Distance covered - One puzzle a day - Puzzle Buddies

Puzzle:There is a 50 m long army platoon marching ahead. The last person in the platoon wants to give a letter to first person leading the platoon. So while the platoon is marching he runs ahead, reaches the first person and hands over the letter to him and without stopping he runs and comes back to his original position.

In the mean time the whole platoon has moved by 50 m.

The question is how much distance did the last person cover in that time assuming that he ran the whole distance with uniform speed.
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Solution:The last person covered 120.71 meters.

It is given that the platoon and the last person moved with  uniform speed. Also, they both moved for the identical  amount of time. Hence, the ratio of the distance they  covered - while person moving forward and backword - are  equal.   Let's assume that when the last person reached the first  person, the platoon moved X meters forward.   Thus, while moving forward the last person moved (50+X)  meters whereas the platoon moved X meters.   Similarly, while moving back the last person moved [50-(50- X)] X meters whereas the platoon moved (50-X) meters.   Now, as the ratios are equal,  (50+X)/X = X/(50-X)  (50+X)*(50-X) = X*X   Solving, X=35.355 meters   Thus, total distance covered by the last person  = (50+X) + X  = 2*X + 50  = 2*(35.355) + 50  = 120.71 meters  

Winner:

Coin Tossing: One puzzle a day - Puzzle Buddies

Puzzle: There's a coin toss game in a casino. You are friends with the dealer and can set up a deal with him. Dealer can get what he wants 80% of times. The bet is that if you win you win $1 and if you loose then you loose $1. You can make X attempts only. What should be your strategy to always win.

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Source: CSE Blog

Camel and Bananas - One puzzle a day - Puzzle Buddies

Puzzle:

A banana plantation is located next to a desert. The plantation owner has 3000 bananas which he wants to transport to the market by camel. The market is located 1000 kilometre away from the plantation with desert in between. The owner has only one camel, which carries a maximum of 1000 bananas at any moment in time. The camel eats one banana every kilometre it travels.

What is the largest number of bananas that can be delivered at the market?

Answer:

The camel needs to make more than one trip to deliver all bananas to the market. No doubt he needs to make at least one stop in between otherwise he can never have enough bananas left for the comeback journey. And this point needs to be less than 500 if he wants any bananas to be left when he reaches the market. Let’s say it’s A.

Plantation======A=====================Market

The trick is to make sure that the camel always travels with his full capacity of bananas. This is true initially as there are total 3000 bananas and the camel carries 1000 bananas at a time. So, he will make 3 trips with 1000 bananas to point A. To make sure he carried his full capacity after point A, we should have a multiple of 1000 bananas left when all bananas are at point A. This number can thus be 2000 at best. So he is eating 1000 bananas from the plantation to point A, making 5 trips in between. So, he eats 200 bananas in 1 trip. Point A is 200 km from the plantation.

Again we need a stopping point in between A and the plantation where there will be 1000 bananas left after 3 trips. So the camel eats 333 1/3 bananas in each trip, point B being 333 1/3 km away from A.

Plantation======A=========B===========Market

In the end, the camel needs to make only 1 trip from B to the market which is 466 2/3 km away, leaving 533 1/3 bananas at the market.

Winner: Ashish and Prabharan

How many Days- One Puzzle a Day -Puzzle Buddies

Puzzle: There is a frog lying under the bottom of a well of depth 40 meters.every day morning the frog leaps up 4 meters and it sleeps till next morning.while the frog is asleep he falls 3 meters down.how many days it will take the frog to escape from the well?

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Solution: on first day the frog jumps from 0 t0 4 meters and by night it slips 3 meters

so day 2 it starts at 1

day 3 it starts at 2 likewise

day 37 it starts at 36 and it jumps 4 meters and it comes out

So it takes 37 days.

Winner: Gaurave

Pair game - One puzzle a day - Puzzle Buddies

Puzzle:Ganesh has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, for each value from 1 to 6, there are two cards in the deck with that value. Ganesh likes to play a game in which he shuffles the deck, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Ganesh finds at least one pair of cards that have the same value?
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Solution:The chance of getting AT LEAST one pair of cards with the same value out of 4 dealt cards should be computed using the 1-x technique. That is, you should figure out the probability of getting NO PAIRS in those 4 cards (an easier probability to compute), and then subtract that probability from 1.

