| PUZZLE CORNER |
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Week of the Year
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Amla, Bamla, Camla, Damla and Emala are good friends. Only two of them are always honest. The other three persons are sometimes honest and sometimes not. When they were asked about who is a liar. Here are the answers from them:
1. Amla said: Bamla is not a lier.
2. Bamla said: Camla is a liar.
3. Camla said: Damla is a liar.
4. Damla said: Emala is a liar.
5. Emala said: Bamla is a liar.
6. Amla said: Emala is not a liar.
7. Emala said: Camla is a liar.
Find out which two persons are always honest?
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If Amla is honest, then the statement 1 and 6 are true. In this case three of them, Amla, Bamla and Emala would be honest which has a conflict with the fact. So Amla must be a liar.
If Camla is honest, then according to statement 3 - Damla is a liar. Statement 4 will tell us Emala should be honest. Statement 7 will tell us Camla is a liar, which will have a conflict with our assumption. So Camla must be a liar.
If Emala is honest, then according to statement 5 Bamla is a liar. Statement 2 will tell us Camla is honest. In the meantime, statement 7 tells us Camla is a liar that is a conflict. So Emala is also a liar.
We can verify Bamla and Damla are honest by going through each statement.
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Out of two coins, one is fair (the probability of getting head is 1/2) and the other is biased with probability of getting a tail 1/3. One of the coins is tossed once, resulting in heads. The other is tossed three times, resulting in two heads. Which coin is more likely to be the biased one?
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Let's find the probability of the given outcomes in two different cases: the first coin being the fair one, and the second-the biased one, and vice versa.
If we assume that the first coin is fair, then the probability of the heads is 1/2. The second coin must be the biased one, and the probability of it coming up with 2 heads and 1 tail in three tosses is 3*2/3*2/3*1/3 = 4/9. Note that there are three ways to get 2 heads: HHT, HTH, THH, the probability of each is being 4/27. Thus, the probability of both coins coming up with the given results is 2/9.
If, on the other hand, the first coin is the biased one, and the second coin is fair the probability of them resulting in the combination given in the problem is 3 * 2/3 * 1/2 * 1/2 * 1/2 = 1/4, or 2/8 which is greater than the 2/9. Therefore it is more probable that the first coin is the biased one.
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A clock gains 2% time during the first week and then loses 2% time during the next one week. If the clock was set right at 12 noon on a Wednesday, what will be the time that the clock will show exactly 14 days from the time it was set right?
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The clock gains 2% time during the first week. In a day there are 24 hours and in a week there are 7 days. Therefore, there are 7 * 24 = 168 hours in a week.
If the clock gains 2% time during the first week, then it will show a time which is 2% of 168 hours more than 12 Noon on next Wednesday (end of first week) = 3.36 hours more.
Subsequently, the clock loses 2% during the next week. The second week again has 168 hours and the clock loses 2% time = 2% of 168 hours = 3.36 hours less than the actual time.
As it gained 3.36 hours during the first week and then lost 3.36 hours during the next week, the net result will be a -3.36 + 3.36 = 0 hour net gain in time.
So the clock will show a time, which is exactly 12 Noon on Wednesday two weeks from the time it was set right.
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A number can be constructed from any date of the year by adding the number of the month to the number of the day. For example: November 22nd would become the number 33, since November is the eleventh month, and 22 + 11 = 33. May 14 would become 19, and so on. How many different numbers can you make, using dates of the normal calendar?
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Following is the list of the numbers that can be constructed each month.
January: 2 to 32 (From 1st January - 31st January)
February: 3 to 30 (From 1st February - 28th February)
March: 4 to 34 (From 1st March - 31st March)
April: 5 to 34 (From 1st April - 30th April)
May: 6 to 36 (From 1st May - 31st May)
June: 7 to 36 (From 1st June - 30th June)
July: 8 to 38 (From 1st July - 31st July)
August: 9 to 39 (From 1st August - 31st August)
September: 10 to 39 (From 1st September - 30th September)
October: 11 to 41 (From 1st October - 31st October)
November: 12 to 41 (From 1st November - 30th November)
December: 13 to 43 (From 1st December - 31st December)
Total 42 numbers are possible, every number between 2 and 43 inclusive.
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In a casino, Anurag bets on number 23 on a spinning wheel 12 times and loses each time. On the 13th spin he does a quick calculation and finds out that the number 17 had appeared three times in the last 12 spins and is therefore, unable to decide whether to bet on 23 or 17 in the 13th spin. Which number (23 or 17) will give him the best chance of winning and what are the odds of winning on the bet that he takes? (Wheel has numbers 1 to 36)
(1) 23; 2:1 (2) 17; 19:17 (3) Either; 18:1 (4) Either; 35:1
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Since each of the spin is an independent event, thus the outcome of the 13th spin will not depend on the outcome of the previous spins. Hence, the odds of winning on the number that Anurag bets on, is 35 : 1 in each case.
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