Finding Horses For Courses

There are 25 horses among which you need to find out the fastest 3 horses. You can conduct race among at most 5 to find out their relative speed. At no point you can find out the actual speed of the horse in a race. Find out how many races are required to get the top 3 horses. 

Click here for logical answer! 

Source 

Races To Find Horses For Courses

What was the task? 

First we need to make 5 groups of 5 horses each. Let's put each group on race & note down rank of each horse & corresponding group. This requires 5 races.

Taking out winner of each group aside. In a next race, winner of each group run & fastest winner is found. Total 6 races are conducted till now.

Now let's name group of fastest winner as A, second fastest winner as B & so on as C, D, E. Each horse would be identified by it's group name & rank as Group.Rank. For example, B.3 means the horse that came 3rd in group B & D.5 is horse that came last in group D.

For next race, we can eliminate few. Let's remind we have to find 3 fastest horses. Now let's logically eliminate few horses. Those horse for which we are sure that at least other 3 are faster than those will be eliminated.

1. Now A.1, B.1 & C.1 are faster than D.1 & E.1. So D.1 & E.1 & respective groups eliminated straightaway.

2. The C.2 (& other members of C) eliminated as A.1, B.1, C.1 are faster ones.

3. The B.3 (& other member of B) eliminated as at least A.1, B.1, B.2 are faster ones.

4. Similarly, A.4 & A.5 have 3 horses ahead as A.1, A.2 & A.3.

5. A.1 is already proved it's fastest among all, so no need to re race by taking it.

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.