First card: The probability of getting NO pairs so far is 1 (since only one card has been dealt).

Second card: There is 1 card left in the deck with the same value as the first card. Thus, there are 10 cards that will NOT form a pair with the first card. We have 11 cards left in the deck.

Probability of NO pairs so far = 10/11.

Third card: Since we have gotten no pairs so far, we have two cards dealt with different values. There are 2 cards in the deck with the same values as those two cards. Thus, there are 8 cards that will not form a pair with either of those two cards. We have 10 cards left in the deck.
Probability of turning over a third card that does NOT form a pair in any way, GIVEN that we have NO pairs so far = 8/10.

Cumulative probability of avoiding a pair BOTH on the second card AND on the third card = product of the two probabilities above = (10/11) (8/10) = 8/11.

Fourth card: Now we have three cards dealt with different values. There are 3 cards in the deck with the same values; thus, there are 6 cards in the deck that will not form a pair with any of the three dealt cards. We have 9 cards left in the deck.
Probability of turning over a fourth card that does NOT form a pair in any way, GIVEN that we have NO pairs so far = 6/9.

Cumulative probability of avoiding a pair on the second card AND on the third card AND on the fourth card = cumulative product = (10/11) (8/10) (6/9) = 16/33.

Thus, the probability of getting AT LEAST ONE pair in the four cards is 1 - 16/33 = 17/33.
Winner: Nonone

How many weights? Part3 - One puzzle a day - Puzzle Buddies

Puzzle: This is the 3rd part of weighing puzzles. You can check Part 2 and Part 1.

You have to weigh 125 packets of sugar with first one weighing 1 lb, second 2 lb, third 3 lb and so on.You can only use one pan of the common balance for measurement for weighing sugar, the other pan has to be used for weights.

The weights have to put on to the balance and it takes time. Time taken to move weights is equal to the number of weights moved.

Eg. So if you are weighing 5 lb using 3lb and 2 lb weight then you have to lift both these weights one by one and thus it will take 2 unit time. If you have to measure only 3lbs it will take only 1 unit time. Now if we have to measure sugar packets of weights 1,3,4 using weights 1 and 3 only. For this first we measure 1 lb, with 1 unit of time, we place 3 lb along with 1 lb and measure 4kg with again 1 unit of time, and finally we move 1kg out of pan to measure 3kg in 1 unit of time. So in 3 units of time we could measure 1,3 and 4lb using weights 1 and 3 only.

Now you have to make sugar packets of all weights from 1 to 125 in minimum time, in other words in minimum movement of weights. The question here is to find out the minimum number of weighs needed and the weight of each the weights used and the strategy to be followed for the creation of 125 packets of sugar.

Note: Don't forget to visit us again. Answer to the puzzle will be posted tomorrow with winner's name. You can provide your answer in comments.
Source: Classic Puzzles

Solution: Amazing explanation provided by Kasturi.
You will need the weights 1, 2, 4, 8, 16, 32 and 64 i.e. 7 weights. You can measure 125 packets in 125 movements. The strategy to be followed is this:
Put 1: measure 1, weights in pan 1

To measure further you need to use a new weight.
Put 2: measure 3, weights in pan 1,2
Remove 1: measure 2, weights in pan 2

Thus, by adding the next smallest weight to the pan i.e. 2, we can measure all weights below 4 in total 3 movements.

To measure further you need to use a new weight.
Put 4: measure 6, weights in pan 2,4
Remove 2: measure 4, weights in pan 4
Put 1: measure 5, weights in pan 1,4
Put 2: measure 7, weights in pan 1,2,4

Thus, by adding the next smallest weight to the pan i.e. 4, we can measure all weights below 8 in total 7 movements.

Following the same strategy, whenever we have measured all weights below n, add n and we can measure measure all the weights below 2n in 2n-1 movements.
Eventually, we can measure 125 packets in 125 movements.


Winner: Kasturi again is the winner.

