SOLUTION: Max has a small jar of coins containing pennies, nickels, and dimes. The total number of coins is 44 and the value of the coins is $2.20. If the number of dimes is twice the number

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Question 1045712: Max has a small jar of coins containing pennies, nickels, and dimes. The total number of coins is 44 and the value of the coins is $2.20. If the number of dimes is twice the number of nickels, how many of each type of coin does Max have?

Found 2 solutions by addingup, MathTherapy:
Answer by addingup(3677)   (Show Source): You can put this solution on YOUR website!
let the pennies be p, nickels be n, and the dimes d:
:
p+n+d = 44
0.01p+0.05n+0.10d = 2.20
:
n = 2d
p = ?
Information missing. Please re-upload the problem with All information.
Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!
Max has a small jar of coins containing pennies, nickels, and dimes. The total number of coins is 44 and the value of the coins is $2.20. If the number of dimes is twice the number of nickels, how many of each type of coin does Max have?
Ignore the other person's response. No additional info is required.



Let number of pennies, nickels, and dimes be P, N, and D, respectively

Then we get: P + N + D = 44 ------- eq (i) 

Also, .01P + .05N + .1D = 2.2 ----- eq (ii)

And, D = 2N ------ eq (iii)



P + N + 2N = 44 ------- Substituting 2N for D in eq (i)

P + 3N = 44______P = 44 - 3N ------- eq (iv)



.01P + .05N + .1(2N) = 2.2 ------- Substituting 2N for D in eq (ii)

.01P + .05N + .2N = 2.2

.01P + .25N = 2.2 ------ eq (v)



.01(44 - 3N) + .25N = 2.2 ------- Substituting 44 - 3N for P in eq (v) 

.44 - .03N + .25N = 2.2

.44 + .22N = 2.2

.22N = 2.2 - .44

.22N = 1.76

N, or 



D = 2(8) ------- Substituting 8 for N in eq (iii)

D, or 



P = 44 - 3(8) -------- Substituting 8 for N in eq (iv)

P, or  

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