A cashier has 25 coins consisting of nickels dimes and quarters with the value of 4.90. If the number of dimes is 1 less than twice the number of nickels many of each type of coin

Answer 1

Let n = nickels, d = dimes, q = quarters

0.05n + 0.10d + 0.25q = 4.9

d = 2n - 1

n + d + q = 25

We can rewrite everything in terms of n (nickels) by solving the last expression for q (quarters) and using substitution

q = 25 - d - n

q = 25 - (2n -1) - n = 25 -2n + 1 - n = 26 - 3n

Now we can solve the equation that originally had 3 unknowns:

0.05n + 0.10d + 0.25q = 4.9

0.05n + 0.10(2n - 1) + 0.25 (26 - 3n) = 4.9

0.05n + 0.20n - 0.1 + 6.5 - 0.75n = 4.9

-0.5n + 6.4 = 4.9

-0.5n = -1.5

n = 3

d = 2n - 1 = 2*3 - 1 = 5

q = 26 - 3n = 26 - 3*3 = 17

The cashier has 3 nickels, 5 dimes and 17 quarters.