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A cylindrical tank has a base with a circumference of

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New post Updated on: 09 Jun 2013, 08:33
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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. \(2\)

Originally posted by bhushan288 on 28 Nov 2010, 02:04.
Last edited by Bunuel on 09 Jun 2013, 08:33, edited 2 times in total.
Edited the question and added the OA
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Re: Probability Triangle(700 lvl Qn )  [#permalink]

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New post 28 Nov 2010, 02:43
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bhushan288 wrote:
hi guys...
can u help me out with this 1....
thnks in advance

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.

Please read and follow: how-to-improve-the-forum-search-function-for-others-99451.html

So please:
Provide answer choices for PS questions.
Make sure you type the question in exactly as it was stated from the source.


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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 --> \(area_{equilateral}=a^2*\frac{\sqrt{3}}{4}=\sqrt{3}\) --> \(a=2\).

Answer: E.
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New post 28 Nov 2010, 02:51
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Thnks a lot Bunuel...
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Re: Probability Triangle(700 lvl Qn )  [#permalink]

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New post 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 --> \(area_{equilateral}=a^2*\frac{\sqrt{3}}{4}=\sqrt{3}\) --> \(a=2\).

Answer: E.[/quote]

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]

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New post 16 Mar 2011, 07:32
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tinki wrote:

Answer: E.


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]

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New post 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

Answer is E.
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Re: A cylindrical tank has a base with a circumference of  [#permalink]

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New post 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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New post 16 Apr 2013, 03:27
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Re: A cylindrical tank has a base with a circumference of  [#permalink]

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New post 09 Jun 2013, 08:10
Hi Bunuel,
can you explain
how can you consider P(out) as fraction of total base?


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Re: A cylindrical tank has a base with a circumference of  [#permalink]

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New post 11 Jun 2013, 07:29
Hi,

Let P(E) = Probability of grain landing inside triangle = 1 - 3/4 = 1/4;--------(1)
Also P(E) = Area of Equilateral Triangle/Area of Base(i.e. Circle) ---------- (2)

Area (Triangle) = (3^1/2 / 4 )*a^2
Area (Circle) = pi*r^2 = pi * (2(3^1/2/pi)^1/2)^2 = 4*3^1/2

By using 1 & 2

a = 4 (Ans.)

Thanks & Regards,
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Re: A cylindrical tank has a base with a circumference of  [#permalink]

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New post 11 Jun 2013, 08:35
pratsh123 wrote:
Hi,

Let P(E) = Probability of grain landing inside triangle = 1 - 3/4 = 1/4;--------(1)
Also P(E) = Area of Equilateral Triangle/Area of Base(i.e. Circle) ---------- (2)

Area (Triangle) = (3^1/2 / 4 )*a^2
Area (Circle) = pi*r^2 = pi * (2(3^1/2/pi)^1/2)^2 = 4*3^1/2

By using 1 & 2

a = 4 (Ans.)

Thanks & Regards,
Prateek Sharma


a=2. Check here: a-cylindrical-tank-has-a-base-with-a-circumference-of-105453.html#p824188
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Re: A cylindrical tank has a base with a circumference of  [#permalink]

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New post 03 Nov 2013, 10:15
Is there any quicker way to do this one? I started down a similar path and was only about 1/4 done by the time I hit 2 minutes
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Re: A cylindrical tank has a base with a circumference of  [#permalink]

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New post 15 Nov 2013, 21:43
Hey guys, I feel like the answer to my problem is something super obvious, but why is the area of the triangle 1/4 of the base (from subtracting the probability 3/4 from 1), resulting in an area of 3.
I got an area of 4, resulting from (Area of Circle)/(Area of Circle + Area of Triangle) = 3/4 (with Area of Circle = 12). I want to say if I was given a problem asking for the probability of red balls when there are 12 red balls and 4 blue balls, i would say the probability is 12/(12+4) = 3/4.

Thank you for the help........I'm slowly losing it with all these fractions, and positives and negatives, and less than greater than's, and that and it not having references, and primary purpase of passages.......
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Re: A cylindrical tank has a base with a circumference of  [#permalink]

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New post 16 Nov 2013, 03:02
Hey bunuel! same answer same approach! I have been posting answers to some questions but m unaware of how to post formulas in the standard form. please give me a link so i can learn to do the same.
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Re: A cylindrical tank has a base with a circumference of  [#permalink]

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New post 16 Nov 2013, 11:02
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New post 16 Nov 2013, 11:03
dwalker0219 wrote:
Hey guys, I feel like the answer to my problem is something super obvious, but why is the area of the triangle 1/4 of the base (from subtracting the probability 3/4 from 1), resulting in an area of 3.
I got an area of 4, resulting from (Area of Circle)/(Area of Circle + Area of Triangle) = 3/4 (with Area of Circle = 12). I want to say if I was given a problem asking for the probability of red balls when there are 12 red balls and 4 blue balls, i would say the probability is 12/(12+4) = 3/4.

Thank you for the help........I'm slowly losing it with all these fractions, and positives and negatives, and less than greater than's, and that and it not having references, and primary purpase of passages.......


Check here: a-cylindrical-tank-has-a-base-with-a-circumference-of-105453.html#p1234048
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Re: A cylindrical tank has a base with a circumference of  [#permalink]

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New post 26 Jun 2014, 14:57
Bunuel wrote:
bhushan288 wrote:
hi guys...
can u help me out with this 1....
thnks in advance

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.

Please read and follow: how-to-improve-the-forum-search-function-for-others-99451.html

So please:
Provide answer choices for PS questions.
Make sure you type the question in exactly as it was stated from the source.


Also:
Please post PS questions in the PS subforum: gmat-problem-solving-ps-140/
Please post DS questions in the DS subforum: gmat-data-sufficiency-ds-141/

No posting of PS/DS questions is allowed in the main Math forum.

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 --> \(area_{equilateral}=a^2*\frac{\sqrt{3}}{4}=\sqrt{3}\) --> \(a=2\).

Answer: E.

Hi Bunuel,

I missed the part where the triangle = 1/4 of the circle.
I did, however, get all the other results.
So what I did was (1-Area of triangle)/area of circle = 3/4
But I seem to get a different answer that you.
Have any idea why?
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Re: A cylindrical tank has a base with a circumference of  [#permalink]

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New post 27 Jul 2014, 07:11
Bunuel wrote:
... also the ares of the equilateral triangle is \(area_{equilateral}=a^2*\frac{\sqrt{3}}{4}\)...

Answer: E.


Hi Bunuel,
The equation for the area of a triangle is only for an equilateral inscribed in a circle, is it not? is it for any triangle painted within a circle?
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Re: A cylindrical tank has a base with a circumference of  [#permalink]

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New post 27 Jul 2014, 15:27
ronr34 wrote:
Bunuel wrote:
... also the ares of the equilateral triangle is \(area_{equilateral}=a^2*\frac{\sqrt{3}}{4}\)...

Answer: E.


Hi Bunuel,
The equation for the area of a triangle is only for an equilateral inscribed in a circle, is it not? is it for any triangle painted within a circle?


\(area_{equilateral}=side^2*\frac{\sqrt{3}}{4}\) is for ANY EQUILATERAL triangle.

Check Triangles chapter of our Math Book: math-triangles-87197.html
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Re: A cylindrical tank has a base with a circumference of   [#permalink] 27 Jul 2014, 15:27

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