Question

A jar contains $9.70 in dimes, quarters, and half dollars. if there are 8 fewer dimes than quarters and 5 fewer half dollars than dimes, then how many coins of each type does the jar contain?

Answer 1

The coin problem is solved by setting up and solving a system of equations. By defining the number of quarters as q, and using the values of the coins, it is determined that there are 6 dimes, 14 quarters, and 1 half dollar in the jar.

Step-by-step explanation:

The student's question asks for the number of dimes, quarters, and half dollars in a jar that contains $9.70, with certain conditions regarding the quantities of each coin.

To answer this, we can set up a system of equations based on the values and relationships of the coins. A dime is worth 10 cents, a quarter is worth 25 cents, and a half dollar is worth 50 cents.

We define the number of quarters as q, then the number of dimes will be q - 8, and the number of half dollars will be q - 8 - 5 or q - 13.

Using the values for each type of coin, we can write the following equation:

10(q - 8) + 25q + 50(q - 13) = 970

Solving this equation, we find that the number of quarters q is equal to 14, which then allows us to calculate the number of dimes and half dollars. So, we have:

• Number of dimes: 14 - 8 = 6
• Number of quarters: 14
• Number of half dollars: 14 - 13 = 1

Thus, the jar contains 6 dimes, 14 quarters, and 1 half dollar.