80

21 Q 7 R 9 S 10 P 13
= 21 / 7 + 9 * 10 – 13
= 3 + 90 – 13
= 80

As per the given data,
6 * number of chocolates + 5 * number of sodas = 32

Now, we know that chocolates and sodas are definitely whole numbers; their respective values will be 2 and 4.

If we assume that Rihana and Peeta pay the same amount, they can decide with “rock, paper, scissor” to determine who will pay the extra dollar. Any other method may be used as well. Suppose that Rihana pays $1 extra.

In this manner, Rihana pays $5, Peeta pays $4 and Logan pays $2.

If the pizza would have cost $11, this method will then prove to be great because both Rihana and Peeta will pay $5 each and Logan will pay $1. This will keep the offset same as previous. Rihana is paying 6% more, Peeta is paying 15% more and Logan is paying 27% more.

The numbers of races are 7

Three. At least two are same color.

It is going left because we can not see the the door.

There are 6 options to distribute 4 cookies between 2 kids (4 above 2):

• One kid and two expired cookies, there are 2 possibilities: kid 1 has 2 good ones or the 2 expired ones.

• If each kid is to get one good and one expired there are 4 possibilities: each can get one of two expired and one of two good – (2×2).

In real life though the less likely possibility is sometimes more likely to happen :o)

You only need one.

You draw one marble from a box with a sticker “red and white” – let’s say its red. That means the box has to be full of red marbles (if there were white ones too the sticker would have been on a correct box). Box marked as “white” must be mixed (red is already taken and white would be correctly).

The last box marked as “red” has to contain white marbles.

If you draw white first the solution is almost exactly same only with opposite colors.

Have you figured it out?

1. Goat there. Nothing back.
2. Wolf there. Goat back.
3. Cabbage there. Nothing back.
4. Goat there.

Prisoners agree that one color means odd number and the other means even number. The last prisoner in row counts (lets say) white stripes and says the color according to their agreement (one color even number of whites other color odd number). He has 50% chance to survive.

The second last prisoner will also count white stripes and if he counts the same parity (odd or even) he knows he has white. If he counts different parity he knows he has red stripe.

Other prisoners will do the same only they have to pay attention to how many times they’ve heard “white”…

So all but one will be released for sure and even the one has a good chance!