A World Class Swimmer's Puzzle

A world-class swimmer can swim at twice the speed of the prevailing tide.

She swims out to a buoy and back again, taking four minutes to make the round-trip.

How long would it take her to make the identical swim in still water?

Click here for the answer!

Solution of A World Class Swimmer's Puzzle

What was the puzzle?

 Let C be the speed of water current then the speed of a world class swimmer will be 2C.

When she swims out to a buoy located at a distance say D and back again, the time needed is - 

D/(C+2C) + D/(2C-C) = 4 minutes

D/3C + D/C = 4

Multiplying both the sides by 3, 

D/C + 3D/C = 12

4D/C = 12

D/C = 3  

Now when she swims out to a buoy and back to shore again in still water, she needs only time -

D/2C + D/2C 

Since D/C = 3, then D/2C = 3/2

Hence, 
 
D/2C + D/2C  = 3/2 + 3/2 = 3.

That is, she needs only 3 minutes to swim out to a buoy and back to shore again in still water.

 
 

Puzzle : The Fake Coin Challenge!

There are 101 coins out of which 1 is fake, the fake coin is identical to a genuine coin but differs in weight. Using weight balance only twice how can we determine whether the fake coin is heavier or lighter than a genuine coin?

 

 

This is how to do it! 

Fake Vs Genuine Coin Weigh Comparison

This was the challenge!

Remember we are asked to determine whether the fake coin is lighter or heavier when compared with the genuine coin and not to identify the fake coin itself.

Keep aside any one coin. Divide remaining 100 coins into 2 groups of 50 coins each. Put these 2 groups on 2 pans of the balance.

1. If they weigh equal the the coin that is kept aside is fake. Weigh it against any genuine among 100 coins to know whether fake coin is lighter or heavier than genuine.

2. If they are not equal then that means the fake coin either made one side heavier or the other side lighter.

3. Take the heavier group of 50 coins for the next test. Divide them into 2 groups of 25 coins each. 

4. Put 25 - 25 coins on weighing balance. If they weigh equal then that means no fake coin among them which also means the fake coin was in the other group of 50 coins which was lighter in the first weighing. 

Hence, the fake coin is lighter present in the other group of 50 coins making the group slightly lighter compared to group of 50 genuine coins.

4.1. And if the result of weighing 25 - 25 coins is unequal then it's clear that the fake coin is among these 50 coins. Also, it must be heavier making this group to weigh more than the other group of 50 genuine coins in the first weighing.

This way, we can determine whether the fake coin is heavier or lighter than genuine one using the weighing balance only twice.

The Poisonous Glass?

You are given 4 identical glasses, completely filled with transparent, odorless liquids. Three of the liquids are pure water, and the fourth is poison, which is slightly heavier. If the water glasses weigh 250 grams each, and the poisoned glass weighs 260 grams, how can you figure out which one is which, using a measuring scale just once?

Here is the process to identify the poisonous glass!

Identifying The Poisonous Glass!

What was the challenge?

Let's number the 4 glasses as 1,2,3 and 4.

Here are steps to identify the glass with poisonous liquid in it. 

1. Empty the glass 1 into another empty glass.

2. Take about 1/4th liquid from glass 2 and pour it into emptied glass 1.

3. Take 3/4th of liquid from glass 3 and pout it into glass 1.

Now, glass 1 has 1/4th of liquid from glass 2 and 3/4th from glass 3.

4. Now put glass 1 and glass 4 on the measuring scale.

    4.1 - If it weighs exactly 500 gm then it suggests that glass 1 had poison.
  
    4.2 - If it weighs between 500 - 505 grams (precisely 502.5 if exactly 1/4th of liquid was taken from glass 2 to glass 1) then obviously glass 2 is with poisonous liquid.

    4.3 - If it weighs between 505 - 510 grams (precisely 507.5 if exactly 3/4th of liquid was taken from glass 3 to glass 1) then obviously glass 3 is with poisonous liquid.

    4.4 - If it weighs exactly 510 gm then it suggests that glass 4 has poison.

 

Case of 3 Identical Notebooks

Three people all set down their identical notebooks on a table. On the way out, they each randomly pick up one of the notebooks. What is the probability that none of the three people pick up the notebook that they started with?

That's correct probability!

Probability in Case of 3 Identical Notebooks

What was the case?

Let's name peoples as Person - 1, Person - 2, Person - 3 and their notebooks as Notebook - 1, Notebook - 2 and Notebook 3 respectively.

Now there can be 6 ways 3 notebooks can be distributed among 3 persons like below.

(Here, for convenience, 3 different colors are assigned to the notebooks of 3 persons.)

As we can see, there are only 2 cases, where the each of person not getting own notebook. In rest of cases, at least 1 person got own notebook & notebooks of others are shuffled between 2.

So the probability that none of the three people pick up the notebook that they started with is 2/6 = 1/3. 

Which Coin is Counterfeit?

You are given eight coins and told that one of them is counterfeit. The counterfeit one is slightly heavier than the other seven. Otherwise, the coins look identical. Using a simple balance scale, how can you determine which coin is counterfeit using the scale only twice?

Identify The Fake Coin

  
This is how it can be found! 

Source 

To Identify The Counterfiet Coin!

Why it was asked to? 

1. Make a group of 3 coins each. Weigh these groups against each other.

2. If both groups weigh equal then weigh remaining 2 against each other. Heavier coin would be easily identified.

3. If one of group is heavier, then take out coins of this group.The counterfeit coin is among them.

4. Now weigh 2 of 3 coins of this group against each other.

5. If both are equal then obviously third one is counterfeit.

6. If one of them is heavier then obviously that one is counterfeit.