A cashier has a total of 30 bills consisting of ones, fives and twenties. the number of twenties is 5 less than the number of ones. the total value of money is $229. how many of each denomination of bill are there?

Let us say that:

a = ones

b = fives

c = twenties

So that the total money is:

1 * a + 5 * b + 20 * c = 229

=> a + 5b + 20c = 229 - - > eqtn 1

We are also given that:

c = a - 5 - - > eqtn 2

a + b + c = 30 - - > eqtn 3

Rewriting eqtn 3 in terms of b:

b = 30 - a - c

Plugging in eqtn 2 into this:

b = 30 - a - (a - 5)

b = 35 - 2a - - > eqtn 4

Plugging in eqtn 2 and 4 into eqtn 1:

a + 5 (35 - 2a) + 20 (a - 5) = 229

a + 175 - 10a + 20a - 100 = 229

11a = 154

a = 14

So,

b = 35 - 2a = 7

c = a - 5 = 9

Therefore there are 14 ones, 7 fives, and 9 twenties.