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

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dirty dozen

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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. </p>

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Anything to do with Gravity has to be guided by Newton's Laws

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Similar to problem of Cheating Wives / Husbands ... Josephine

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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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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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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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nak added a post in a topic
that is because

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nak added a post in a topic
Indeterminate case for 2 item prices for each Mr. A and Mr. B.

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@ 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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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>
<p> </p>
<p> </p>

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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 (N1) people (365N+1)/365
For all persons 2 to N to have each a diff B'Day P = FACT(364) / FACT (365N)/ 365^(N1)
Prob for at least 2 to share B'day = 1  {FACT(364) / FACT (365N)/ 365^(N1)}
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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