nak

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About nak

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  1. nak added a post in a topic   

  2. nak added a post in a topic   

    Can't get it :
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  3. nak added a post in a topic   

    dirty dozen
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  4. nak added a post in a topic   


    Would tend to agree. Squashing the bug involves Work. And the log with the maximum energy can do the most work. The kinetic energy of the log at impact will get transmitted to the bugs innards and make them fly in all directions. So the log with the most energy (in this case the 31st from top and 20th from bottom) will make the bug particles fly the maximum distance and cause the flattest bug
    My earlier hypothesis that the log with maximum momentum (impulse), which incidentally is the 12th from bottom and 39th from the top will cause the maximum force to be applied on the bug stands corrected. The log with the max energy will transfer the max force on the bug to squash it.

    Instead of the squashing bugs if the logs were to fall on nails or spikes stuck into the ground, then a similar poser would have been which log drives the nail the deepest. This would be the principle on which industrial pile drivers are designed. And these take into account the Kinetic energy transferred by the Hammer to the pile head.&nbsp;</p>
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  5. nak added a post in a topic   

    Anything to do with Gravity has to be guided by Newton's Laws
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  6. nak added a post in a topic   

    Similar to problem of Cheating Wives / Husbands ... Josephine
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  7. nak added a post in a topic   

    To note: Bee paid by Credit Card i.e. he has to really pay out when his Card Statement falls due. But Ay paid in cash that is today.

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  8. nak added a post in a topic   

    If the person is an ETPS (extremely tenacious poor smoker) and does not mind working to get infinitesimal fractions of a cigarette then the answer truly is
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  9. nak added a topic in New Logic/Math Puzzles   

    I walk South by my compass for a distance of 2 KM. I then walk West (again by my compass) for a distance of 2 KMs. I walk North (as you can guess using the same compass) , a distance of 2 Kms. I hoped to Walk East for a distance of 2 Kms to complete my prescribed 5 mile walk and be home. But I find that I already reached home having walked 6 Kms.

    Where do I live?
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  10. nak added a post in a topic   

    that is because
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  11. nak added a post in a topic   

    Indeterminate case for 2 item prices for each Mr. A and Mr. B.


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  12. nak added a post in a topic   

    @ skpoudyal

    Thks for the simplified 3rd degree polynomial. Got stuck in reducing that.

    Resorted to goal seek in excel to get answer. Hope that is not against the law.
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  13. nak added a post in a topic   

    The original words were :BEND SOON SEND BOND SINE BIDS NINE NOSE BINS SEEN NONE DONE DINE NOON

    Stage 1: After dealing out letters of one of the words, Ajit says : No one knows how many vowels in the chosen word.
    Now if any of the friends had received the letter B then this would not be true. Because all words containing B have only ONE vowel.
    So Remove the words with B and then we are left with:
    SOON SEND SINE NINE NOSE SEEN NONE DONE DINE NOON (note that only one word has ONE vowel, rest all have TWO)
    Stage 2:Again Ajit says "no one knows how many vowels in chosen word"
    in this reduced set notice that all words with O or I have two vowels each. So if anyone had an O or an I, Ajit would have been wrong. So no one had an O or an I and so we can also further drop all the words with an O or an I. That would leave :
    SEND SEEN

    Now at this stage if anyone had a D - he would have known what the chosen word was. So that meant each of 4 friends had letters S, E, N among themselves.
    Also at this point the 4 wise friends (supposing each exactly knew how long it would take the others to figure out the above), after a wait could have concluded that none had D and so the word would have been SEEN. Perhaps they were keeping quiet because they were not asked OR they wanted the confirmation to come from Ajit - which he provided with his next statement

    Stage 3: Ajit who knew that choice had shortened to SEND and SEEN, stated again "none knows how many vowels". This would hold true if the friends had only the letters S,E,N among themselves.
    This sealed the answer to SEEN. In fact all of the friends - no mater what letter they were holding would have come to the conclusion. It is just that Rohit was the quicker and responded or maybe he was impatient or bored with the manner in which Ajit was dragging the game
    <p> </p>
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  14. nak added a post in a topic   

    First take case of K= 2 (this has already been posted before):

    What is the chance that out of N people in a Room AT LEAST 2 people will have same birthday

    Let us choose Person 1 out of the N, with a birthday on a give date

    Now for person 2 : what is the chance he will have a different birthday 364/365
    For person 3: what is the chance he will have a different birthday than 1 & 2 363/365
    For person N : what is the chance he will have a different B'day than all other (N-1) people (365-N+1)/365

    For all persons 2 to N to have each a diff B'Day P = FACT(364) / FACT (365-N)/ 365^(N-1)
    Prob for at least 2 to share B'day = 1 - {FACT(364) / FACT (365-N)/ 365^(N-1)}


    Work out the probs like above for increasing N but K =2:


    N Probability Prob of ALL Prob of atleast
    of Nth person 1 to N having TWO having same
    having a diff diff B'Days B'Day
    B'Day
    from others
    2 99.7260% 99.7260% 0.2740%
    3 99.4521% 99.1796% 0.8204%
    4 99.1781% 98.3644% 1.6356%
    5 98.9041% 97.2864% 2.7136%
    6 98.6301% 95.9538% 4.0462%
    7 98.3562% 94.3764% 5.6236%
    8 98.0822% 92.5665% 7.4335%
    9 97.8082% 90.5376% 9.4624%
    10 97.5342% 88.3052% 11.6948%
    .
    .
    23 93.9726% 49.2703% 50.7297%
    24 93.6986% 46.1656% 53.8344%
    25 93.4247% 43.1300% 56.8700%
    26 93.1507% 40.1759% 59.8241%
    27 92.8767% 37.3141% 62.6859%
    28 92.6027% 34.5539% 65.4461%
    29 92.3288% 31.9031% 68.0969%
    30 92.0548% 29.3684% 70.6316%

    So for N= 30 and K = 2 result is 70.63%

    Lets go for N= 30 and K = 3

    Out of the 30 above where at least 2 have same B'Day- take out one of the 2. leaving 29. Now the problem becomes find the prob of 2 having same b'day out of 29.

    so teh answer would be P[n=30, k=2] x P {n=29, k=2) = 70.63 x 68.09 = 48.10% .... and likewise.
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