Just two puzzles?

Question

First Puzzle:- There are 2 Threads which are uneven although their lengths. Each will burn for 60 minutes. You need to calculate 45 minutes using both those threads. You had no watch or anything, just only the threads and the match box.
Second Puzzle:- You are given 8 balls which are common in size, but 1 is defective in them Remaining 7 are equal in weights except that 1. You had find out the defective using the common balance in only 2 iterations. How will u do that.

Nice thinkin' mate....can U gimme ur yahoo ID and come online at yahoo messenger? so tat we can discuss abt this....b/w I can't understand how U calculated the first puzzle....I understand U managed to find tat the half of the first rope gets burnt within the first half hr....but do U determine the 15 mins??? it is slightly confusing!

Answer 1

First puzzle:

The logic of burning half rope = 30 minutes has a flaw because the ropes don't burn at a uniform rate, so here is my solution:

Light both ends of rope #1 and one end of rope #2 simultaneously. In the instant that rope #1 burns completely (30 minutes elapsed time), light the other end of rope #2. The remaining part of rope #2 will burn for 15 more minutes completing the 45 minutes required.

Second Puzzle:

After working for a while on this puzzle with no success I did some research and it turned out that the problem as stated has no possible solution.

From the article shown in the link below it follows that in this puzzle, which is known as “General Counterfeit Penny Problem” or “How do you find a single counterfeit coin [either heavier or lighter than a good one] among n coins in only k weighings on a two-pan balance”, the maximum number of balls allowing detection of a defective ball in 2 weighings or less is 4.

If the defective ball is known to be heavier (or lighter), two weighings may be sufficient.

Answer 2

Answer of first puzzle : Fold each thread twice to make 4 parts. Each part will take 15 minutes to burn. Therefor 3 parts will take 45 minutes.
Answer for second puzzle : Assume that the defective ball is lighter than other seven balls. Put 3/3 balls in each side of common balance. If any side is going up. The defective ball is in it. Out of it put one/one ball in balance & can find out from those two balls. If they have same weight, then the third one is defective. Suppose when we put 3/3 balls in each side of common balance, & found that both sides are equal. Means out of two remaining balls, one is faulty & it can be easily find out.

Answer 3

1) Light one thread, when half burnt (half length = half time = 30 mins) light second, when it's half burnt, first will be half burnt, which will be 15 more minutes, 15 + 30 = 45...I think that's the general idea, but I think you've omitted a necessary bit of info.

2) as stated, it can't be done, we need to know whether the defective ball is lighter or heavier than the rest, since the idea is you weigh some balls and have some in your hand. If they balance, the odd ball is in your hand, if they don't it's on the balance, but you need to know whether to reweigh the high side (odd ball is lighter) or low side (odd ball is heavier).

Answer 4

huh

Answer 5

IAM POOR IN MATHAMATICS CALCULATIONS