There is no missing Php10. The correct statement to account for the Php150 is as follows: Php100 paid for the meal + Php 30 returned to each of the 3 + Php20 in the tip box = Php150. To account for the Php120 expense (3 x Php40), use Php100 paid to the cashier + Php20 in the tip box.

In this problem, you only need to open one box knowing that all labels are incorrect. If you open "Apples and Grapes" and you know that it cannot have both apples and grapes, the only conclusion is that it can have apples only or grapes only. If you get apples inside, get the label "Apples" and put it on this box. The box which used to have the label for apples will necessarily have grapes in it. Put the label "Grapes" on that box and the box that used to have the label for grapes should end up with the label "Apples and Grapes".

This is a classic problem, with variants only on who the characters are. If we assume the

*delawan*to be X and the*dede-es*to be Y, while the current side of the river they are on is side A and the other side is side B, the river crossing will be as follows:
(1) XXX Y at side A | YY crossing to side B

(2) XXX Y at side A | Y crossing back to side A | Y at side B

(3) XXX at side A | YY crossing to side B | Y at side B

(4) XXX at side A | Y crossing back to side A | YY at side B

(5) XY at side A | XX crossing to side B | YY at side B

(6) XY at side A | XY crossing back to side A | XY at side B

(7) YY at side A | XX crossing to side B | XY at side B

(8) YY at side A | Y crossing back to side A | XXX at side B

From point (8) above, Y will just cross back and forth to shuttle the rest of Y's, while all X's are already at side B. In no time from point (1) to point (8) has there been a number of X less than Y, so the group can cross the river this way without a fight ensuing, Y's are kept at bay by being equal to or less in number than X.

With a little Math, you'd find out that most probably, the father is currently working overtime with the mother for purposes of procreation. Here's the algebra behind it:

Let x be the mother's age and y be the child's age.

If the mother is 21 years older than her child, this means that x = y + 21.

If the mother will be exactly 5 times as old as the child in exactly 6 years from now, then x + 6 = 5 (y + 6). So we have:

(Equation 1) x = y + 21

(Equation 2) x + 6 = 5 (y + 6)

Substituting the value of x from Equation 1 in Equation 2:

y + 21 + 6 = 5 (y + 6)

y + 27 = 5y + 30

27 - 30 = 5y - y

- 3 = 4y

y = -(3/4)

We all know that 3/4 of a year is 9 months. The negative sign in front means that it is to be in future.

If we know that one of the guards always tell the truth while the other always tell a lie, then we can ask this: "If I were to ask the other guard, what will he/she tell me?" Whatever answer is given, go for its opposite.

In the case above, let's say that the correct choice is Door A.

If you spoke to the liar, the liar will tell you that the other guard will say "Door B" because although the other guard will actually tell you "Door A" the liar will change it and lie to you. So you go for the opposite of the answer which is Door A.

On the other hand, if you spoke to the guard that always tells the truth, he/she would tell you "Door B" because although he/she knows the correct choice is Door A, he/she knows that the other guard will lie, so he will tell you truthfully that the other guard will say "Door B". So you go for the opposite, which is Door A.