Solve Basic Word Problem: Find Coins & Combos for $3.00

[war485]

Homework Statement

I have nickels, dimes and quarters in a jar. There are 20 coins altogether and exactly twice as many dimes as nickels. The total value of the coins is $3.00
Find the number of coins of each type. Then find all possible combinations of 20 coins (nickels, dimes and quarters) that will make exactly $3.00

Homework Equations

everything is in cents
N + D + Q = 20
-N+ 2D = 0 ---> from 2D = N --> "twice as many dimes as nickels"
5N + 10D + 25Q = 300

The Attempt at a Solution

I put this into a matrix form
1 1 1 20
1 2 0 0
5 10 25 300

reduced it to:

1 0 0 16
0 1 0 -8
0 0 1 12

Funny thing is I have -8 dimes?
So I think I don't know how to put this statement into an equation:
"twice as many dimes as nickels"

As for the number of possible combinations of 20 coins, I really don't know how to start that in a matrix form.

 

[Dick]

It's not that funny. The second line of your matrix is 1 2 0 0 which says N+2D=0. Or D=(-N/2). -8 is -16/2. As you started saying you want the equation to be N=2D. Or wait. Is that right? Doesn't that say the number of nickels is TWICE the number of dimes?

 

[war485]

It's not that funny. The second line of your matrix is 1 2 0 0 which says N+2D=0. Or D=(-N/2). -8 is -16/2. As you started saying you want the equation to be N=2D. Or wait. Is that right? Doesn't that say the number of nickels is TWICE the number of dimes?

When you say "the number of nickels is TWICE the number of dimes" is much clearer. I thought that's what is meant by "twice as many dimes as nickels", guess not. And thanks for pointing out about me forgetting the negative sign in my second line of the matrix. It works much better now.

As for the number of possible combinations of 20 coins, I just multiply Q D N together since they are independent events right? Or is there a different method that is specifically for matrices?

 

[Dick]

I think you are done with matrices. I think they just expect you to count all possible solutions. I would break it into cases, like:

Q=12, N=0, D=0. (1 solution).
Q=11 then the number of dimes is 0,1 or 2. The rest are nickels. (3 solutions).
Q=10 then the number of dimes is 0,1,2,3,4 or 5. (6 solutions)
Q=9 now you can't have zero dimes, because then you would need to have 15 nickels. That's too many coins.

Get the idea? For each number of quarters figure out the possibilities for the number of dimes. It sounds tedious, but I can't think of any other elementary way to count them.

 

[war485]

thanks for your help