Question

7. A coin bank contains $12.45 in quarters and dimes. If there are 16 more dimes than quarters,

how many are there of each?

Answer 1

There are 31 quarters.

There are 47 dimes.

Explanation:

let q = number of quarters

let d = number of dimes

d = 16 + q

.10d + .25q = 12.45 Multiply through by 100

10d + 25q = 1245 Substitute 16 + q for d

10(16 + q) + 25q = 1245 Distribute the 10

160 + 10q + 25q = 1245 Combine like terms

160 + 35q = 1245 Subtract 160 from both sides

160 - 160 + 35q = 1245 - 160

35q = 1085 Divide both sides by 35

q = 31

There are 31 quarters.

Substitute 31 for q

d = 16 + 31

d = 47

There are 47 dimes.

Check:

d = 16 + q

47 = 16 + 31

47 = 47 Checks

.10d + .25q = 12.45

.10(47) + .25(31) = 12.45

4.70 + 7.75 = 12.45

12.45 = 12.45 Checks.

Answer 2

There are 31 quarters and 47 dimes.

Explanation:

To solve this problem, we can create and solve a system of equations.

Define the variables:

• Let q be the number of quarters.
• Let d be the number of dimes.

Values of the coins:

• The value of a quarter is $0.25.
• The value of a dime is $0.10.

Given a coin bank contains $12.45 in quarters and dimes:

• 
  `0.25q + 0.10d = 12.45`

Given there are 16 more dimes than quarters:

• 
  `d = q + 16`

Therefore, the system of equations that represents the problem is:

`\begin{cases} 0.25q + 0.10d = 12.45\\d = q + 16\end{aligned}`

Substitute the second equation into the first equation to eliminate d:

`0.25q+0.10(q+16)=12.45`

Solve the equation for q to find the number of quarters:

`\begin{aligned}0.25q+0.10(q+16)&=12.45\\0.25q+0.10q+1.6&=12.45\\0.35q+1.6&=12.45\\0.35q&=10.85\\q&=31\end{aligned}`

Therefore, there are 31 quarters.

Substitute the found value of q into the second equation and solve for d:

`\begin{aligned}d &= q + 16\\&= 31 + 16\\&= 47\end{aligned}`

Therefore, there are 47 dimes.