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?

Rhonda has 7 dimes, 14 quarters, 13 nickels

Let the number of dimes be d

Let the number of quarters be q

Let the number of nickels be n

Rhonda has 6 more nickels than dimes

n  =  d  +  6

She has twice as many quarters as dimes

q  =  2d

n = d + 6

1 dime = $0.1

1 quarter = $0.25

1 nickel = $0.05

Total amount of money Rhonda has is $4.85

0.1d  +  0.25q  +  0.05n   =  4.85

0.1d  +  0.25(2d)  +  0.05(d+6)  =  4.85

0.1d + 0.5d  + 0.05d + 0.3 = 4.85

0.65d = 4.55

d  =  4.55/0.65

d =  7

Substitute d = 7 into q = 2d

q = 2(7)

q  =  14

Substitute d = 7 into n = d + 6

n = 7 + 6

n = 13

Rhonda has 7 dimes, 14 quarters, 13 nickels

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nickels = 13, dimes = 7, quarters = 14

Step-by-step explanation:

First, set up the equations for each coin:

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

5/12, or 5: 12

step-by-step explanation: