Question

Bob has $2.50 is dimes and nickels. There

are four more than 10 times as many dimes
as nickels. How many of each coin are
there?

Answer 1

The problem is solved by creating and solving a system of equations. It is determined that Bob has 2 nickels and 24 dimes.

Step-by-step explanation:

Let's denote the number of nickels as n and the number of dimes as d. The problem states there are four more than 10 times as many dimes as nickels, which can be formulated as d = 10n + 4. Since each nickel is worth 5 cents and each dime is worth 10 cents, we can write the total amount of money ($2.50 or 250 cents) as a value equation: 5n + 10d = 250.

Substitute the value of d from the first equation into the second equation:

5n + 10(10n + 4) = 250.

This simplifies to:

5n + 100n + 40 = 250.

Combine like terms:

105n + 40 = 250.

Subtract 40 from both sides:

105n = 210.

Divide both sides by 105:
n = 2.

Now, find the number of dimes using n:
d = 10 \times n + 4

d = 10 \times 2 + 4

d = 20 + 4
d = 24.

Bob has 2 nickels and 24 dimes.