SOLUTION: A collection of 24 coins, consisting of nickels, dimes, and quarters, has a value of $3.40. If there are twice as many quarters as nickels, and one-third as many nickels as dimes,

Question 1046005: A collection of 24 coins, consisting of nickels, dimes, and quarters, has a value of $3.40. If there are twice as many quarters as nickels, and one-third as many nickels as dimes, how many coins of each kind are there?
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39620) (Show Source): You can put this solution on YOUR website!
Variables, n,d,q

Account for the coins: n+d+q=24

Account for the money: 0.05n+0.1d+0.25q=3.4

Transcribe the other coin count descriptions:

Understand, you SHOULD need only three equations because you have only three unknown variables. Currently you have four equations.

To move things along, your system of equations is this:

Solve the system. In case any inconsistencies, recheck the problem description and your work.
Answer by MathTherapy(10553) (Show Source): You can put this solution on YOUR website!

A collection of 24 coins, consisting of nickels, dimes, and quarters, has a value of $3.40. If there are twice as many quarters as nickels, and one-third as many nickels as dimes, how many coins of each kind are there?

Let the number of nickels, dimes, and quarters, be N, D, and Q, respectively
Then we get: N + D + Q = 24 -------- eq (i)
Also, Q = 2N
As there are as many nickels as dimes, it follows that: D = 3N
N + 3N + 2N = 24 ------ Substituting 2N for Q, and 3N for D in eq (i)
6N = 24
N, or number of nickels = , or
Number of dimes: 3(4), or
Number of quarters: 2(4), or
As seen, the value of these 24 coins was not needed. In other words, it was totally unnecessary to include it!
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