Amazon Interview Puzzles Questions With Answers

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

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.

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.

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

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.

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.

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.

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

1,2, & 3

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

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.