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Coin Word Problem

Jenny received \$6.10 in tips one afternoon. All of her tips were in quarters, dimes, and nickels. there five less dimes than quarters and seven less nickels than dimes. How many of each kind of coin was there?

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6.10 = .25Q + .10(Q-5) + .05 (Q-5-7)
6.10 = .25 Q +.10Q-.50 + .05Q -.60
6.10 = .40Q -1.10
7.20 = .40Q
18=Q
so 18 quarters (4.50)
18-5=13 dimes (1.30)
13-7=6 nickels (.30)
4.50+1.30+.30= 6.10

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This coin word problem is worked out in the following manner: \$6.10 = 0.25Q + 0.10(Q-5) + 0.05 (Q-50 -7); 6.10 = .25 Q +.10Q-.50 + .05Q -.60; 6.10 = .40Q -1.10. You will get 7.20 = .40Q, which translates to 18=Q, meaning there were 18 quarters, or about \$4.50. The dimes were 13 (18-5=13) dimes and the nickels were 6 (13-7=6).

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