Boat and Anchor - One Puzzle a Day - Puzzle Buddies

Puzzle:

A fisherman is sitting in his boat in the center of a small lake. He throws his anchor into the water. Does the water level rise, fall or stay the same?

Solution:

When the anchor is in the boat, the water displaced is equal to the weight of the anchor. When the anchor is dropped in the water, the water displaced is equal to the volume of the anchor. Since the anchor is denser than water, its weight is more than its volume as compared to water. Hence, it displaces less water when it is thrown in the lake. Thus, the water level falls.

Winner: Vinay

Puzzle: Palindrome - One Puzzle a Day -Puzzle Buddies

Puzzle:

What is smallest possible palindrome number whose cube root is not a palindrome?

For example 1331 is a palindrome. Cube root of 1331 is 11 which is also a palindrome.so this does not satisfy the condition.

Note:a palindrome is a word,phrase or a number that can be read same in either direction.

Note: Don't forget to visit us again. Answer to the puzzle will be posted tomorrow with winner's name. You can provide your answer in comments.
Solution:10662526601 = 2201³.It is not yet known that if there are any other palindromic numbers whose cube roots are not palindromes.

As of now there are no palindromes of the form X ⁿ where X is not a palindrome and n>3.

the only exception is 2201.

Winner:

Passenger seating - One puzzle a day - Puzzle Buddies

Puzzle:A line of 100 airline passengers is waiting to board the plain. They each hold a ticket to one of the 100 seats on that flight. (For convenience, let's say that the Nth passenger in line has a ticket for the seat number N.)
Unfortunately, the first person in line is crazy, and will ignore the seat number on their ticket, picking a random seat to occupy. All the other passengers are quite normal, and will go to their proper seat unless it is already occupied. If it is occupied, they will then find a free seat to sit in, at random.
What is the probability that the last (100th) person to board the plane will sit in their proper seat (#100)?
Note: Don't forget to visit us again. Answer to the puzzle will be posted tomorrow with winner's name. You can provide your answer in comments.
Solution:The solution to this one is pretty mathimatically involved.

The crazy passenger that boards first has an equal chance of getting on any of the 100 seats. Therefore, the probability of this person sitting in a particular seat k is 1/100.

Now for any value of k, all the passengers up to k will board and take their seats as normal. Then the k-th passenger will find themselves out of a seat, and pick a random seat, in effect, becoming the crazy passenger for a problem with 101 - kpassengers.

(For example if the crazy person takes the tenth seat, passengers 2 through 9 will be seated normally, and when passenger 10's turn comes, they can be considered the crazy passenger first in line of 91 people boarding a plane with 91 seats.)

Let us define P(n) as the probability that the last of n passengers boards their proper place under the conditions of the problem. So for solution, we are looking to find the value of P(100).

In a situation with n passengers, the first, "crazy", passenger can take any of the seats 1 .. n with the probability of 1/n. Also, given some seat number k that the crazy passenger may occupy, the problem will become that of 1+n-k passengers. Therefore, we can say that

P(n) = 1/n * (1 + P(n-1) + P(n-2) + ... + P(2))

(This problem does not make sense with less than 2 passengers. Also, it is easy to see that P(2) is 1/2)

What we would like to do is see what the value of P(n) is in terms of P(n-1).

P(n-1) = 1/(n-1) * (1 + P(n-2) + P(n-3) + ... + P(2) P(n-1) * (n-1) = (1 + P(n-2) + P(n-3) + ... + P(2) P(n) = 1/n * (1 + P(n-1) + P(n-2) + ... + P(2)) (P(n) * n) - P(n-1) = (1 + P(n-2) + ... + P(2))

So,

(P(n) * n) - P(n-1) = P(n-1) * (n-1) (P(n) * n) - P(n-1) = (P(n-1) * n) - P(n-1) (P(n) * n) = (P(n-1) * n) P(n) = P(n-1)

This shows us that the problem's answer does not change if we add one more passenger. Starting at 2 passengers with the anser of 1/2, we can add passengers indefinitely, and still arrive at the same answer. Thus for the instance with 100 passengers, the probability of the last one getting to their proper seat is also 1/2, or 50%.
Winner: