# 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? ##### 1 Answer Sep 4, 2017 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 $\left(q + 1\right)$for $d$and we can substitute $\left(q - 3\right)$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 = \frac{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\$