Question

Paul has $4.75 in coins. He has some quarters, one more dime than quarters, and 3 less nickels than quarters. How many dimes does he have?

Answer 1

See a solution process below:

Explanation:

First, let's name some variables:

Let's call the number of quarters Paul has: `q`

Let's call the number of dimes Paul has: `d`

Let's call the number of nickles Paul has: `n`

We know:

`d = q + 1`

`n = q - 3`

`$0.25q + $0.10d + $0.05n = $4.75`

We can substitute `(q + 1)` for `d` and we can substitute `(q - 3)` for `n` and solve for `q`:

`$0.25q + $0.10(q + 1) + $0.05(q - 3) = $4.75`

`$0.25q + ($0.10 * q) + ($0.10) + ($0.05 * q) - ($0.05 * 3) = $4.75`

`$0.25q + $0.10q + $0.10 + $0.05q - $0.15 = $4.75`

`$0.25q + $0.10q + $0.05q + $0.10 - $0.15 = $4.75`

`($0.25 + $0.10 + $0.05)q + ($0.10 - $0.15) = $4.75`

`$0.40q - $0.05 = $4.75`

`$0.40q - $0.05 + color(red)($0.05) = $4.75 + color(red)($0.05)`

`$0.40q - 0 = $4.80`

`$0.40q = $4.80`

`($0.40q)/color(red)($0.40) = ($4.80)/color(red)($0.40)`

`(color(red)(cancel(color(black)($0.40)))q)/cancel(color(red)($0.40)) = (color(red)(cancel(color(black)($)))4.80)/color(red)(color(black)(cancel(color(red)($)))0.40)`

`q = 4.80/0.40`

`q = 12`

We can find the number of dimes by substituting `12` for `q` into the first equation and calculating `d`:

`d = q + 1` becomes:

`d = 12 + 1`

`d = 13`