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## Cylindrical tank

GmatKiss GMAT Titan
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Cylindrical tank Sat May 19, 2012 11:07 am
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Anurag@Gurome GMAT Instructor
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Sat May 19, 2012 11:49 am
The probability of the grain of sand landing on the portion of the base outside the triangle is 3/4. Hence, The probability of the grain of sand landing on the portion of the base inside the triangle is (1 - 3/4) = 1/4.

Therefore, the ratio of area of the triangle to the area of the base of the cylinder = 1/4

Say, length of each side of the equilateral triangle is a.
Hence, area of the equilateral triangle = (√3/4)*a²

Now, circumference of the base = 4√(π√3)
So, the radius of the base = 4√(π√3)/(2π)
So, area of the base = π[4√(π√3)/(2π)]² = π[16*(π√3)/(4π²)] = 4√3

Hence, area of the equilateral triangle = √3

Hence, (√3/4)*a² = √3 ---> a = √4 = 2

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Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

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Sat May 19, 2012 11:54 am
This problem feels to be very difficult but if read patiently then it is very easy. Here is what I did:

Given circumference 4√[∏√3]

we know 2∏r = 4√[∏√3]
=> r = 2√[(√3)/∏]

so Area of the base = ∏r^2 => ∏ * (4√3)/∏
Area of the base = 4√3

the problem says that the sand grain has a probability of falling outside the EQUILATERAL triangle is 3/4.
This means the area of the triangle is 1/4 of the total area of the base.

so area of the equilateral triangle is √3

But we know the area of the equilateral triangle is (√3/4)a^2

so (√3/4)a^2 = √3
a^2 = 4
a = 2

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