1.) Four teachers bought an old boat for $60.The first teacher paid one-half of the sum of the amounts paid by the other teachers.The second teacher paid one-third of the sum of the amounts paid by the other teachers.The third teacher paid one-fourth of the sum of the amounts paid by the other teachers.How much did the fourth teacher pay? 2.) What is the smallest number of coins that you need so that you can give exact change for any amount from 1 cent to 99 cents? Show your work please. Any improper answers will be reported and removed- thanks!:) Have fun!

For problem number 1:-
1)Let's assume the payment which each one of them paid are A,B,C, and D, for the 1st, 2nd, 3rd and 4th respectively.
so the total payment is the sum of all of them  
A+B+C+D= 60 ----->(eq 1)
The 1st paid 0.5 of the sum of the  other teachers payments
A= 0.5 (B+C+D) ---->(eq 2)

The 2nd paid 1/3 of the others payment
[tex] B= \frac{A+C+D}{3}[/tex] ------>(eq 3) 

The 3rd paid 1/4 of the others payment
[tex] C = \frac{A+B+D}{4} [/tex] ---->(eq 4)
we have now 4 equations in 4 unknowns.

From equation 1 and 2 :-
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By multiplying (eq 2) by 2 and substitute in (eq 1)
2 A = B+C+D (sub in eq 1)
A+ 2A =60
then 3A= 60
then A =$20 (The first teacher's payment)

From equations 1 and 3 :-
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By multiplying (eq 3) by 3
3B = A+C+D (sub in eq 1)
B+3B=60
4B=60
B= $15 (The second teacher's payment)

From equations 1 and 4:-
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By multiplying (eq 4) by 4
then 4 C=A+B+D (sub in eq 1)
4C+ C = 60
5C=60
C= $6 (The third teacher's payment)

By substituting the values of A,B and C in (eq 1).
20 + 15 + 6 + D=60
41+ D =60 (By subtracting both sides by 41)
D = $19.(The forth teacher's payment)

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2)unfortunately i don't your coins cuz im not american :D