# Puzzle: Of wine and water…

There is a wine glass filled with 1000 spoonful of wine. There is a water beaker with 2000 spoonful of water.

You take a spoon of water from the beaker and put it in the wine glass and mix it thoroughly. Then you take one spoon of the stuff (mostly wine with a little water) from the wine glass and put it in the water beaker and mix it thoroughly.

You do the whole thing over again. And one more time (so total of three times).

Of course increasingly the water beaker is getting more wine and vice versa.

Question is at the end of the third full round, do you have more wine (by volume) in the water beaker or more water (by volume) in the wine glass and by how much?

Here is a tougher version… what if after the above three cycles, you did the water beaker to wine glass one more time and stopped? (You did not do the wine glass to water beaker to complete the cycle; essentially three and a half cycles). Now, do you have more wine (by volume) in the water beaker or more water (by volume) in the wine glass and by how much?

Instead of putting your answer (and how you arrived at it) in the Comments section, write to me directly in messenger.

# A hard puzzle from India!

Tuesday afternoon, I was all by myself. Both dad and mom were sleeping. Not knowing anything better, I went around going thru the books that still adorn the book shelves in my parents’ house. These are books from my school days. They have kept them still. Most of them are disintegrating – but they are still there. After going thru some of the books that I had studied in high school, I chanced upon this book I still remember for very hard math/physics problems. Written by Irodov, it was a book I had bought from Kolkata Book Fair in 1984.

I glanced thru a few pages. Sure enough, they are as hard as I remember them. There was one that caught my attention as a very interesting one. You can see the one in the picture that I have circled with the red line.

This is a very tough problem until you hit the solution and realize how elegant the solution is.

Here is a version of the problem:

There is an equilateral triangle – each of side length “a”. There are three ants at the three corners. Let’s call the ants A,B and C. At the same moment the three ants start moving at exactly the same speed. Ant A keeps moving towards wherever Ant B is. Ant B keeps moving towards wherever Ant C is. And Ant C keeping moving towards wherever Ant A is. The question is : eventually when they meet, how much distance would each ant have traveled?

Now realize that for every ant, the target ant is moving continuously. So, every ant is continually changing its direction. It is not as simple as an ant goes from one vertex to the other. That is what makes the problem hard.

After thinking about the problem for some time – and not getting anywhere – I posed it to my brother when I met him two days later. Together we spent about an hour in our drive to Kalyani from Kolkata discussing the problem. Eventually, we reached Kalyani and asked mom for some paper and pen. Another half an hour later, we did manage to solve the problem. Excitedly, we pulled out the Irodov book again from the shelf to see if the answer it had given matched ours. It did!!!

This was only Chapter 1 of the book and the 12th problem in it!! That chapter alone had another hundred plus problems. And then there were many more chapters!!

Man, I am way past my prime when it comes to ability to solve these kind of problems.

Anyways, see if this excites you to give it a crack.

If you get it, try the same problem with a square of side “a” instead of an equilateral triangle.

# Saturday morning puzzle

You might have heard various variations of this problem before.

You have 2 ostrich eggs in your hand (fairly hard shells) and you are in front of a 50 story building. The basic challenge is to figure out which is the floor from and above which the eggs will break on impact to the ground when you drop them from a window of that floor. Of course, this can be done with only one egg. You start with first floor – if it does not break, go to second floor. And keep doing this till you find the floor where it does break. The minimum number of tries you will need to guarantee finding out that floor is 50. (Basically, in the worst case, you have to keep trying all the way to the top)

But you have 2 eggs. Now, the challenge is to find out the minimum number of tries with which you can guarantee finding out that threshold floor at or above which an egg will break (one try means one dropping of an egg). Needless to say, once an egg breaks, you cannot use it again.

# A puzzle for the weekend

In a remote part of an African jungle once lived a bunch of lions. No other animal ever came there. Except one day, somehow a deer showed up. Of course, the lions immediately wanted to eat it. Now, these lions were very smart and very courteous. Some might even say they took special “pride” in it 🙂

Puns aside, here are the rules of the puzzle:

(*) the lions have an understanding that only one lion – whoever is first to touch – can kill and eat a deer. (when one attacks, nobody else does; that is where the courteous bit comes in)

(*) they also know that this deer has a property that whichever lion attacks and eats it, itself becomes a deer the next day (which can be attacked and killed by other lions then)

(*) Every lion is desperate to eat a deer. They are okay to live the rest of their life as a deer itself – but will not do so if they know they might get killed the next day by the lions.

Question: Will the deer survive?

# A quick puzzle

# Basketball throws: a rather intriguing puzzle

I had run into this problem long time back. I ran into it again last week. Thought will post it. See if you can solve it without using Google to find the answer. If not, then Google it – pretty interesting, huh? (I will post the answer later).

A basketball player keeps track of his throws for a full calendar year. He, of course, misses a few shots and succeeds with a few more. He missed the very first shot of the year. But he ended the year with 83% successful shots. You have to prove that there had to be a point where is success rate was exactly 75%.

Hint: this is not necessarily true for any number – e.g. 60%. But it is always true for 75%.

In fact, can you guess what are the other % (other than 75%) for which this is also true?

# Puzzle time!

Here is one that will have you think about shapes…

Tomorrow is 4th of July and we will be seeing a lot of flags flying proudly all over the USA. Did you know that there is only one country in the world whose flag is not rectangular in shape?

In any case, imagine a country that has a rectangular flag (See my poor rendering in red) with a smaller rectangle of different color inside it somewhere. (See my blue rectangle). Note that the blue rectangle can be anywhere and in any orientation.

Question is, can you cut the flag to make two identical pieces?

# Take a guess…

How many of these statements are true?

1. You can see the Great Wall of China from the moon on clear days (my sixth grade teacher had taught us this)

2. Bulls get excited by red color (which explains why they charge the matador waving the red flag)

3. Napolean was rather diminutive (by French standards those days anyways)

4. Only the royals among the Vikings wore the horned helmets

5. Einstein was weak in math (and failed once) as a school kid.

6. We have 5 senses (sight, sound, taste, touch and smell). (I was taught this pretty early in life).

7. Speaking of senses, the tongue has different parts where we taste different tastes (sweet, salt etc etc)

8. Continuing with our body, artistic folks are more active on the right side of the brain and vice versa for the science and math oriented ones.

9. There is no such thing as a “scientific proof”

10. A steep learning curve implies you will have great difficulty learning it.

If you have Googled, which ones surprised you?

# Sunday morning puzzle

Here is a puzzle we were solving this morning.

Each letter stands for a digit. Those digits are 0,1,2,3,4,5

One twist – The letters below the line do NOT match the letters above the line in terms of the digits they represent. In fact, the digit represented by a letter above the line is separated from the digit represented by the same letter below the line by 1. So if C below the line is 2, then C above the line has to be either 1 or 3.

Can you solve the following subtraction? Send me a message with the answer.