Question

Rhonda has $4.85 in coins. If she has six more nickels than dimes and twice as many quarters as dimes, how many coins of each type does she have?

Answer 1

Answer: nickels = 13, dimes = 7, quarters = 14

Step-by-step explanation:

First, set up the equations for each coin:

$\begin {array}{c|c|c|l}\underline{\quad Type\quad }&\underline{Value}&\underline{Quantity}&\underline{\qquad \qquad \qquad Equation\qquad }\\ Nickel& \ 0.05&6+d&0.05(6+d) = 0.30 + 0.05d\\Dime& \ 0.10&d&\qquad 0.10(d)=0.10d\\Quarter& \ 0.25&2d&\qquad 0.25(2d)=0.50d\\\end{array}$

Next, the sum of the coins is $4.85 so substitute and solve for the variable:

   Nickels         +   Dimes  +   Quarters = Sum

(0.30 + 0.05d)  +    0.10d   +     0.50d   =  4.85

0.30 + 0.65d = 4.85

            0.65d = 4.55

                    d = 7

Lastly, plug the d-value into the quantity equation for the nickels and quarters to find their quantity.

nickels: 6 + d   = 6 + (7)   = 13

quarters: 2d   = 2(7)   = 14