### #1 - River Riddle

Four people need to cross a dark river at night.They have only one torch and the river is too risky to cross without the tourch. if all people cross simultanoesly then torch light wont be sufficient.Speed of each person of crossing the river is different.cross time for each person is 1 min, 2 mins, 7 mins and 10 mins. What is the shortest time needed for all four of them to cross the river ?

The initial solution most people will think of is to use the fastest person as an usher to guide everyone across. How long would that take? 10 + 1 + 7 + 1 + 2 = 21 mins. Is that it? No. That would make this question too simple even as a warm up question.

Let's brainstorm a little further. To reduce the amount of time, we should find a way for 10 and 7 to go together. If they cross together, then we need one of them to come back to get the others. That would not be ideal. How do we get around that? Maybe we can have 1 waiting on the other side to bring the torch back. Ahaa, we are getting closer. The fastest way to get 1 across and be back is to use 2 to usher 1 across. So let's put all this together.

1 and 2 go cross
2 comes back
7 and 10 go across
1 comes back
1 and 2 go across (done)

Total time = 2 + 2 + 10 + 1 + 2 = 17 mins

### #2 - Famous Probability puzzle SHOOT

Mr. Black, Mr. Gray, and Mr. White are fighting in a truel. They each get a gun and take turns shooting at each other until only one person is left. Mr. Black, who hits his shot 1/3 of the time, gets to shoot first. Mr. Gray, who hits his shot 2/3 of the time, gets to shoot next, assuming he is still alive. Mr. White, who hits his shot all the time, shoots next, assuming he is also alive. The cycle repeats. If you are Mr. Black, where should you shoot first for the highest chance of survival?

He should shoot at the ground.

If Mr. Black shoots the ground, it is Mr. Gray's turn. Mr. Gray would rather shoot at Mr. White than Mr. Black, because he is better. If Mr. Gray kills Mr. White, it is just Mr. Black and Mr. Gray left, giving Mr. Black a fair chance of winning. If Mr. Gray does not kill Mr. White, it is Mr. White's turn. He would rather shoot at Mr. Gray and will definitely kill him. Even though it is now Mr. Black against Mr. White, Mr. Black has a better chance of winning than before.

### #3 - Logic Problem

There is a bag which have 21 blue balls and 23 red balls. You also have 22 red balls outside the bag.

Randomly remove two balls from the bag.
* If they are of different colors, put the blue one back in the bag.
* If they are the same color, take them out and put a red marble back in the bag.

Repeat this until only one marble remains in the bag.

What is the color of the sole marble left in the bag ?

The last marble will be blue.

Since balls can only be taken out in pairs and you started off with an odd number of blues there is always going to be one blue left over that you'll keep putting back in the box until it's left on it's own.

### #4 - Logical Clock Problem

Time in a digital clock can be palindromic like 12:12 (same when read forwards or backwards).

Whats the minimum interval between 2 times that are palindromic ?

2 minutes (between 9:59 and 10:01)

### #5 - Aeroplane Hardest Quiz

The puzzle question is : On Bagshot Island, there is an airport. The airport is the homebase of an unlimited number of identical airplanes. Each airplane has a fuel capacity to allow it to fly exactly 1/2 way around the world, along a great circle. The planes have the ability to refuel in flight without loss of speed or spillage of fuel. Though the fuel is unlimited, the island is the only source of fuel.
What is the fewest number of aircraft necessary to get one plane all the way around the world assuming that all of the aircraft must return safely to the airport? How did you get to your answer?
Notes:
(a) Each airplane must depart and return to the same airport, and that is the only airport they can land and refuel on ground.
(b) Each airplane must have enough fuel to return to airport.
(c) The time and fuel consumption of refueling can be ignored. (so we can also assume that one airplane can refuel more than one airplanes in air at the same time.)
(d) The amount of fuel airplanes carrying can be zero as long as the other airplane is refueling these airplanes. What is the fewest number of airplanes and number of tanks of fuel needed to accomplish this work? (we only need airplane to go around the world)

As per the puzzle given ablove The fewest number of aircraft is 3! Imagine 3 aircraft (A, B and C). A is going to fly round the world. All three aircraft start at the same time in the same direction. After 1/6 of the circumference, B passes 1/3 of its fuel to C and returns home, where it is refuelled and starts immediately again to follow A and C.

