Solution:

Light both the ends of the first string and one end of the second string. 30 minutes will have passed when the first string is fully burned, which means 30 minutes have burned off the second string. Light the end of the second string and when it is fully burned, 45 minutes will have passed.

Solution:

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)

Solution:

0%

Explanation:
1) why cant be 1/4 : If the answer is 1/4, then as we know two out of four answer choices is '1/4', the answer has be 1/2.
This is a contradiction, so the answer cannot be 1/4.

2) why cant be 1/2 : If the answer is 1/2 then because answer:"1/2" is 1 out of 4 answer choices, the answer must be 1/4. This is also a contradiction. So the answer cannot be 1/2.

3) why cant be 1 : If the answer is 1 then because answer:"1" is 1 out of 4 answer choices, the answer must be 1/4. Again the same contradiction and therefore answer cannot be 1.

Solution:

17 min

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 minutes. 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 minutes

Solution:

Let us rename the Five gallons basket and Three gallons basket as F and T.
Fill F completely and then pour to fill T completely. By doing this you have 2 gallons remaining in F. Pour out T entirely and then fill it with the remaining liquid from F. Now, T has 2 gallons of liquid.
Fill F fully again. Now pour the liquid to fill T which has just 1 gallons of capacity remaining. By doing this, you are left with the required 4 gallons of liquid in F.

Solution:

50%

Explanation:
When the coins are tossed there can be eight possible outcomes (2*2*2)
HHH
HTT
HTH
HHT
TTT
THH
THT
TTH

If you find out, we have only four favorable outcomes. Also, the events here are independent and thus we can calculate the probability by using these only.

Thus the probability is 50 percent.

Solution:

Assume that P, Q and R are the three friends. Now P is earning X amount, Q is earning Y amount and R is earning Z amount.

First, P will add his salary to a random amount A and will tell the combined sum to Q.
X + A

Now, Q will add his salary and a random amount B to the amount passed by P. He will then pass it to R.
X + A + Y + B

Now R will add his salary and a random amount C to the total amount passed by Q.
X + A + Y + B + Z + C.

Now, R will pass this amount back to P. P will deduct the random amount added by him in the beginning.
X + Y + B + Z + C

He will now pass this amount to Q who will subtract the random amount he added in his turn.
X + Y + Z + C

He will pass this amount to R who will deduct his random amount.
X + Y + Z

Now this is the total sum of their salaries. R will tell it to P and Q and then each of them can find the average of their salaries by dividing the total amount with 3.

Solution:

Playing chess

Game needs two players so Brother-8 is playing chess with Brother-2

Solution:

So that the following plan can be followed, let us number the coins from 1 to 12. For the first weighing let us put on the left pan coins 1,2,3,4 and on the right pan coins 5,6,7,8.

There are two possibilities. Either they balance, or they don't. If they balance, then the different coin is in the group 9,10,11,12. So for our second one possibility is to weigh 9,10,11 against 1,2,3

(1) They balance, in which case you know 12 is the different coin, and you just weigh it against any other to determine whether it is heavy or light.
(2) 9,10,11 is heavy. In this case, you know that the different coin is 9, 10, or 11, and that that coin is heavy. Simply weigh 9 against 10; if they balance, 11 is the heavy coin. If not, the heavier one is the heavy coin.
(3) 9,10,11 is light. Proceed as in the step above, but the coin you're looking for is the light one.

That was the easy part.

What if the first weighing 1,2,3,4 vs 5,6,7,8 does not balance? Then any one of these coins could be the different coin. Now, in order to proceed, we must keep track of which side is heavy for each of the following weighings.

Suppose that 5,6,7,8 is the heavy side. We now weigh 1,5,6 against 2,7,8. If they balance, then the different coin is either 3 or 4. Weigh 4 against 9, a known good coin. If they balance then the different coin is 3, otherwise it is 4. The direction of the tilts can tell us whwther the offending coin is heavier or lighter.

Now, if 1,5,6 vs 2,7,8 does not balance, and 2,7,8 is the heavy side, then either 7 or 8 is a different, heavy coin, or 1 is a different, light coin.

For the third weighing, weigh 7 against 8. Whichever side is heavy is the different coin. If they balance, then 1 is the different coin. Should the weighing of 1,5, 6 vs 2,7,8 show 1,5,6 to be the heavy side, then either 5 or 6 is a different heavy coin or 2 is a light different coin. Weigh 5 against 6. The heavier one is the different coin. If they balance, then 2 is a different light coin.

Solution:

22 times

am
12:00
1:05
2:11
3:16
4:22
5:27
6:33
7:38
8:44
9:49
10:55

pm
12:00
1:05
2:11
3:16
4:22
5:27
6:33
7:38
8:44
9:49
10:55