Algebra II question?

Question

A jar contains 41 coins consisting of nickels,dimes,and quarters and having a total value of $4.75. How many of each kind of coins are there if there are twice as many dimes as quarters

Answer 1

Let n = # of nickels and q = # of quarters

Since there are as many dimes as quarters, # of dimes = 2q

n + 2q + q = 41 (eq1) --> total # of coins = 41

0.05n + 0.10(2q) + 0.25(q) = 4.75 (eq2)

From eq1, n + 3q = 41 ==> n = 41 - 3q

From eq2, 0.05n + 0.20q + 0.25q = 4.75

0.05n + 0.45q = 4.75

Sub n = 41 - 3q into eq2,

0.05(41-3q) + 0.45q = 4.75

2.05 - 0.15q + 0.45q = 4.75

0.30q = 2.7

q = 9

# of quarters = 9

# of dimes = 2q = 2x9 = 18

# of nickels = 41 - 3q = 41 - 27 = 14

Your answer is correct.

Answer 2

quarter 9

dime  18  

nickel 14