Joe has a collection of nickels and dimes that is worth $6.10. If the number of dimes was tripled and the number of nickels was decreased by 49, the value of the coins would be $14.05. How many nickels and dimes does he have?
Let: n = number of nickles d = number of dimes we know from the problem statement that: 0.05n + 0.1d = 6.10 and 0.05(n-49) + 0.1*(3d) = 14.05 To solve, we have to solve one equation for one variable (doesn't matter which): 0.05n + 0.1d = 6.10 n = (6.10 - 0.1d)/0.05 n = 122 - 2d then substitute that into the other equation and solve: 0.05(n-49) + 0.1*(3d) = 14.05 0.05([122-2d]-49) + 0.1*(3d) = 14.05 6.10 - 0.1d - 2.45 + 0.3d = 14.05 .2d = 10.4 d = 52 then plug that into any of the previous equations to find n: n = 122 - 2d n = 122 - 2(52) n = 18
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