Cars Around Interesting Circular Track

Around a circular race track are n race cars, each at a different location. At a signal, each car chooses a direction and begins to drive at a constant speed that will take it around the course in 1 hour. When two cars meet, both reverse direction without loss of speed. Show that at some future moment all the cars will be at their original positions.

Here is that proof!


Source 

Analysing Interesting Circular Race Track

What was the interesting fact about?

Just imagine that each car carries a flag on it and on meeting pass on that flag to the next car. Obviously, this flag will move at the constant speed around the track as cars carrying it are also moving at the constant speed. So, the flag will be back at the original position after 1 hour.

Let's assume there are only 2 cars on the track at diagonally opposite points as shown below. 

After 15 minutes, on meeting with Car 2, Car 1 will pass on the flag & both will reverse their own direction. 30 minutes later (i.e. 45 minutes after start) both cars again meet each other and Car 2 will pass on flag back to Car 1 & directions are reversed again. Again in another 15 minutes (i.e. after 1 hour from start), both cars are back at the original positions. 

Now, let's suppose that there are 4 cars on the track positioned as below.

The above image shows how cars will be positioned after different points of time & how they reverse direction after meeting.

Again, all are back to the original position after 1 hour including the flag position. One more thing to note that the orders in which cars are never changes. It remains as 1-2-3-4 clockwise. 

To conclude, for 'n' number of cars, at some point of time all the cars will be in original positions in future.   
 

"Are you holding true or fake coin?"

You have 101 coins, and you know that 50 of them are counterfeit. Every true coin has the same weight, an unknown integer, and every false coin has the same weight,which differs from that of a true coin by 1 gram. You also have a two-pan pointer scale that will show you the difference in weight between the contents of each pan. You choose one coin. 

Can you tell in a single weighing whether it’s true or false?

Well, this trick will help you to identify that coin! 

Knowing The Truth of the Coin in Hand!

What was the task given?

Yes, you can tell that whether the coin is true or false with single weighting.Just divide 100 coins into 2 groups of 50 coins each & put into 2 pans of weighing balance.

Let's assume true coin weighs 1 gram (or 2 gram) & fake coin weighs 2 gram (or 1 gram). Remember, if the sum of 2 integers is even then difference between two is bound to be even. And if the sum of those is odd then difference between them has to be odd.

CASE 1 :

If the coin that you are holding is true then the total weight on the balance will be
50 + (50x2) = 50 + 100 = 150  (or 50x2 + 50 = 150). So, the total sum of weights in 2 pans is even, hence difference between them has to be even. For example, if those 150 grams are distributed as 80 vs 70 then difference between them is 10 which is even.


CASE 2 :

If the coin you are holding is fake then the total weight on the balance will be
51 + (49x2) = 51 + 98 = 149 (or 51x2 + 49x1 = 153).

Here, total is odd hence the difference must be odd too. For example, if above 149 grams are distributed as 90 vs 59 then pointer of balance will point at 31 which is odd.

Conclusion:

In short, you have to notice the difference between 2 weights on the pans. If it's even then the coin you are holding is true and if difference is odd then you are holding a fake coin.
 

Effect of Average Speed on Time

If a car had increased its average speed for a 210 mile journey by 5 mph, the journey would have been completed in one hour less. What was the original speed of the car for the journey?

Here is the calculation of averages speed!

Calculation of the Original Speed!

What was the question?

Let S1 be the original speed and S2 be the modified speed and T1 be the time taken with speed S1 and T2 be the time taken with speed S2.

As per given data,

T1 - T2 = 1 hr.

D/S1 - D/S2 = 1 hr.

Here, D = 210 miles, S2 = S1 + 5,

210/S1 - 210/(S1+5) = 1

210(S1+5) - 210s = 1S1(S1+5)

S1^2 + 5S1 - 1050 = 0

(S1-30)(S1+35) = 0 

S1 = 30 or S1 = -35.

Since speed can't be negative, S1 = 30 mph.

Hence, the original speed is 30 mph and average speed is 30 + 5 = 35 mph.

 
With the original speed it would have taken 210/30 = 7 hours but with average speed it took only 210/35 = 6 hours saving 1 hour of time.