A) Fill 5 ml gallon ( 5mlGallon - 5, 3mlGallon - 0)
B) Transfer to 3 ml gallon (5mlGallon - 2, 3mlGallon - 3)
C) Empty 3 ml gallon ( 5mlGallon - 2, 3mlGallon - 0)
D) Transfer 2 ml from 5 ml gallon to 3 ml gallon (5mlGallon - 0, 3mlGallon - 2)
E) Fill 5 ml gallon(5mlGallon - 5, 3mlGallon - 2)
F) Transfer 1 ml from 5 ml gallon to 3 ml gallons(5mlGallon - 4, 3mlGallon - 3)
Suppose there is a Christmas tree and four angels are sitting on it amidst the other ornaments. Two of them have black halos and two of them have white halos. No body among them can see above their head. Angel A is sitting on the top branch and can see angels B and C sitting below him. B can see C who is sitting in a branch lower than his. Angel D is at the base of the tree and can't be seen due to the branches in between. Also, he can't see anybody as well.
If they are asked to guess the color of their own halo (they dont know that), who do you think will be able to deduce and speak up first with a right answer?
Now there can be two solutions to the given situation because there can be two situations:
Suppose if B and C have same colors, A will know that his color is the other one and he will be able to speak up the first.
Now if B and C do not have same colors, A will stay silent. This will tell B that his and C's colors are different. Thus he will speak up first.
So either A or B will speak up first depending on the situation.
4 + 9 = 1
The trick to solve is to assume the first number as the time in AM and the second number is the number of hours you have to add to it. You'll find the time in PM when you do that.
You have ten boxes and an electronic weighing machine. In those ten boxes, you have chocolates. Each chocolate weigh 20 grams. But in one box the chocolates are defective and each weigh 19 grams exactly.
Now you can weigh in the electronic weighing machine but you can use that machine just once. How will you find out which box has the defected chocolates.
If you are thinking to hold one chocolate from each box in hand and then balancing weight in bare hands, you are thinking all wrong.
Let us begin by labelling boxes as 1, 2, 3 and so on till 10.
Now pick one chocolate from box 1, two chocolates from box 2, three from box 3 and so on. In total, you will have 55 chocolates now. (1 + 2 + 3 + ..... + 10)
The ideal weight of the chocolates should be 55 * 20 = 1100. However, somewhere in there are the defected chocolate/s.
You can judge that clearly by noting down the result of 1100 - total weight of chocolates. If the weight is less than 1 gram, the defected box is box 1, if the weight is less than 2 grams, the defected box is box 2 and so on.
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?
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.