#1 - 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?
(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.

#2 - Missing Piece Puzzle

Below the four parts have been reorganized. The four partitions are exactly the same in both arrangements. Why is there a hole?

Missing Piece Puzzle

he gradient of the teal hypotenuse is different than the gradient of the red hypotenuse.

#3 - Building Puzzle

A certain street contains 100 buildings. They are numbered from 1 to 100. How many 9's are used in these numbers?

Just count the nines in the numbers: 9, 19, 29, 39, 49, 59, 69, 79, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99.
Note: 99 contains TWO nines!

#4 - Wise Cipher Riddle

yyyy u r, yyyyy u b, i c u r y+y 4 ? by sharks. How would you survive?

Wise you are, wise you be, I see you are too wise for: ME!

#5 - Challenging Mind puzzles

You are the ruler of a medieval empire and you are about to have a celebration tomorrow. The celebration is the most important party you have ever hosted. You've got 1000 bottles of wine you were planning to open for the celebration, but you find out that one of them is poisoned.

The poison exhibits no symptoms until death. Death occurs within ten to twenty hours after consuming even the minutest amount of poison.

You have over a thousand slaves at your disposal and just under 24 hours to determine which single bottle is poisoned.

You have a handful of prisoners about to be executed, and it would mar your celebration to have anyone else killed.

What is the smallest number of prisoners you must have to drink from the bottles to be absolutely sure to find the poisoned bottle within 24 hours?

10 prisoners must sample the wine. Bonus points if you worked out a way to ensure than no more than 8 prisoners die.

Number all bottles using binary digits. Assign each prisoner to one of the binary flags. Prisoners must take a sip from each bottle where their binary flag is set.

Here is how you would find one poisoned bottle out of eight total bottles of wine.

Bottle 1 Bottle 2 Bottle 3 Bottle 4 Bottle 5 Bottle 6 Bottle 7 Bottle 8
Prisoner A X X X X
Prisoner B X X X X
Prisoner C X X X X
In the above example, if all prisoners die, bottle 8 is bad. If none die, bottle 1 is bad. If A & B dies, bottle 4 is bad.

With ten people there are 1024 unique combinations so you could test up to 1024 bottles of wine.

Each of the ten prisoners will take a small sip from about 500 bottles. Each sip should take no longer than 30 seconds and should be a very small amount. Small sips not only leave more wine for guests. Small sips also avoid death by alcohol poisoning. As long as each prisoner is administered about a millilitre from each bottle, they will only consume the equivalent of about one bottle of wine each.

Each prisoner will have at least a fifty percent chance of living. There is only one binary combination where all prisoners must sip from the wine. If there are ten prisoners then there are ten more combinations where all but one prisoner must sip from the wine. By avoiding these two types of combinations you can ensure no more than 8 prisoners die.

One viewer felt that this solution was in flagrant contempt of restaurant etiquette. The emperor paid for this wine, so there should be no need to prove to the guests that wine is the same as the label. I am not even sure if ancient wine even came with labels affixed. However, it is true that after leaving the wine open for a day, that this medieval wine will taste more like vinegar than it ever did. C'est la vie.

#6 - Cipher Age Puzzle

If Susan is 10, Arabella is 20, and Jim and Neal are both 5, but Richard is 10, how much is Jennifer by the same system?

Cipher Age Puzzle

Jennifer is fifteen, in a system that awards five for each syllable.

#7 - Hardest Math Riddle

Create a number using only the digits 4,4,3,3,2,2,1 and 1. So i can only be eight digits. You have to make sure the ones are separated by one digit, the twos are separated by two digits the threes are separated with three digits and the fours are separated by four digits

Hardest Math Riddle


#8 - Puzzle Boat

At the local model boat club, four friends were talking about their boats.

There were a total of eight boats, two in each colour, red, green, blue and yellow. Each friend owned two boats. No friend had two boats of the same colour.

Alan didn't have a yellow boat. Brian didn't have a red boat, but did have a green one. One of the friends had a yellow boat and a blue boat and another friend had a green boat and a blue boat. Charles had a yellow boat. Darren had a blue boat, but didn't have a green one.

Can you work out which friend had which coloured boats?

Alan had a red boat and a green boat.
Brian had a green boat and a blue boat.
Charles had a yellow boat and a red boat.
Darren had a blue boat and a yellow boat.

#9 - Weighing Balance Puzzle

You can place weights on both side of weighing balance and you need to measure all weights between 1 and 1000. For example if you have weights 1 and 3,now you can measure 1,3 and 4 like earlier case, and also you can measure 2,by placing 3 on one side and 1 on the side which contain the substance to be weighed. So question again is how many minimum weights and of what denominations you need to measure all weights from 1kg to 1000kg.

Weighing Balance Puzzle

For this answer is 3^0, 3^1, 3^2... That is 1,3,9,27,81,243 and 729.

#10 - Pyramid Number Puzzle

The below is a number puzzle. It should be read left to right, top to bottom.
1 1
2 1
1 2 1 1
1 1 1 2 2 1
? ? ? ? ? ?
? ? ? ? ? ? ? ?
Question 1: What is the next two rows of numbers?
Question 2: How was this reached?

Popular Number Sequence Puzzle
3 1 2 2 1 1
1 3 1 1 2 2 2 1

Line 1 is 'Two ones' (2 1)
Line 2 then becomes 'One two, and one one' (1 2 1 1)
Line 3 therefore is 'One one, one two and two ones' (1 1 1 2 2 1)
Line 4 is 'Three ones, two twos and one one' (3 1 2 2 1 1)
Line 5 is 'One three, one one, two twos and two ones' (1 3 1 1 2 2 1 1)