Identified The Order of The Cards

How cards were shuffled? 

Since he puts the leftmost card (from our point of view) to it's opposite location that card must be 2 (that's the smallest one & needs to be first in ascending order).

So the last card (from his point of view) must be 2. We still don't know exact positions of 3,4,5. Now 2 is in first position

Now what he does is that, moves third card from his left (our right) to the last position. Now to be in ascending order the last card must be 5. So the card that he used in this move must be 5.

Since these 2 moves completes the ascending order, rest must be in ascending from left to right already i.e as 3 & 4 but 5 in between them.

Hence the order before he moved cards must be as 3,5,4,2.

  Previous Order of The Cards

 

Time Of Arrival?

One day Rohit decided to walk all the way from city Bangalore to Tumkur. He started exactly at noon. And Samit in city Tumkur decided to walk all the way to Bangalore from Tumkur and he started exactly at 2 P.M. on the same day.

Both met on the Bangalore - Tumkur Road at five past four, and both reached their corresponding destination at exactly the same time.

At what time did we both arrive?

      Beginning of Journey

Find here their arrival time! 

Finding The Time Of Arrival

What was the puzzle? 

Let 'x' km/minute be the speed of Rohit's walk. He started to walk at 12 PM & met Samit at 4:05 PM. That means he has walked for 245 minutes. 

Distance traveled by Rohit = 245x km

Let 'y' km/minute be the speed of Samit's walk. He started to walk at 2 PM & met Samit at 4:05 PM. That means he has walked for 125 minutes.

Distance traveled by Samit = 125y km 

    Time of Meeting

Now after meeting each other they resumed their journey further. That means Rohit continues to Tumkur & covers distance of 125y km at his speed of x km/minute. Time taken by him further to complete the journey is 125y / x minutes (Time = Distance/Speed).

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?

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. 

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. 

Round Table Coin Game

You are sitting with one opponent at an empty, round table. Taking turns, you should place one dollar on the table, in such a way that it touches none of the coins that are already on the table. The first player that is not able to place a dollar on the table has lost. By tossing a coin, it has been decided that you may start.

Which strategy will you follow to make sure you are guaranteed to win?

  
Trick to win this game always! 

Never Loose Round Table Coin Game

What was the game? 

There is little trick with which you will always end on winning side in this Round Table Coin Game. Since you have got first chance to place the coin you should place the coin right at the center of the round table. Now for every next coin placed by opponent you need to place coin in such a way that it 'mirrors' opponent's coin.

Imagine line from the center to opponent's coin. Place your coin exactly opposite to that coin at distance equal to distance between center & opponent's coin. Or imagine a circle (with the center fixed at the round table) with opponent's coin lying on it's border.  And place your coin at diagonally opposite point of point where opponent placed coin on that imaginary circle. (Assume imaginary circle though it's not appearing perfectly in the image above)

In this way for every move of your opponent, you will have 'answer' in form of space for placing the coin. This will continue until last place left on the table with your turn of placing the coin in the end. 

This is how to make sure you always on winning side in this 'Round Table Coin Game'!