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Cylindrical tank

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GmatKiss GMAT Titan Default Avatar
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Cylindrical tank Post Sat May 19, 2012 11:07 am
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  • Lap #[LAPCOUNT] ([LAPTIME])
    Please help!
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    Anurag@Gurome GMAT Instructor
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    Post 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

    The correct answer is E.

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    coolhabhi Rising GMAT Star Default Avatar
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    Post 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
    Answer is E

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