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Angels Fly

Re: Puzzles/Riddles Time
« Reply #15 on: May 04, 2014, 03:13:30 »
After much discussion in a call Doug came up with the right answer. :Cheer:
 Answer: If the officials thought he was jumping with a hostage, they would never risk giving him a faulty parachute.
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Angels Fly

Re: Puzzles/Riddles Time
« Reply #16 on: May 04, 2014, 03:14:23 »
What word am I

WHAT WORD AM I

I am eight letters long - “12345678”
My 1234 is an atmospheric condition.
My 34567 supports a plant.
My 4567 is to appropriate.
My 45 is a friendly thank-you.
My 678 is a man’s name.
Q: What word am I?
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Offline George

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Re: Puzzles/Riddles Time
« Reply #17 on: May 04, 2014, 03:15:17 »
4 = T
5 = A
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Offline Emma

Re: Puzzles/Riddles Time
« Reply #18 on: May 04, 2014, 03:16:21 »
mistaken
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Angels Fly

Re: Puzzles/Riddles Time
« Reply #19 on: May 04, 2014, 03:17:39 »
lol goooo emma :Charge:
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Angels Fly

Re: Puzzles/Riddles Time
« Reply #20 on: May 04, 2014, 03:19:02 »
Two old friends, Jack and Bill, meet after a long time.

Three kids
Jack: Hey, how are you man?
Bill: Not bad, got married and I have three kids now.
Jack: That’s awesome. How old are they?
Bill: The product of their ages is 72 and the sum of their ages is the same as your birth date.
Jack: Cool… But I still don’t know.
Bill: My eldest kid just started taking piano lessons.
Jack: Oh now I get it.

How old are Bill’s kids?
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Offline Sunshine Kid

Re: Puzzles/Riddles Time
« Reply #21 on: May 04, 2014, 03:23:17 »
3, 3 & 8
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Angels Fly

Re: Puzzles/Riddles Time
« Reply #22 on: May 04, 2014, 03:25:43 »
Solution
3,3,8
Congrats both of  you  :Cheer: :Cheer:
Lets break it down. The product of their ages is 72. So what are the possible choices?

2, 2, 18 sum(2, 2, 18) = 22
2, 4, 9 sum(2, 4, 9) = 15
2, 6, 6 sum(2, 6, 6) = 14
2, 3, 12 sum(2, 3, 12) = 17
3, 4, 6 sum(3, 4, 6) = 13
3, 3, 8 sum(3, 3, 8 ) = 14
1, 8, 9 sum(1,8,9) = 18
1, 3, 24 sum(1, 3, 24) = 28
1, 4, 18 sum(1, 4, 18) = 23
1, 2, 36 sum(1, 2, 36) = 39
1, 6, 12 sum(1, 6, 12) = 19

The sum of their ages is the same as your birth date. That could be anything from 1 to 31 but the fact that Jack was unable to find out the ages, it means there are two or more combinations with the same sum. From the choices above, only two of them are possible now.

2, 6, 6 sum(2, 6, 6) = 14
3, 3, 8 sum(3, 3, 8 ) = 14

Since the eldest kid is taking piano lessons, we can eliminate combination 1 since there are two eldest ones. The answer is 3, 3 and 8.
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Angels Fly

Re: Puzzles/Riddles Time
« Reply #23 on: May 04, 2014, 03:26:57 »
Mr. Black, Mr. Gray, and Mr. White are fighting in a Duel. 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?
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Offline George

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Re: Puzzles/Riddles Time
« Reply #24 on: May 04, 2014, 03:36:52 »
mr black uses his gun as a boomarang and knocks mr gray and white out, then runs far far away to live hapily ever after in a gun, and violence free world with the deers and rabbits...
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Offline Riddle

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Re: Puzzles/Riddles Time
« Reply #25 on: May 04, 2014, 03:54:02 »
He shoots somewhere else not at Mr. Gray or Mr. White. It is then Mr. Grays turn, he would shoot Mr. White as he is more of a threat.
IF Gray kills White then it is Blacks turn and he has a higher chance of survival than before.

