### #1 - 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.

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

### #2 - Trick Question

Can you circle exactly four of these numbers such that the total is twelve?

1 6 1
6 1 6
1 6 1
6 1 6

Turn the grid upside down. Circle the numbers 1, 9, 1, 1 When you flip it upside down the 6 becomes 9

### #3 - 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.

### #4 - Building Puzzle

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

20
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!

### #5 - Find next number in series

What are the next two numbers in this sequence?

7, 14, 17, 21, 27, 28, 35, 37, ?, ?

42, 47.

They are the numbers that either contain the digit 7 or are divisible by 7.

### #6 - 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.
(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.

### #7 - Sound Series Pattern

Can you figure out the logic I used to decide the order of the following words: gun, shoe, spree, door, hive, kicks, heaven, gate, line, den

Each word rhymes with its numeric position in the list. (e.g. 'gun' rhymes with 'one', etc.)

### #8 - Missing Number in this Sequence

Can you discover the missing number in this series?

37, 10, 82
29, 11, 47
96, 15, 87
42, ?, 15

The missing number is 6.
The number in the middle of each triple is the same as the digits of either end's number when added together. 3+7=10=8+2 and so on.

### #9 - Simple Simple Google Interview Puzzle

The puzzle is if the shopkeeper can only place the weights in one side of the common balance. For example if shopkeeper has weights 1 and 3 then he can measure 1, 3 and 4 only. Now the question is how many minimum weights and names the weights you will need to measure all weights from 1 to 1000. This is a fairly simple problem and very easy to prove also.  Answer for this puzzle is given below.

This is simply the numbers 2^0,2^1,2^2 ... that is 1,2,4,8,16... So for making 1000 kg we need up to 1, 2, 4, 8, 16, 32, 64, 128, and 512

### #10 - Paradox Probability Puzzle

This is a famous paradox which has caused a great deal of argument and disbelief from many who cannot accept the correct answer. Four balls are placed in a hat. One is white, one is blue and the other two are red. The bag is shaken and someone draws two balls from the hat. He looks at the two balls and announces that at least one of them is red. What are the chances that the other ball he has drawn out is also red?

There are six possible pairings of the two balls withdrawn, RED+RED, RED+WHITE, WHITE+RED, RED+BLUE, BLUE+RED, WHITE+BLUE. We know that the WHITE + BLUE combination has not been drawn. This leaves five possible combinations remaining. Therefore the chances that the RED + RED pairing has been drawn are 1 in 5. Many people cannot accept that the solution is not 1 in 3, and of course it would be, if the balls had been drawn out separately and the color of the first ball announced as red before the second had been drawn out. However, as both balls had been drawn together, and then the color of one of the balls announced, then the above solution, 1 in 5, must be the correct one.