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A collection of 30 coins worth $5.50 consists of nickels, dimes, and quarters. There are twice as many dimes as nickels. How many quarters are there?
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Q + D + N = 30 there are more dimes than nickels D=2N $5.50 = 550 cents each Q is worth 25 cents, D 10 cents, N 5 cents -> 25Q+10D+5N = 550 solve using substitution sub 2N for D in top equation Q +2N +N = 30 Q+3N = 30 3N = 30-Q N = 10-(1/3)Q sub this in for N ->25Q +10(2*(10-(1/3)Q))+5(10-(1/3)Q) = 550 distribute and add like terms (50/3)Q +250 = 550 Q = (550-250)*(3/50) Q = 90/5 = 18
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