If the answer is C then i can try to explain otherwise otherwise i am also in cue for the solution.tlt2372 wrote:This problem is making my head explode! Please say there is an easy way to do it!
Thanks!
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Probability and Geometry
Source: Beat The GMAT — Problem Solving |
Saurabh Goyal
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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. root (2 (root 6))
b. (root 6 (root 6))/2
c. root (2 root 3)
d. root 3
e. 2
Let P = pi
circumference = 2Pr
2Pr = 4*√(P√3)
r = 4*√(P√3) / (2P)
= 2*√(P√3) / P
area = Pr^2
= P * [2*√(P√3) / P]^2
= P * 4P√3 / P^2
= 4√3
Since P(outside triangle) = 3/4, P(triangle) = 1/4
triangle = 1/4 * 4√3 = √3
area of equilateral triangle = 1/4 * b^2 *√3
1/4 * b^2 *√3 = √3
b^2 = 4
b=2
The correct answer is E.
a. root (2 (root 6))
b. (root 6 (root 6))/2
c. root (2 root 3)
d. root 3
e. 2
Let P = pi
circumference = 2Pr
2Pr = 4*√(P√3)
r = 4*√(P√3) / (2P)
= 2*√(P√3) / P
area = Pr^2
= P * [2*√(P√3) / P]^2
= P * 4P√3 / P^2
= 4√3
Since P(outside triangle) = 3/4, P(triangle) = 1/4
triangle = 1/4 * 4√3 = √3
area of equilateral triangle = 1/4 * b^2 *√3
1/4 * b^2 *√3 = √3
b^2 = 4
b=2
The correct answer is E.
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area of equilateral triangle = 1/4 * b^2 *√3
Only formula i know for the area of a triangle is = (Base * Height )/2
Can you please explain how to reach to that formula that you mentioned above.
Because as i feel its easy to drive formula's in comparison remember them.
thanks.
Only formula i know for the area of a triangle is = (Base * Height )/2
Can you please explain how to reach to that formula that you mentioned above.
Because as i feel its easy to drive formula's in comparison remember them.
thanks.
Saurabh Goyal
[email protected]
-------------------------
EveryBody Wants to Win But Nobody wants to prepare for Win.
[email protected]
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EveryBody Wants to Win But Nobody wants to prepare for Win.
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GMATGuruNY
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The attached drawing shows how the formula for the area of an equilateral triangle is derived.goyalsau wrote:area of equilateral triangle = 1/4 * b^2 *√3
Only formula i know for the area of a triangle is = (Base * Height )/2
Can you please explain how to reach to that formula that you mentioned above.
Because as i feel its easy to drive formula's in comparison remember them.
thanks.
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- (23.74 KiB) Downloaded 177 times
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Thanks guru......GMATGuruNY wrote:The attached drawing shows how the formula for the area of an equilateral triangle is derived.goyalsau wrote:area of equilateral triangle = 1/4 * b^2 *√3
Only formula i know for the area of a triangle is = (Base * Height )/2
Can you please explain how to reach to that formula that you mentioned above.
Because as i feel its easy to drive formula's in comparison remember them.
thanks.
Saurabh Goyal
[email protected]
-------------------------
EveryBody Wants to Win But Nobody wants to prepare for Win.
[email protected]
-------------------------
EveryBody Wants to Win But Nobody wants to prepare for Win.
thanks a lot,GMATGuruNY wrote: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. root (2 (root 6))
b. (root 6 (root 6))/2
c. root (2 root 3)
d. root 3
e. 2
Let P = pi
circumference = 2Pr
2Pr = 4*√(P√3)
r = 4*√(P√3) / (2P)
= 2*√(P√3) / P
area = Pr^2
= P * [2*√(P√3) / P]^2
= P * 4P√3 / P^2
= 4√3
Since P(outside triangle) = 3/4, P(triangle) = 1/4
triangle = 1/4 * 4√3 = √3
area of equilateral triangle = 1/4 * b^2 *√3
1/4 * b^2 *√3 = √3
b^2 = 4
b=2
The correct answer is E.
however i dont understand the circumsference part.
i was thinking of the circumference of cyn linrical tank then got confused...
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GMATGuruNY
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The problem says that the circumference of the base -- which is a circle = 4*√(P√3).diebeatsthegmat wrote:thanks a lot,GMATGuruNY wrote: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. root (2 (root 6))
b. (root 6 (root 6))/2
c. root (2 root 3)
d. root 3
e. 2
Let P = pi
circumference = 2Pr
2Pr = 4*√(P√3)
r = 4*√(P√3) / (2P)
= 2*√(P√3) / P
area = Pr^2
= P * [2*√(P√3) / P]^2
= P * 4P√3 / P^2
= 4√3
Since P(outside triangle) = 3/4, P(triangle) = 1/4
triangle = 1/4 * 4√3 = √3
area of equilateral triangle = 1/4 * b^2 *√3
1/4 * b^2 *√3 = √3
b^2 = 4
b=2
The correct answer is E.
however i dont understand the circumsference part.
i was thinking of the circumference of cyn linrical tank then got confused...
The formula for the circumference of a circle = 2Pr.
So 2Pr = 4*√(P√3)
Divide each side by 2P:
r = 4*√(P√3) / (2P)
Divide the 4 in the numerator by the 2 in the denominator:
r = 2*√(P√3) / P
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
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As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