C continues to fly alongside A until they are 1/4 of the distance around the world. At this point C completely fills the tank of A which is now able to fly to a point 3/4 of the way around the world. C has now only 1/3 of its full fuel capacity left, not enough to get back to the home base. But the first 'auxiliary' aircraft reaches it in time in order to refuel it, and both 'auxiliary' aircraft are the able to return safely to the home base.

Now in the same manner as before both B and C fully refuelled fly towards A. Again B refuels C and returns home to be refuelled. C reaches A at the point where it has flown 3/4 around the world. All 3 aircraft can safely return to the home base, if the refuelling process is applied analogously as for the first phase of the flight.

### #6 - Gold Bar Fewest Cut Puzzle

A worker is to perform work for you for seven straight days. In return for his work, you will pay him 1/7th of a bar of gold per day. The worker requires a daily payment of 1/7th of the bar of gold. What and where are the fewest number of cuts to the bar of gold that will allow you to pay him 1/7th each day?

Just 2
Day One: You make your first cut at the 1/7th mark and give that to the worker.
Day Two: You cut 2/7ths and pay that to the worker and receive the original 1/7th in change.
Day three: You give the worker the 1/7th you received as change on the previous day.
Day four: You give the worker 4/7ths and he returns his 1/7th cut and his 2/7th cut as change.
Day Five: You give the worker back the 1/7th cut of gold.
Day Six: You give the worker the 2/7th cut and receive the 1/7th cut back in change.
Day Seven: You pay the worker his final 1/7th.

### #7 - Difficult Brain Twister

There are 100 bulbs in a room. 100 strangers have been accumulated in the adjacent room. The first one goes and lights up every bulb. The second one goes and switches off all the even numbered bulbs - second, fourth, sixth... and so on. The third one goes and reverses the current position of every third bulb (third, sixth, ninth… and so on.) i.e. if the bulb is lit, he switches it off and if the bulb is off, he switches it on. All the 100 strangers progresses in the similar fashion.

After the last person has done what he wanted, which bulbs will be lit and which ones will be switched off ?

Ponder over the bulb number 56, people will visit it for every divisor it has. So 56 has 1 & 56, 2 & 28, 4 & 14, 7 & 8. So on pass 1, the 1st person will light the bulb; pass 2, 2nd one will switch it off; pass 4, light it; pass 7, switch it off; pass 8, light it; pass 14, switch it off; pass 28, light it; pass 56, switch it off.
For each pair of divisors the bulb will just end up back in its preliminary state. But there are cases in which the pair of divisor has similar number for example bulb number 16. 16 has the divisors 1 & 16, 2 & 8, 4&4. But 4 is recurring because 16 is a perfect square, so you will only visit bulb number 16, on pass 1, 2, 4, 8 and 16… leaving it lit at the end. So only perfect square bulbs will be lit at the end.

### #8 - Famous Logic Bag Puzzle

There are 3 bags and all are labeled incorrectly as follow:
Bag1 : Silver
Bag2 : Gold
Bag3 : Silver and Gold.

You need to label all bags correctly by opening just one bag.

How come ?

It have already mention that all bag have labeled in correctly
let consider
bag A-silver
bag B-gold
bag C-gold and silver
if i open bag C ..i:e gold and silver
then i should get somthing else(silver or gold) ....becouse it have already mention that all bag are labeled in correctly
so the rest 2 bag A and B should also cantain wrong item.
So.... If i assume that bag c ....cantain silver.
Then bag A(labeled silver) should cantain gold
and bag b(labeled gold)should cantain gold and silver
becouse it have mention that all bag are labeled incorrectly

### #9 - Math Riddle

Find three whole, positive numbers that have the same answer when multiplied together as when added together.

1,2, & 3

1 x 2 x 3 = 6 and 1 + 2 + 3 = 6

### #10 - Monday Probability Problem

What is the probability of getting five Monday in a 31-days month ?

If a 31day month starts on a Saturday, Sunday or Monday it will have five Monday, if not it will have 4 Monday.
Probability is 3/7.