233k views
5 votes
A pepper carrier has $6.90 in change. He has six more quarters than dimes and three × as many nickels as quarters. How many coins of each type does he have?

a) 7 quarters, 1 dime, 18 nickels
b) 8 quarters, 2 dimes, 15 nickels
c) 9 quarters, 3 dimes, 12 nickels
d) 10 quarters, 4 dimes, 9 nickels

1 Answer

3 votes

Final answer:

To solve this problem, we can set up a system of equations. Let's assume the number of dimes is x. Since the pepper carrier has six more quarters than dimes, the number of quarters would be x + 6. And since the pepper carrier has three times as many nickels as quarters, the number of nickels would be 3(x + 6). The total value of the coins can be expressed as: 0.1x (dimes) + 0.25(x + 6) (quarters) + 0.05(3(x + 6)) (nickels) = 6.90.

Step-by-step explanation:

To solve this problem, we can set up a system of equations.

Let's assume the number of dimes is x.

Since the pepper carrier has six more quarters than dimes, the number of quarters would be x + 6.

And since the pepper carrier has three times as many nickels as quarters, the number of nickels would be 3(x + 6).

The total value of the coins can be expressed as: 0.1x (dimes) + 0.25(x + 6) (quarters) + 0.05(3(x + 6)) (nickels) = 6.90.

Simplifying the equation, we get: 0.1x + 0.25x + 1.5 + 0.15x + 0.9 = 6.90.

Combining like terms, we have: 0.5x + 2.4 = 6.90.

Subtracting 2.4 from both sides, we get: 0.5x = 4.50.

Dividing both sides by 0.5, we get: x = 9.

Therefore, the pepper carrier has 9 dimes, 9 + 6 = 15 quarters, and 3(15) = 45 nickels.

User Fadmaa
by
7.5k points