The man must have been born at midday on February 19, 1873, and at midday on November 11, 1928, he had lived 10, 176.5  days in each century.
Of course, the century ended at midnight on December 31, 1900, which was not a leap year, and his age on November 11, 1928, was 55 years and nearly 9 months.

The explanation is simply this. The two ways of selling are only identical when the number of apples sold at three for a penny and two for a penny is
in the proportion of three to two.  Thus, if the first woman had handed over 36 apples, and the second woman  24, they would have fetched 24¢, whether sold separately or at five for 2¢.  But if they each held the same number of apples there would be a loss when sold together of l¢ in every 60 apples. So if they had 60 each there would be a loss of2¢. If there were 180 apples (90 each) they would lose 3¢, and so on.  The missing penny in the case of 60 arises from the fact that the three a penny woman gains 2¢, and the two a penny woman loses 3¢.  Perhaps the fairest practical division of the 24¢ would be that the first woman receives 9.5¢ and the second woman 14.5 , so that each loses 0.5¢ on the transaction.

He must have had 168 each of dollar bills, half dollars, and quarters,  making a total of $294.00. In each of the six bags there would be 28 of each
kind; in each of the seven bags 24 of each kind; and in each of the eight bags, 21 of each kind.

The smallest number originally held by one person will be (in cents) one more than the number of persons. The others can be obtained by continually
doubling and deducting one. So we get their holdings as 10, 19,37, 73, 145, 289,577, 1,153, and 2,305. Let the largest holder start the payment and work backwards, when the number of cents in the end held by each person will be 29 or 512-that is, $5.12.

The contents of the ten bags (in dollar bills) should be as follows: $1,2,4, 8, 16,32,64, 128,256, 4S9. The first nine numbers are in geometrical progression,  and their sum, deducted from 1,000, gives the contents of the tenth bag.

The seven men, A, B, C, D, E, F, and G, had respectively in their pockets before play the following sums: $4.49, $2.25, $1.13, 57¢, 29¢, 15¢, and 8¢. The
answer may be found by laboriously working backwards, but a simpler method is as follows: 7 + 1 = 8; 2 X 7 + 1 = 15; 4 X 7 + 1 = 29; and so on, where the multiplier increases in powers of 2, that is, 2, 4, 8, 16, 32, and 64.

The third man. This is because he knows there are only two of each color cap. If the man behind him (the fourth man) saw two caps that were the same color in front of him, he would know that his own must be the opposite. However, because the caps alternate in color. The fourth man has only a 50% chance of getting his hat color correct, so therefore he stays quiet. The third man realizes that the fourth man is quiet because he must not see two caps of the same color in front of him, otherwise the fourth man would say the opposite of the caps in front of him. Therefore, the third man presumes his own cap must be the opposite of the mans in front of him, and his presumption is correct. Under this same logic, after the third man speaks his color hat, the second man, even though he sees only wall, would be the next to go free, because he knows his cap must be the opposite of whichever color the third mans cap was.

He can propose a plan that he gets 98 pieces of gold, the 3rd pirate gets 1 piece, and the 5th pirate gets 1 as well.

If there were just 2 pirates the younger pirate would definitely deny the plan so he could get all of the gold.

If there were 3 pirates the first pirate can offer the second pirate 1 piece of gold and take the rest himself because the second pirate wouldn’t get anything if he has to propose a plan himself.

If there were 4 pirates the first pirate could take 99 for himself and offer 1 to the youngest pirate. They would both agree. If the youngest disagrees then he won’t get any gold in the next plan.

So when there are 5 pirates it is in the interest of the 3rd and 5th pirate to accept 1 piece, because if they don’t they won’t get anything in the next plan.

If you have come up with the fact that the bugs can avoid the collision only when they decide to move in the same direction be it clockwise or anti clockwise, you have already solved half of it. In any other way, the bugs will definitely run into each other.

Now the bugs can move either clockwise or anti-clockwise. The probability of choosing a specific direction is then 1/2.

We will utilize the simple equation of probability to find out our answer

P(No Collision) = P (The bugs move in clockwise direction) + P(The bugs move in anti-clockwise direction)
= 0.5 * 0.5 * 0.5 + 0.5 * 0.5 * 0.5
= 0.25

Thus the probability that the bugs won’t collide into each other is 0.25 or 1/4.

This one definitely sounds too unusual. But there is an easy way to look at things.

Let us say that there is one unfaithful husband in the town. In such a case, 99 women will know about him. The wife of that man will be the only one who will not know about him and would have always thought that his husband is truly faithful to her. But since the mayor announced that there is at least one cheater among them, she will come to a realization that her husband must be cheating on her. So her husband will be executed on the first day of the announcement.

Now let us take up a case that there are two cheating husbands in the town. Now 98 women of the town must be knowing about them both. The two wives of those men will think that there is only one cheater in the town. Both of them believes that their respective husbands are loyal and this would not report about them

But when both of them will find out on the next day that no one was executed, they will realize that it could only mean that both of their husbands are cheaters. Thus both the husbands will be executed on the second day.

Thus if we have X number of cheating husbands, they all will die on the Xth day after the announcement is being made.