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
MGMAT Prep Test 2
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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
A cylindrical tank has a base with a circumference of meters
Also can you post the 5 answer choices as well
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.
Anyone have any shortcuts other than just brute force algebraic manipulation? thanks.
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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
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
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- Master | Next Rank: 500 Posts
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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
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