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Re: [ascent CAT] new problem

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  • hai crux
    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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    • dhruv
      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
      • anuj singla
        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
          >Yahoo! Groups Sponsor
          >ADVERTISEMENT
          >Yahoo! Groups Links
          >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
          >Yahoo! Terms of Service
          >.
          >





          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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