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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. 

Identifying The Heavier/Lighter Balls


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.
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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.  

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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.
    
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     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. 

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Case 3 : Red + White < Blue + White.

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

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Mathematical Talk Between Horse And Camel

A horse and a camel were carrying boxes on their backs. The horse started complaining to the camel that his load is too heavy.

The camel replied 'Why are you complaining? If you gave me one of your boxes I would have double what you have and if I give you one of my boxes we two would have an even load.'


How many boxes do each of the animal (horse & camel) is carrying ?


Camel's clues about loads - Maths Puzzle




Find here the load on each of them! 
 

Suggestion From Mathematical Talk!


What was the talk?

Let's assume C be the number of boxes that camel is carrying and H be that being carried by horse.

As per first part of camel's statement i.e. if you gave me one of your boxes I would have double what you have  

C + 1 = 2 (H - 1) 

C + 1 = 2H -2

C = 2H - 3 ........(1)

Now in second part (i.e.if I give you one of my boxes we two would have an even load) of camel's statement suggests,

C - 1 = H + 1 


C = H + 2 


Putting (1) in above,

2H - 3 = H + 2

H = 5

Again putting this value in (1) gives,

C = 2*5 - 3 = 10 - 3 = 7 

C = 7. 

Horse is carrying 5 boxes and the camel is carrying 7 boxes.  

Camel's clues about loads - Maths Puzzle

 

Journey From Top To Ground

Galileo dropped balls of various weights from the top of the Leaning Tower of Pisa to refute an Aristotelian belief that heavier objects fall faster than lighter objects. 

If the balls were dropped from a height of 54 meters, how long did it take for the balls to hit the ground?

How does gravity affect on objects of different weights?

Click here to know the answer! 

Time Needed To Reach The Ground


What was the problem? 

To solve the problem, we need to remember what we have learned in our early days of school. The gravity of the earth accelerates the falling object at the rate of 9.8 meter per second square. We know, the kinematic equation,

s = ut + (1/2) a t^2   .......(1)

where,

s = distance covered,
u = initial velocity,
a = acceleration,
t = time taken to travel distance s.

In this case, initial velocity must be 0 as it is dropped from height 54m. Again height here is the distance covered by the object. And the acceleration in this case is nothing but the acceleration due to gravity which is 9.8 m/s^2.

So putting s = 54 m, u = 0 m/s, a = 9.8 m/s^2 in equation (1) above,

54 = 0 x t + (1/2) x 9.8 x t^2

54 = 4.9 t^2

t^2 = 11.021

t = 3.3 seconds. 

Objects need same time to reach the ground if....

So theoretically both balls should take 3.3 seconds to reach the ground. But resistance due to air will make the difference in time taken by balls to hit the ground. 
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