<PRE><B>Question 185117</B>
A cashier has 30 bills, all of which are $10 or $20 bills. The total value of the money is $330. How many of each type does the cashier have?


Let the number of $10 bills be m and the number of $20 bills be n.


We know m + n = 30
and 10m + 20n = 330 (m $10 bills + n $20 bills totalling $330).


So we have two simultaneous equations with the third equation a multiple (10) of the first:

(1)      m +     n =   30

(2)	10m + 20n = 330

(3)	10m + 10n = 300


The difference of last two equations
10n = 30
Therefore n = 3 and m = 27
27 x $10 + 3 * $20 = $330 using 30 bills.
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