Identifying The Heavier/Lighter Balls

What was the task given?

Actually, we need only 2 weighing. 

Weigh 1 red & 1 white ball against 1 blue & 1 white ball. Weighing result between these balls helps us to deduce the relative weights of other balls.
  ----------------------------------------------------------------------------------------------------

Case 1 : 

If this weighing is equal then the weight of red ball and blue ball are different.
That's because the weighing is equal only when there is combination of 
H (Heavy) + L (Light) against L (Light) + H (Heavy). 

Since, one white ball must be heavier than the other white ball placed in other pan, the red & blue balls place along with them must be of 'opposite' weights with respect to the white balls place along with them.

So weigh this red against blue will give us the heavier red (or blue) leaving other red (or blue) as lighter. 

If red (or blue) weighs more in this weighing then the white ball placed with this red (or blue) in first weighing must be lighter than the other white ball placed with blue (or red) ball.  

----------------------------------------------------------------------------------------------------

Case 2 : Red + White > Blue + White.

The white ball in Red + White must be heavier than the white ball in Blue + White.

The reason is if this white was lighter then even heavier red in the Red + White can't weigh more than Blue + White having heavier white. That is H + L can't beat H + H or obviously would equal with H + L!

So we have got the heavier and lighter white balls for sure.

Now, weigh red from Red + White and blue from Blue + White against other red and blue balls left.

     Case 2.1/2.2 : Red + Blue > or < Red + Blue 

     Obviously, red and blue balls in left pan are heavier (or lighter) than those in other.
    
     ------------------------------------------------------------------------------------

     Case 2.3 :  Red + Blue = Red + Blue 

     Obviously, that's because of H + L = L + H.

      But the red is taken from Red + White which was heavier than the Blue + White. 
      Since, L + H (white is heavier as concluded) can't weigh more than H + L
      (other white is lighter as concluded) or L + L, the red in Red + White 
      must be heavier making H + H combination in that pan in first weighing (case 2). 

      So, we got heavier red and lighter blue obviously leaving lighter red 
      and heavier blue in other pan. 

----------------------------------------------------------------------------------------------------

Case 3 : Red + White < Blue + White.

Just replace Red with blue & vice versa in the deduction made in case 2. 

----------------------------------------------------------------------------------------------------

Larger, Smaller or Similar?

Which of the yellow areas is larger?

Here is comparison of areas! 

Both Sharing Equal Area!

Which areas are into comparison?

Actually, areas of both are equal. The diagonal divide the rectangle into 2 halves. So triangle A and A' or B and B' have equal areas.

When diagonal divides the area of main rectangle into 2 halves, area of triangles A (or A') and area of triangle B (or B') are further subtracted from each half to get the areas of the shaded region.

Since equal areas are subtracted from triangles formed by diagonal to get the shaded area, the area of shaded parts are equal. 

 That is from each half area subtracted = A + B = A' + B'.

The Numbered Hats Test!

One teacher decided to test three of his students, Frank, Gary and Henry. The teacher took three hats, wrote on each hat an integer number greater than 0, and put the hats on the heads of the students. Each student could see the numbers written on the hats of the other two students but not the number written on his own hat.

The teacher said that one of the numbers is sum of the other two and started asking the students:

— Frank, do you know the number on your hat?

— No, I don’t.

— Gary, do you know the number on your hat?

— No, I don’t.

— Henry, do you know the number on your hat?

— No, I don’t.

Then the teacher started another round of questioning:

— Frank, do you know the number on your hat?

— No, I don’t.

— Gary, do you know the number on your hat?

— No, I don’t.

— Henry, do you know the number on your hat?

— Yes, it is 144.

What were the numbers which the teacher wrote on the hats?

Here are the other numbers!

Source 

Cracking Down The Numbered Hats Test

What was the test?

Even before the teacher starts asking, the student must have realized 2 facts.

1. In order to identify numbers in this case, the numbers on the hats has to be in proportion i.e. multiples of other(s). Like if one has x then other must have 2x,3x etc.

2. Two hats can't have the same number say x as in that case third student can easily guess the own number as 2x since x-x = 0 is not allowed.

--------------------------------------------------------------------------------------------------

Now, if numbers on the hats were distributed as x, 2x, 3x then the student wearing hat of number 3x would have quickly responded with correct guess. That's because he can see 2 number as x and 2x on others hats and he can conclude his number as x + 2x = 3x since 
2x - x = x is invalid combination (x, x, 2x) where 2 numbers are equal.

Other way, he can think that the student with hat 2x would have guessed own number correctly if I had x on my own hat. Hence, he may conclude that the number on his hat must be 3x.

But in the case, all responded negatively in the first round of questioning. So x, 2x, 3x combination is eliminated after first round.

-------------------------------------------------------------------------------------------------- 

That means it could be x, 3x, 4x combination of numbers on the hats.

In second round of questioning, Henry guessed his number correctly.

If he had seen 3x and 4x on other 2 hats then he wouldn't have been sure with his number whether it is x or 7x.

Similarly, he must not have seen x and 4x as in that case as well he couldn't have concluded whether his number is either 5x or 3x.

But when he sees x and 3x on other hats he can tell that his number must be 4x as 2x (x,2x,3x combination) is eliminated in previous round!

So Henry can conclude that his number must be 4x.

Since, he said his number is 144,

4x = 144

x = 36

3x = 108.

Hence, the numbers are 36, 108, 144.

Divide The Cake Into Equal Parts!

I have just baked a rectangular cake when my wife comes home and barbarically cuts out a piece for herself. The piece she cuts is rectangular, but it’s not in any convenient proportion to the rest of the cake, and its sides aren’t even parallel to the cake’s sides. 

 
I want to divide the remaining cake into two equal-sized halves with a single straight cut. How can I do it?



This is how it can be cut! 

Cutting The Cake Into Equal Parts!

What was the problem? 

Generally, a line drawn through the center of rectangle divides it into 2 equal parts.
Hence, a line drawn through the centers of both rectangles would divide each of them into 2 equal parts as shown below. (To get the center of each rectangle, all we need to do is draw diagonals of both).