Question 1108131: a newspaper carrier has $4.90 in change. He has five more quarters than dimes but two times as many nickles as quarters. How many coins of each type does he have? How many dimes?
Found 3 solutions by josgarithmetic, algebrahouse.com, ikleyn:
Answer by josgarithmetic(39629)   (Show Source):
You can put this solution on YOUR website! Literally translate the description; use obvious coin count variables, n, d, q.
First, simplify the money count equation; and then use the next two equations to write ONE single equation in the ONE variable q. Solve for q.,
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Answer by algebrahouse.com(1659)  (Show Source):
You can put this solution on YOUR website! x = number of dimes
x + 5 = number of quarters {he has five more quarters than dimes}
2(x + 5) = number of nickels {he has two times as many nickels as quarters}
0.1x + 0.25(x + 5) + 0.05(2x + 10) = 4.9 {value of coin times number of coins equals total value}
0.1x + 0.25x + 1.25 + 0.1x + 0.5 = 4.9 {used distributive property}
0.45x + 1.75 = 4.9 {combined like terms}
0.45x = 3.15 {subtracted 1.75 from each side}
x = 7 {divided each side by 0.45}
x + 5 = 12 {substituted 7, in for x, into (x + 5)}
2(x + 5) = 24 {substituted 7, in for x, into 2(x + 5)}
7 dimes
12 quarters
24 nickels
For more help like this, visit: www.algebrahouse.com
Answer by ikleyn(52873)  (Show Source):
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