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# From MitDavidDv: A 700-800 Level GMAT Quant Question

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Joined: 28 Jan 2011
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Kudos [?]: 9 [1] , given: 55

From MitDavidDv: A 700-800 Level GMAT Quant Question [#permalink]  29 Mar 2011, 13:01
1
KUDOS
00:00

Difficulty:

45% (medium)

Question Stats:

55% (03:13) correct 44% (02:38) wrong based on 49 sessions
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
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Kudos [?]: 385 [2] , given: 25

Re: From MitDavidDv: A 700-800 Level GMAT Quant Question [#permalink]  29 Mar 2011, 13:10
2
KUDOS
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

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Re: From MitDavidDv: A 700-800 Level GMAT Quant Question   [#permalink] 29 Mar 2011, 13:10
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