Logical Response By Master of Logic!

What was the test?

Let A, B and C be the names of three logicians and C be the logician who correctly guessed the color of dot on forehead. 

Now, this could be the C's logic behind his correct guess - 

"If I had blue dot on my forehead then A and B must had raised hands after looking red dots on the foreheads of other. In case, what A (or B) would have thought? His logic would be -

    "If C is with the blue dot then B (or A) must have raised hand after noticing 
    red dot on my forehead, hence I must have red dot."

So A (or B) would have successfully guessed color of dot on own forehead easily.

But neither A or B not responding that means I must have red dot on my forehead!"  

The logician who guess it correctly could be either A or B not necessarily be C; here it is assumed C is wisest for the sake of convenience.

Solve The Picure Equation!

Can you find correct number for '?' ?

Here is the solution!

Solution of Picture Equation!

What was the equation?

From picture, it is clear that,

Pair of shoes = 20, single shoe = 10

Man = 5

Goggles = 2

Single Glove = 20

Now, equation in question has man with 2 shoes, 2 gloves and 1 goggles hence his value in equation = 20 + 40 + 5 + 2 = 67

Hence, final equation appears as

10 + 67 x 2 = 144.

 
The answer is 144.

Heavier Vs Lighter Balls

We have two white, two red and two blue balls. For each color, one ball is heavy and the other is light. All heavy balls weigh the same. All light balls weigh the same. How many weighing on a beam balance are necessary to identify the three heavy balls? 

You need only 2 weighings! Click here to know how!

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'.