Maze Challenge For a Rat?

A rat is placed at the beginning of a maze and must make it to the end. There are four paths at the start that he has an equal chance of taking: path A takes 5 minutes and leads to the end, path B takes 8 minutes and leads to the start, path C takes 3 minutes and leads to the end, and path D takes 2 minutes and leads to the start.

What is the expected amount of time it will take for the rat to finish the maze?

This could be the average time that rat needed!
 

A Rat Finishing off The Maze!

The challenge ahead of rat?

For rat, there are 2 paths viz A (5 minutes) and C (3 minutes) leading to the end while paths B (8 minutes) and D (2 minutes) lead to the start again.

Since, there are 4 paths & each having equal chance of being chosen by rat, there is 1/4 th chance for each path for to be chosen by rat.

Let's assume T be the time needed for rat to finish the maze. 

But if rat selects path B or D then rat need T more time again as these paths lead to the start of the maze again.

Hence,

T = (1/4) x A + (1/4) x B + (1/4) x C + (1/4) x D

T =  (1/4) x 5 + (1/4) x (8 + T) + (1/4) x 3 + (1/4) x (2 + T)

T = (5/4) + (2) + (T/4) + (3/4) + (1/2) + (T/4)

T =  (9/2) + (T/2)

T/2 = 9/2

T = 9

That is rat needs 9 minutes to finish the maze. 

 

A World Class Swimmer's Puzzle

A world-class swimmer can swim at twice the speed of the prevailing tide.

She swims out to a buoy and back again, taking four minutes to make the round-trip.

How long would it take her to make the identical swim in still water?

Click here for the answer!

Solution of A World Class Swimmer's Puzzle

What was the puzzle?

 Let C be the speed of water current then the speed of a world class swimmer will be 2C.

When she swims out to a buoy located at a distance say D and back again, the time needed is - 

D/(C+2C) + D/(2C-C) = 4 minutes

D/3C + D/C = 4

Multiplying both the sides by 3, 

D/C + 3D/C = 12

4D/C = 12

D/C = 3  

Now when she swims out to a buoy and back to shore again in still water, she needs only time -

D/2C + D/2C 

Since D/C = 3, then D/2C = 3/2

Hence, 
 
D/2C + D/2C  = 3/2 + 3/2 = 3.

That is, she needs only 3 minutes to swim out to a buoy and back to shore again in still water.