Re: [ascent CAT] new problem

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• let us divide the circle in to 2 semicircles.the prob for a selected point to be on one of the semicircle is equal to the prob of being on the other.so prob
Message 1 of 8 , Jun 24, 2004
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let us divide the circle in to 2 semicircles.the prob for a selected point to be on one of the semicircle is equal to the prob of being on the other.so prob for 8 selected points to be on any one of semicircles = 2*1/256 = 1/128.

On Thu, 24 Jun 2004 josephite_shashi wrote :

>hii friends,
>i have a problem? plzz help me to solve it.
>8 points are selected at random on a circle, what is the probability
>they lie on a semicircle?
>regards
>shashi
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• hi rahul it should be minimum outside surface area the surface area for a cylindrical bucket with one side open is: pi*(r^2)[bottom]+2*pi*r*h[curved surface]
Message 2 of 8 , Jun 25, 2004
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hi rahul
it should be minimum outside surface "area"
the surface area for a cylindrical bucket with one side open is:
pi*(r^2)[bottom]+2*pi*r*h[curved surface] say S
the volume
pi*r^2*h=30*1000 c.c [given]
this gives h=30000/(pi*r^2) cm
substituting in eq for area gives:
S= pi*r^2+ 60000/r cm sq
for this to be min dS/dr=0 [calculas funda] and d2S/dr2 be +ve
which in this case is 6000*pi so
which gives r^3=30000/pi;
which gives
S= 9000*(pi/30)^(1/3)
S=4242.14 cm sq

--- In ascent4cat@yahoogroups.com, <rahul.soni@w...> wrote:
> hi,
> solve this question....
>
>
> Calculate the minimum outside surface of a cylindrical bucket
with an open upper side and which can hold 30 liters of water.
>
> The Question: What is the minimum outside surface?
>
>
> Thanks & Regards,
> Rahul Soni
• I feel that the probability should be (1/2)^8 i.e. 1/256. Since the two semi-circles are imaginary and so cannot be differntiated as such and so we shuld not
Message 3 of 8 , Jun 30, 2004
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I feel that the probability should be (1/2)^8 i.e. 1/256. Since the two semi-circles are imaginary and so cannot be differntiated as such and so we shuld not multiply by 2.

hai crux <dgb_3@...> wrote:

let us divide the circle in to 2 semicircles.the prob for a selected point to be on one of the semicircle is equal to the prob of being on the other.so prob for 8 selected points to be on any one of semicircles = 2*1/256 = 1/128.

On Thu, 24 Jun 2004 josephite_shashi wrote :
>hii friends,
>i have a problem? plzz help me to solve it.
>8 points are selected at random on a circle, what is the probability
>they lie on a semicircle?
>regards
>shashi
>Ascent Education
>An IIM Alumni Venture
>Class for CAT, XAT, GRE, GMAT
>http://www.ascenteducation.com
>Archives of past CAT questions can be viewed at
>http://www.ascenteducation.com/india-mba/iim/cat/questionbank/questionbank.shtml
>To visit your group on the web, go to:
>http://groups.yahoo.com/group/ascent4cat/

>
>To unsubscribe from this group, send an email to:
>ascent4cat-unsubscribe@yahoogroups.com

>
>Your use of Yahoo! Groups is subject to the
>.
>

Ascent Education
An IIM Alumni Venture
Class for CAT, XAT, GRE, GMAT
http://www.ascenteducation.com

Archives of past CAT questions can be viewed at http://www.ascenteducation.com/india-mba/iim/cat/questionbank/questionbank.shtml

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