IF Gray fails to kill white it is then Whites turn, White would shoot Gray as Gray is more of a threat and would kill Gray as White never misses. It is then Blacks turn again and he would have a much higher chance of survival. BOOM!
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Angels Fly

Re: Puzzles/Riddles Time
« Reply #26 on: May 04, 2014, 03:56:00 »
 :Cheer: :Cheer: :Cheer:Listening to you work that out  in a call was amazing  Doug.  :Cheer: :Cheer: :Cheer:

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.
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Angels Fly

Re: Puzzles/Riddles Time
« Reply #27 on: May 04, 2014, 04:00:18 »
You are given 2 eggs.
 You have access to a 100-storey building.
 Eggs can be very hard or very fragile means it may break if dropped from the first floor or may not even break if dropped from 100 th floor.
Both eggs are identical.
 You need to figure out the highest floor of a 100-storey building an egg can be dropped without breaking.
 Now the question is how many drops you need to make. You are allowed to break 2 eggs in the process
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Angels Fly

Re: Puzzles/Riddles Time
« Reply #28 on: May 05, 2014, 06:09:51 »
 :Cheer: :Cheer: :Cheer:Well done Doug for getting the answer to  the egg Riddle :Cheer: :Cheer: :Cheer:

The answer is: 14
Let x be the answer we want, the number of drops required.

So if the first egg breaks maximum we can have x-1 drops and so we must always put the first egg from height x. So we have determined that for a given x we must drop the first ball from x height. And now if the first drop of the first egg doesn’t breaks we can have x-2 drops for the second egg if the first egg breaks in the second drop.

Taking an example, lets say 16 is my answer. That I need 16 drops to find out the answer. Lets see whether we can find out the height in 16 drops. First we drop from height 16,and if it breaks we try all floors from 1 to 15.If the egg don’t break then we have left 15 drops, so we will drop it from 16+15+1 =32nd floor. The reason being if it breaks at 32nd floor we can try all the floors from 17 to 31 in 14 drops (total of 16 drops). Now if it did not break then we have left 13 drops. and we can figure out whether we can find out whether we can figure out the floor in 16 drops.

Lets take the case with 16 as the answer

1 + 15 16 if breaks at 16 checks from 1 to 15 in 15 drops
1 + 14 31 if breaks at 31 checks from 17 to 30 in 14 drops
1 + 13 45 .....
1 + 12 58
1 + 11 70
1 + 10 81
1 + 9 91
1 + 8 100 We can easily do in the end as we have enough drops to accomplish the task


Now finding out the optimal one we can see that we could have done it in either 15 or 14 drops only but how can we find the optimal one. From the above table we can see that the optimal one will be needing 0 linear trials in the last step.

So we could write it as
(1+p) + (1+(p-1))+ (1+(p-2)) + .........+ (1+0) >= 100.
Let 1+p=q which is the answer we are looking for
q (q+1)/2 >=100
Solving for 100 you get q=14.
So the answer is: 14
Drop first orb from floors 14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99, 100... (i.e. move up 14 then 13, then 12 floors, etc) until it breaks (or doesn't at 100)

 
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Angels Fly

Re: Puzzles/Riddles Time
« Reply #29 on: May 05, 2014, 06:14:13 »
On Bag shot Island, there is an airport. The airport is the home base of an unlimited number of identical aeroplane. Each aeroplane 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 aeroplane must depart and return to the same airport, and that is the only airport they can land and refuel on ground.
(b) Each aeroplane  must have enough fuel to return to airport.
(c) The time and fuel consumption of refuelling can be ignored. (so we can also assume that one aeroplane can refuel more than one aeroplane in air at the same time.)
(d) The amount of fuel aeroplane carrying can be zero as long as the other aeroplane is refuelling these aeroplane . What is the fewest number of  aeroplane and number of tanks of fuel needed to accomplish this work? (we only need  aeroplane to go around the world)
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