what is the length of a side of the triangle?"

I find the MGMAT Quant part extremely difficult for most questions. I finished the OG quant and even the Kaplan 800 and didn't make many mistakes except for some minor "did not pay attention" errors.

Is it just me or are the MGMAT Questions really that much tougher/different than the other material??


I'm greatful for every kind of help.


PS: Sorry for spelling/grammar errors, I'm a non-native speaker from Germany
Top
Source: — Problem Solving |
Legendary Member
Posts: 683
Joined: Tue Jul 22, 2008 1:58 pm
Location: Dubai
Thanked: 73 times
Followed by:2 members

by mals24 » Tue Nov 18, 2008 8:13 am
Have you copied the question correctly. You haven't mentioned how many meters is the circumference of the circle.
A cylindrical tank has a base with a circumference of meters

Also can you post the 5 answer choices as well
Top
User avatar
Community Manager
Posts: 1049
Joined: Sun Apr 06, 2008 5:15 pm
Location: Pittsburgh, PA
Thanked: 113 times
Followed by:27 members
GMAT Score:710

by dmateer25 » Tue Nov 18, 2008 8:27 am
https://www.manhattangmat.com/forums/pro ... t2672.html
Top
Newbie | Next Rank: 10 Posts
Posts: 5
Joined: Tue Sep 16, 2008 10:40 pm

Thanks

by domi84 » Tue Nov 18, 2008 8:56 am
@mals24: you are right, it didn't copy the circumference, sorry for that one

@dmateer25: thanks for the link :D

looks like other people are troubled too with the question. but even if I know the single steps of the solution i still find it pretty hard to solve the question within 2 min
Top
Master | Next Rank: 500 Posts
Posts: 122
Joined: Mon Sep 22, 2008 5:11 pm
Thanked: 1 times

by abcdefg » Fri Jul 17, 2009 8:27 pm
i just did this problem on the manhattan gmat practice exam also and had the same thoughts as you did. there's no way i could have solved this problem under 2 minutes.

Anyone have any shortcuts other than just brute force algebraic manipulation? thanks.
Top
Master | Next Rank: 500 Posts
Posts: 122
Joined: Fri May 22, 2009 10:38 pm
Thanked: 8 times
GMAT Score:700

by rah_pandey » Sat Jul 18, 2009 12:10 pm
if probability of grain falling outside is 3/4 than the ratio of area outside the triangle to area of the triangle=3/4/1/4=3

C-A/A=3

C=nr^2
n=pi
r=radius
A=area of triangle
=>C=4A
=>A=C/4
now A=(root(3)/4)*a^2
=>rt(3)a^2=nr^2
=>a=(n/rt(3))^(1/2)r

This is a good 700+ question. According to me if you know the concept than a similar question can be solved within 2 mins. Key is to move in the right direction at first go. Obviously no scope for trial and error if you want to finish within 2 mins
Top
Master | Next Rank: 500 Posts
Posts: 338
Joined: Fri Apr 17, 2009 1:49 am
Thanked: 9 times
Followed by:3 members

by kaulnikhil » Sat Jul 18, 2009 1:15 pm
u have the probability ...so u can use it in the following manner
let the length of the side of equilateral triangle be L.
area of the the base be pi r^2
the area outside the triangular region is pi r^2 - V3/4 L^2
so we have (v stands for square root)
(pi r^2 - v3/4 L^2)/pi r2 = 3 /4-----1

we get L in terms of r
the link provide says circumference is 4(root(pi(root 3))) meters

we get r from this .. substitute r in equation 1
we get L = 2
Top
Post new topic Post Reply
• Page 1 of 1