SOLUTION: Mike has 19 coins in his pocket worth a total of $2.35. He only has nickels, dimes and quarters. There are 2 more dimes than nickels. How many of each coin type does Mike have?

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Question 1165757: Mike has 19 coins in his pocket worth a total of $2.35. He only has nickels, dimes and quarters. There are 2 more dimes than nickels. How many of each coin type does Mike have?







Found 2 solutions by Boreal, MathTherapy:
Answer by Boreal(15235)   (Show Source): You can put this solution on YOUR website!
N+D+Q=19
.05N+.10D+.25Q=2.35
but D=N+2
so .05N+.10(N+2)+.25.Q=2.35
.05N+.10N+.20+.25Q=2.35
.15N+.25Q=2.15
Q=19-(2N+2)
so .15N+.25(19-2N-2)=2.15
.15N+4.25-.50N=2.15
-.35N=-2.10
N=6 nickels (0.30)
D=8 (0.80)
Q=5 (1.25)

Answer by MathTherapy(10551)   (Show Source): You can put this solution on YOUR website!

Mike has 19 coins in his pocket worth a total of $2.35. He only has nickels, dimes and quarters. There are 2 more dimes than nickels. How many of each coin type does Mike have?
Let number of nickels be N

Then number of dimes = N + 2, and number of quarters = 19 - (N + N + 2) = 19 - 2N - 2 = 17 - 2N

We then get: .05N + .1(N + 2) + .25(17 - 2N) = 2.35

.05N + .1N + .2 + 4.25 - .5N = 2.35

.05N + .1N - .5N = 2.35 - 4.45

- .35N = - 2.1

Number of nickels, or 

You should now be able to find the number of dimes, and quarters! 

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