A box contains $8.75 in nickels dimes and quarters there are 51 coins in all and the sum of numbers of nickels and dimes is three less than the number of quarters how many coins of each kind are there?
Hint: Here are the equations
Hope this helped!
q + q-3 = 51
q = 27 27 quarters = 6.75 there is 24 coins (n +d) and 2.00 left
.05n + .10 (24-n) = 2.00
-.05n = -.4 n = 8 nickels then 24-n = 16 dimes