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A cylindrical tank has a base with a circumference of

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A cylindrical tank has a base with a circumference of [#permalink] New post 29 Mar 2011, 13:01
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A cylindrical tank has a base with a circumference of 4\sqrt{\pi{\sqrt{3}} 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?

A. \sqrt{2{\sqrt{6}}

B. \frac{\sqrt{6{\sqrt{6}}}}{2}

C. \sqrt{2{\sqrt{3}}

D. \sqrt{3}

E. 2

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-cylindrical-tank-has-a-base-with-a-circumference-of-105453.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 04 Jun 2014, 07:01, edited 1 time in total.
Renamed the topic and edited the question.
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Re: From MitDavidDv: A 700-800 Level GMAT Quant Question [#permalink] New post 29 Mar 2011, 13:10
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MitDavidDv wrote:
A cylindrical tank has a base with a circumference of 4sqrt^(pi * sqrt^3) meters and an equilateral triangle painted on the interior side of the base. If a stone is dropped inside the tank and the probability of the stone hitting the portion of the base outside the triangle is 3/4, what is the length of a side of the triangle?


a.) Sqrt^(2*sqrt^6)

b.) Sqrt^(6*sqrt^6)/2

c.) Sqrt^(2*sqrt^3)

d.) Sqrt^3

e.) 2


Let the radius of the base be r
Right so the circumference = 4 \sqrt{\pi * \sqrt{3}}=2 \pi r
Simplifying we get r = \frac{4\sqrt{3}}{\sqrt{\pi}}
Area of base = \pi r^2 = 4 \sqrt{3}

Let the side of the triangle be x, the area of the triangle is \frac{\sqrt{3}x^2}{4}

Probably that the stone will fall inside the triangle = 1 - (3/4) = (1/4) = (Area of triangle)/(Area of base)

Substituting the areas calculated above : \frac{\frac{\sqrt{3}x^2}{4}}{4\sqrt{3}} = \frac{1}{4}
Which simplifies to give : x^2=4

Hence x=2

Answer = E
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Re: From MitDavidDv: A 700-800 Level GMAT Quant Question [#permalink] New post 04 Jun 2014, 05:30
MitDavidDv wrote:
A cylindrical tank has a base with a circumference of 4sqrt^(pi * sqrt^3) meters and an equilateral triangle painted on the interior side of the base. If a stone is dropped inside the tank and the probability of the stone hitting the portion of the base outside the triangle is 3/4, what is the length of a side of the triangle?


a.) Sqrt^(2*sqrt^6)

b.) Sqrt^(6*sqrt^6)/2

c.) Sqrt^(2*sqrt^3)

d.) Sqrt^3

e.) 2


Since we know the probability of hitting outside of the triangle is 3/4, we then know the area of hitting anywhere inside the triangle must be 1/4. Thus, the Area of triangle = 1/4 the area of the circle. Since we can find the area of the circle from the circumference, we just have to solve for the area of the triangle, and solve for the side from there.

C = 4 \sqrt{\pi * \sqrt{3}}=2 \pi r
2 \sqrt{\pi * \sqrt{3}}= \pi r

Square both sides:

4\pi\sqrt{3}= \pi^2 r^2
4\sqrt{3} = \pi r^2

We don't even need to solve for r; we already have the area here.

4\sqrt{3} = 4*Area of Triangle, and thus the Area of the Triangle must be \sqrt{3}.

Since we know the formula for the Area of an Equilateral Triangle with side s is \frac{\sqrt{3}}{4} s^2, we see that s^2/4 = 1, and thus that s = 2.

Answer: E
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Re: A cylindrical tank has a base with a circumference of [#permalink] New post 04 Jun 2014, 07:03
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A cylindrical tank has a base with a circumference of 4\sqrt{\pi{\sqrt{3}} 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?

A. \sqrt{2{\sqrt{6}}

B. \frac{\sqrt{6{\sqrt{6}}}}{2}

C. \sqrt{2{\sqrt{3}}

D. \sqrt{3}

E. 2

Given: circumference=4\sqrt{\pi{\sqrt{3}} and P(out)=\frac{3}{4}

Now, as the probability of the grain of sand landing on the portion of the base outside the triangle is 3/4 then the the portion of the base (circle) outside the triangle must be 3/4 of the are of the base and the triangle itself 1/4 of the are of the base.

Next: circumference=4\sqrt{\pi{\sqrt{3}}}=2\pi{r} --> square both sides --> 16\pi{\sqrt{3}}=4{\pi}^2{r}^2 --> 4{\sqrt{3}}={\pi}{r}^2 --> area_{base}=\pi{r^2}=4{\sqrt{3}};

The area of the equilateral triangle is 1/4 of the base: area_{equilateral}=\frac{1}{4}*4{\sqrt{3}}=\sqrt{3} --> also the ares of the equilateral triangle is area_{equilateral}=a^2*\frac{\sqrt{3}}{4}, where a is the length of a side --> area_{equilateral}=a^2*\frac{\sqrt{3}}{4}=\sqrt{3} --> a=2.

Answer: E.

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-cylindrical-tank-has-a-base-with-a-circumference-of-105453.html

3-D Geometry Questions Directory: 3-d-geometry-questions-171024.html
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Re: A cylindrical tank has a base with a circumference of   [#permalink] 04 Jun 2014, 07:03
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