Confusing Ride on the Ferris Wheel

There are 10 two-seater cars attached to a Ferris wheel. The Ferris wheel turns so that one car rotates through the exit platform every minute. 

The wheel began operation at 10 in the morning and shut down 30 minutes later. 

What's the maximum number of people that could have taken a ride on the wheel in that time period?

Here is count of people enjoying the ride!

Peoples Enjoying Ferris Wheel Ride

What was the puzzle?

Let's simplify the situation by naming 10 cars as A, B, C, D, E, F, G, H, I, J.

Suppose the Car A is at the exit platform at 10:00 AM. Obviously it can be 'loaded' with 2 peoples say A1-A2.

At 10:01 AM, the Car B will be at the exit platform which can be 'loaded' with 2 more peoples say B1-B2.

So continuing this way, at 10:09 AM the car J will be loaded with peoples J1-J2.

At 10:10 AM, the Car A will be again at the exit platform and now A1-A2 can get off the board to allow 2 more new peoples A3-A4 to get on the board.

At 10:11 AM, the Car B will be at the exit platform where B3-B4 will replace B1-B2. 

Continuing this way, at 10:19 AM the J1-J2 in the Car J will be replaced by J3-J4.

So far, 10 x 4 = 40 different peoples would have enjoyed the ride. 

It's clear that every car takes 10 minutes to be at the exit platform after once it goes through it. That's how Car A is at the exit platform at 10:10 AM, 10:20 AM and it can be at 10:30 AM as well when the wheel is supposed to be shut down.

Now, since wheel needs to be shut down 10:30 AM, emptying all the cars must be started except Car A for the reason stated above.

Therefore, at 10:20, peoples A5-A6 replace A3-A4. Thereafter, every car should be emptied. So, far 40 + 2 = 42 different people have boarded on the cars of the wheel.

So, at 10:21 AM, the Car B is emptied, at 10:22 AM, the Car C should be emptied and so on.

At 10:29 AM, J3-J4 of Car J get out of the car and finally at 10:30 AM, A5-A6 get out of the Car A. 

Now, the ferris wheel can be shut down with no one stuck at any of cars.

To conclude, 42 different peoples can enjoy the ride in given time frame.

Count The Number of Arrangements

There are 10 parking spaces numbered from 101 to 110. At least one car is parked in these slots. If cars can be parked only at the consecutively numbered parking slots, how many such arrangements can be made. 

Consider that only one car can be parked in one parking slot and all cars are identical.

Here is the possible count! 

Possible Number of Arrangements

What was the puzzle?

Suppose there is only 1 car that is to be parked in 1 of the 10 slots. 

Number of possible arrangement = 10C1 = 10!/1!9 = 10.

That is 1 car can be parked in 10 slots in 10 number of ways.

Now, let's suppose that there are 2 cars that to be parked in 2 of the 10 parking slots. But the condition is that they need to be parked in consecutive slots. 

Among 10 slots for there are 9 possible consecutive slots for 2 cars. That is, 2 cars can be parked in consecutive slots in 9C1 = 9 number of ways. It's like placing 1 group of cars (having 2 cars) in 9 possible slots.

Similarly, in 10 parking slots for parking 3 cars there are 8 possible consecutive slots. Hence, there are 8 such arrangements are possible.

And so on for the rest number of cars.

Hence, there are total 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 55 such arrangements are possible.  

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.   
 

Complex Time, Speed & Distance Maths

There is a circular race-track of diameter 1 km. Two cars A and B are standing on the track diametrically opposite to each other. They are both facing in the clockwise direction. At t=0, both cars start moving at a constant acceleration of 0.1 m/s/s (initial velocity zero). Since both of them are moving at same speed and acceleration and clockwise direction, they will always remain diametrically opposite to each other throughout their motion.

At the center of the race-track there is a bug. At t=0, the bug starts to fly towards car A. When it reaches car A, it turn around and starts moving towards car B. When it reaches B, it again turns back and starts moving towards car A. It keeps repeating the entire cycle. The speed of the bug is 1 m/s throughout.

After 1 hour, all 3 bodies stop moving. What is the total distance traveled by the bug?

Find out complex calculations here!

Simple Solution of Complex Problem

Here is that complex looking problem! 

Everything built, written or designed in the given problem is to distract you from basic physics formula.

Speed = Distance / Time

Hence,

Distance = Speed x Time 

All the details given except speed of bug & time for which it traveled are there to confuse you. Speed of bug is 1 m/s & it traveled for 1 hour = 3600 seconds.

Distance = 1 x 3600 = 3600 m

So the total distance traveled by the bug is 3600 m.