Question

Roger has 3.10 consisting of quarters dimes and nickles. The number of nickles was 3 less than the number of dimes. The number of dimes was 5 more than the number of quarters. How many coins of each type did he use?
Please help me, thank you very much!! <3

Answer 1

Answer:

5 dimes 2 nickels 10 quarters

Step-by-step explanation:

Let n = number of nickels

      d = number of dimes

      q = number of quarters

n = d - 3 (The number of nickles was 3 less than the number of dimes)

d= q - 5 (The number of dimes was 5 more than the number of quarters)

0.05n + 0.1d + 0.25q = 3.1

Use substitution:

n=q-8

0.05n + 0.1d + 0.25q = 3.1

0.05(q-8) + 0.1(q-5) + 0.25q = 3.1

0.05q-.4 + 0.1q -.5 +0.25q = 3.1

0.4q-0.9=3.1

0.4q=4

q = 10

n =q-8    n = 10-8     n=2

d=q - 5     d = 10-5     d = 5