Ariel also has a collection of coins. Her collection is worth $4.80. She has nickels, dimes, and

quarters. The number of nickels that she has is 6 less than the total number of dimes and quarters,

while the number of dimes is equal to the sum of the number of nickels and quarters. Find how many

of each coin Ariel has.

There are 25 nickels, 28 dimes and 3 quarters

Step-by-step explanation:

- Ariel has a collection of coins

- Her collection is worth $4.80

- She has nickels, dimes, and quarters

- The number of nickels that she has is 6 less than the total number of

dimes and quarters

- The number of dimes is equal to the sum of the number of nickels

and quarters

- Assume that the number of nickels is x, the number of dimes is y and

the number of quarters is z

∵ 1 nickel = 5 cents

∵ 1 dime = 10 cents

∵ 1 quarter = 25 cents

∵ 1 dollar = 100 cent

∵ Her collection worth $4.80

∵ 4.80 dollars = 4.80 * 100 = 480 cents

∴ 5x + 10y + 25z = 480 ⇒ (1)

∵ The number of nickels is 6 less than the total number of dimes

and quarters

∴ x = y + z - 6 ⇒ (2)

∵ The number of dimes is equal to the sum of the number of nickels

and quarters

∴ y = x + z

- subtract z from both sides

∴ x = y - z ⇒ (3)

- Equate equation (2) and (3)

∴ y - z = y + z - 6

- Subtract y from both sides

∴ - z = z - 6

- Add 6 to both sides

∴ 6 - z = z

- Add z to both sides

∴ 6 = 2z

- Divide both sides by 2

∴ z = 3

- Substitute the value of z in equation (3)

∴ x = y - 3

- Substitute x and z in equation (1)

∴ 5 (y - 3) + 10y + 25 (3) = 480

∴ 5y - 15 + 10y + 75 = 480

∴ 15y + 60 = 480

- subtract 60 from both sides

∴ 15y = 420

- Divide both sides by 15

∴ y = 28

- Substitute value of y in equation (3)

∴ x = 28 - 3

∴ x = 25

* There are 25 nickels, 28 dimes and 3 quarters