cylindrical tank

Question

A quick way to solve this under 2 minutes?

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?

\sqrt{2 \sqrt{6}}

(\sqrt{6 \sqrt{6}})/2

\sqrt{2 \sqrt{3}}

\sqrt{3 }

2

walker · Answer

Can you explain what it means. Does circumference equals 1 meter? Or how many meters?

gmatnub · Answer

I think it just mean that the measurements/dimensions are in meters.

durgesh79 · Answer

I dont think this questin can be solved. we need radius of the circle, which is not given 
there is a number missing in highlighted part.

walker · Answer

I feel that something is wrong with this question.

the probability 3/4 means that relation between area of a circle and that of a triangle is 1/4. So,

\frac{S_t}{S_c}=\frac14

\frac{L^2*\sqrt3}{4*S_c}=\frac14

L=sqrt{\frac{S_c}{\sqrt3}}

gmatnub · Answer

My bad, the number did not get copied. BTW, what is the math symbol for pi?

walker · Answer

So, we know that  \pi d=4\sqrt{\pi\sqrt3}  --->  d=4\sqrt{\frac{\sqrt3}{\pi}}

S_c=\frac{\pi d^2}{4}=\frac{\pi 16 \sqrt3}{4\pi}=4\sqrt3

L=sqrt{\frac{S_c}{\sqrt3}}=sqrt{\frac{4\sqrt3}{\sqrt3}}=2  - E

by the way, use \pi  \pi

Oski · Answer

The radius r of the circle is  2 \Pi r = 4 \sqrt{\Pi \sqrt{3}}  =>  r = 2 \sqrt{\frac{sqrt{3}}{\Pi}}

Then the area inside the circle is  \Pi r^2 = 4 sqrt 3

The area of an equilateral triangle of side c is  \frac{c^2}{4} sqrt 3  and we want this area to  \frac{1}{4}  of the area inside the circle

We thus have to solve  \frac{\frac{c^2}{4} sqrt 3}{4 sqrt 3} = \frac{1}{4}  that gives  c = 2

scthakur · Answer

I am surely missing something here.

If are of circle is 4sqrt(3) then are of triangle will be 4 / sqrt(3).

Now, if x is the side of equilateral traingle, then its area will be sqrt(3)*x*x/4

Equating the two, I get x as 4/sqrt(3).

What am I doing wrong here?

scthakur · Answer

Forget my comment. My silly mistake was that I was considering 1/3 in place of 1/4.

cylindrical tank

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