Puzzle 21: Complete one round.

Puzzle 21: There is an round island with one Air base. You have supply of "n" air planes. You need to make one plane should complete the circle boundary with the following constraints:

1. One plane can travel only half of the island boundary with one full tank.
2. There is no way a plane can land in between.
3. You can transfer the fuel from one plane to other in air only, but make sure that plane which is transferring the fuel from one plane to other should have sufficient fuel to come back to the base.
4. Planes can fly in clockwise or anti-clockwise direction.
5. Make sure no plane crash due to any reason.

Solution: This can be solved by using 6 planes as follows:

Now as per given data a plane can try with one full tank from A to B max.

Refer the following table to get idea how things work with multiple planes.

To make it easy, let's consider all planes need 4 units of fuel to full the tank.

So fuel required to reach to B from A (Clockwise direction) will be 1 Unit.

So from the above table, it's clear that by using 3 planes we can reach from A to G in clockwise direction.

Similarly by applying symmetry we can make one more plane to reach G in anti-clockwise direction  by using 3 planes used in the same manner.

So the full tank plane, let's say Plane 6 reach with full tank at position G then plane 6 will transfer 2 units fuel to Plane 1 and both can come back to base safely.

Puzzle 20: Survivors Issue

Puzzle 20: A group of Englishmen are travelling on a German plane, piloted by an Norwegian, bound for Finland. If the plane was to crash in Sweden, where would the survivors be buried?
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Solution: You do not usually bury survivors.

Puzzle 19: 1 sheep for 3 animals

Puzzle 19: A lion can consume one sheep in two hours, a wolf can consume three sheep in eight hours and a hyena can consume one sheep in eight hours. If all of the animals were in a cage and were given a sheep, how long would it take them to consume it?

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Solution: 1 hour.
Explanation: A lion consumes 1/2 a sheep per hour, a wolf consumes 3/8 of a sheep per hour, and a hyene consumes 1/8 of a sheep per hour. So in one hour we have 1/2 + 3/8 + 1/8 = 1 sheep in 1 hour.

Puzzle 18: Misfortune Clock

Puzzle 18: You have the misfortune to own an unreliable clock. This one loses exactly 20 minutes every hour. It is now showing 4:00am and you know that is was correct at midnight, when you set it. The clock stopped 4 hours ago, what is the correct time now?

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Solution: 10 am

Explanation: 10:00am: Since the clock is losing 20 minutes every hour, for every real hour that has passed, the clock will only show 40 minutes. Since the clock shows 4:00am, we know that 240 clock minutes have passed. This therefore equals 360 real minutes and hence 6 hours. The clock stopped 4 hour ago and the time must now be 10:00am.

Puzzle 17: Top 3 among 25 horses

Puzzle 17: You have 25 horses. When they race, each horse runs at a different, constant pace. A horse will always run at the same pace no matter how many times it races.

You want to figure out which are your 3 fastest horses. You are allowed to race at most 5 horses against each other at a time. You don't have a stopwatch so all you can learn from each race is which order the horses finish in.

What is the least number of races you can conduct to figure out which 3 horses are fastest?

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Solution: You need to conduct 7 races.
Explanation:

First, separate the horses into 5 groups of 5 horses each, and race the horses in each of these groups. Let's call these groups A, B, C, D and E, and within each group let's label them in the order they finished. So for example, in group A, A1 finished 1st, A2 finished 2nd, A3 finished 3rd, and so on.

We can rule out the bottom two finishers in each race (A4 and A5, B4 and B5, C4 and C5, D4 and D5, and E4 and E5), since we know of at least 3 horses that are faster than them (specifically, the horses that beat them in their respective races).

This table shows our remaining horses:

A1	B1	C1	D1	E1
A2	B2	C2	D2	E2
A3	B3	C3	D3	E3

For our 6th race, let's race the top finishers in each group: A1, B1, C1, D1 and E1. Let's assume that the order of finishers is: A1, B1, C1, D1, E1 (so A1 finished first, E1 finished last).

We now know that horse D1 cannot be in the top 3, because it is slower than C1, B1 and A1 (it lost to them in the 6th race). Thus, D2 and D3 can also not be in the to 3 (since they are slower than D1).

Similarly, E1, E2 and E3 cannot be in the top 3 because they are all slower than D1 (which we already know isn't in the top 3).

Let's look at our updated table, having removed these horses that can't be in the top 3:

A1	B1	C1
A2	B2	C2
A3	B3	C3

We can actually rule out a few more horses. C2 and C3 cannot be in the top 3 because they are both slower than C1 (and thus are also slower than B1 and A1). And B3 also can't be in the top 3 because it is slower than B2 and B1 (and thus is also slower than A1). So let's further update our table:

A1	B1	C1
A2	B2
A3

We actually already know that A1 is our fastest horse (since it directly or indirectly beat all the remaining horses). So now we just need to find the other two fastest horses out of A2, A3, B1, B2 and C1. So for our 7th race, we simply race these 5 horses, and the top two finishers, plus A1, are our 3 fastest horses.

Puzzle 16: 10 Cigarette Butts

Puzzle 16: Bruce is an inmate at a large prison, and like most of the other prisoners, he smokes cigarettes. During his time in the prison, Bruce finds that if he has 3 cigarette butts, he can cram them together and turn them into 1 full cigarette. Whenever he smokes a cigarette, it turns into a cigarette butt.

One day, Bruce is in his cell talking to one of his cellmates, Steve.

"I really want to smoke 5 cigarettes today, but all I have are these 10 cigarette butts," Bruce tells Steve. "I'm not sure that will be enough."

"Why don't you borrow some of Tom's cigarette butts?" asks Steve, pointing over to a small pile of cigarette butts on the bed of their third cellmate, Tom, who is out for the day on a community service project.

"I can't," Bruce says. "Tom always counts exactly how many cigarette butts are in his pile, and he'd probably kill me if he noticed that I had taken any."

However, after thinking for a while, Bruce figures out a way that he can smoke 5 cigarettes without angering Tom. What is his plan?

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Solution: Bruce takes 9 of his 10 cigarette butts and turns them into 3 cigarettes total (remember, 3 cigarette butts can be turned into 1 cigarette). He smokes all three of these, and now he has 4 cigarette butts.
He then turns 3 of the 4 cigarette butts into another cigarette and smokes it. He has now smoked 4 cigarettes and has 2 cigarette butts.
For the final step, he goes and borrows one of Tom's cigarette butts. With this cigarette butt plus the 2 he already has, he is able to make his 5th cigarette to smoke. After smoking it, he is left with 1 cigarette butt, which he puts back in Tom's pile so that Tom won't find anything missing.

Puzzle 15: Racing position.

Puzzle 15: In the final stretch of a road race, you pass the 2nd-place runner right before crossing the finish line. What place do you finish in?
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Solution: You finish in second place.

Explanation : You would have had to pass the first place racer to have finished in first place.