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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]  28 Nov 2010, 02:04
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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 07 May 2012, 04:37, edited 1 time in total.
Edited the question and added the OA
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Re: Probability Triangle(700 lvl Qn ) [#permalink]  28 Nov 2010, 02:43
bhushan288 wrote:
hi guys...
can u help me out with this 1....

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?

Hi bhushan288, and welcome to Gmat Club.

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Original question is:

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.

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Re: Probability Triangle(700 lvl Qn ) [#permalink]  28 Nov 2010, 02:51
Thnks a lot Bunuel...
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A cylindrical tank has a base [#permalink]  09 Jan 2011, 14:18
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?

a. \sqrt{2[square_root]6}[/square_root]
b. \sqrt{6[square_root]6}[/square_root]/2
c. \sqrt{2[square_root]3}[/square_root]
d. \sqrt{3}
e. 2

real tough one.
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Re: A cylindrical tank has a base [#permalink]  09 Jan 2011, 14:31
Merging similar topics.
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Re: Probability Triangle(700 lvl Qn ) [#permalink]  16 Mar 2011, 07:13
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.

Hello Bunuel

ur explanation is perfect . i just cant understand one thing. i know that formula for the side of equilateral triangle inscribed in the circle
should be a= √3 * r where r is radius and a is side of the triangle, but when using this formula i am not getting the right answer in the above exmpl. what could be the problem? is somth. wrong with formula ? thanks
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Re: Probability Triangle(700 lvl Qn ) [#permalink]  16 Mar 2011, 07:32
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tinki wrote:

Hello Bunuel

ur explanation is perfect . i just cant understand one thing. i know that formula for the side of equilateral triangle inscribed in the circle
should be a= √3 * r where r is radius and a is side of the triangle, but when using this formula i am not getting the right answer in the above exmpl. what could be the problem? is somth. wrong with formula ? thanks[/quote]

The triangle is not necessarily inscribed because it is not mentioned in the question. It can be any equilateral triangle drawn within the base. The vertices of the triangle may not touch the circle.
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Re: Probability Triangle(700 lvl Qn ) [#permalink]  16 Mar 2011, 08:18
Ahaaa, got it , thanks
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Re: Probability Triangle(700 lvl Qn ) [#permalink]  17 Mar 2011, 01:40
2*pi*r = 4(sqrt(pi sqrt(3))

=> r = 4(sqrt(pi sqrt(3))/2*pi = 2(sqrt(pi sqrt(3))/pi

Area of circle C = pi * r^2 = 4 * pi* 1/(pi)^2 * pi * sqrt(3) = 4*sqrt(3)

A/C = 1/4

=> A = sqrt(3)

Area of Triangle = sqrt(3) = sqrt(3)/4 * (side)^2

So side = sqrt(4) = 2

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Re: A cylindrical tank has a base with a circumference of [#permalink]  15 Apr 2013, 09:59
could u please explain me why we cant use this formula here ?

The radius of the circumscribed circle is R=a*\sqrt{3}/3

math-triangles-87197.html
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Re: A cylindrical tank has a base with a circumference of [#permalink]  16 Apr 2013, 03:27
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LalaB wrote:
could u please explain me why we cant use this formula here ?

The radius of the circumscribed circle is R=a*\sqrt{3}/3

math-triangles-87197.html

Because we are told that equilateral triangle is painted so not necessarily inscribed on the interior side of the base.
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Re: A cylindrical tank has a base with a circumference of   [#permalink] 16 Apr 2013, 03:27
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