Please explain your answers!

-1. Five bags of nuts are placed in front of you. There are fifty-two nuts altogether in the first and second bag. There are forty-three between the second and third bags, thirty-four between the third and fourth and thirty between the fourth and fifth.

There are a total of 100 nuts, how many in each bag?

-2. What have the numbers 3, 7, 8, 40, 50, and 60 in common that no other whole numbers have?

-3. My first is a number, my second another,
and each, I assure you, will rhyme with the other.
My first you will find is one-fifth of my second,
and truly my whole a long period reckoned.
Yet my first and my second (nay, think not I cozen)
When added together will make but two dozen.

-4. There are 4 wheels with their relative sizes connected as shown. If the leftmost wheel turns one revolution, how many revolutions has the rightmost wheel turned?

2. they all have 5 letters?

Yes! A little non-math math fun.

Number 1:

Let a = number in 1st bag; then:
1: a
2: 52-a
3: 43-(52-a) = a-9
4: 34-(a-9) = 43-a
5: 30-(43-a) = a-13

a + 52-a + a-9 + 43-a + a-13 = 100
a + 73 = 100
a = 100-73 = 27

So:
1: 27
2: 52-27 = 25
3: 27-9 = 18
4: 43-27 = 16
5: 27-13 = 14

27+25+18+16+14 = 100

Quite a "delightfully nutty" problem Sue!

Solved by an equally delightfully nutty guy, Denis.

3. four score

Well solved, manxman!

I think 4 must be 2/3 of a revolution

I think so too!

Nice solve, Sir Lexicus!

Took me a while to finally "see" that them there circles
represent "gear wheels".
And, of course, the "gears" must be all same size on all the "wheels".
Soooo slow me reasoned that (w = wheel):
while w1 rotated once, w2 rotated twice
while w2 merrygorounded twice, w3 spinned 4 times
while w3 spinned 4 times, w4 circled [(1/4)*4]/(3/2) = 1/(3/2) = 2/3 of a revolution...

S'that how it works Sue?
Thanks...I've been losing sleep over this!!

Looks about perfect, Denis. Glad you lost some sleep over it, but not too much, I hope.