College Algebra

Question

A box contains $6.20 in nickels, dimes, and quarters. There are 36 coins in all, and the sum of the number of nickels and dimes is 2 less than the number of quarters. How many coins of each kind are there?

Answer 1

N + D + Q = 36 <-- 36 coins

5N + 10D + 25Q = 620 <-- $6.20 = 620 pennies

N+D + 2 = Q <-- # of quarter is 2 more than the # of nickles and dimes

Substitutes 3rd equation:

N + D + n + d + 2 = 36

5N + 10d + 25(n+d+2) = 620

combines like terms and distributive:

2n + 2d + 2 = 36

5n + 10d + 25n + 25d + 50 = 620

solves:

2n + 2d = 34

30n + 35d = 570

simplifies:

n + d = 17

6n + 7d = 114

n = 17-d

6(17-d) +7d = 114

102 - 6d + 7d = 114

102 + d = 114

d=12

n = 17-d = 17-12 = 5

36 coins in total per 1st equation,

so q=19 quarters

5 nickles, 19 quarters, 12 dimes

0.25 + 4.75 + 1.20 = $6.20