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?

[tex]\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}[/tex]

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

Adetunmbiadekunle Adetunmbiadekunle · Answer 2 · 2021-11-26T11:13:28+01:00

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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