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How many coins are there?

  • #1
1. There are 100 coins on the table. The coins consist of nickels, dimes, and quarters. The total value is $15.85. How many of each coins are there?



3.

I setup a system of equations.
x = nickels
y = dimes
z = quarters

x + y + z = 100
0.05x + 0.10y + 0.25z = 15.85


I then solve and reduce and get:

x + y + z = 100
y + 5z = 217

I move over variables and get:

x = 100 - y - z
y = 217 - 5z

Since there are more variables than equations I assume that there are many solutions, and try using a parametric representation.


x = 100 - y - z
y = 217 - 5t

where t = any real number

I plug in numbers but end up not getting the value 15.85. The most I can get is somewhere near 14.00.

I was wondering if my setup is wrong.
Is there another way to solve this problem?
 

Answers and Replies

  • #2
Borek
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  • #3
SteamKing
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Obviously, t is not any real number. t is going to come from the set of positive integers. You can't have half a nickle, or Sqrt(2) of a quarter.

Instead of introducing yet another variable (t) into your problem, make your parameter one of the coin denominations, either the nickle, the dime, or the quarter. By choosing the number of one of these coins, the original two equations you wrote will have two unknowns. These equations can then be solved, and the solutions inspected to see if they both fall into the set of positive integers.
 
  • #4
phyzguy
Science Advisor
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1. There are 100 coins on the table. The coins consist of nickels, dimes, and quarters. The total value is $15.85. How many of each coins are there?



3.

I setup a system of equations.
x = nickels
y = dimes
z = quarters

x + y + z = 100
0.05x + 0.10y + 0.25z = 15.85


I then solve and reduce and get:

x + y + z = 100
y + 5z = 217

I move over variables and get:

x = 100 - y - z
y = 217 - 5z

Since there are more variables than equations I assume that there are many solutions, and try using a parametric representation.


x = 100 - y - z
y = 217 - 5t

where t = any real number

I plug in numbers but end up not getting the value 15.85. The most I can get is somewhere near 14.00.

I was wondering if my setup is wrong.
Is there another way to solve this problem?
I think you made a mistake. The second equation should be:

y = 217 -4z
 
  • #5
Ray Vickson
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1. There are 100 coins on the table. The coins consist of nickels, dimes, and quarters. The total value is $15.85. How many of each coins are there?



3.

I setup a system of equations.
x = nickels
y = dimes
z = quarters

x + y + z = 100
0.05x + 0.10y + 0.25z = 15.85


I then solve and reduce and get:

x + y + z = 100
y + 5z = 217

I move over variables and get:

x = 100 - y - z
y = 217 - 5z

Since there are more variables than equations I assume that there are many solutions, and try using a parametric representation.


x = 100 - y - z
y = 217 - 5t

where t = any real number

I plug in numbers but end up not getting the value 15.85. The most I can get is somewhere near 14.00.

I was wondering if my setup is wrong.
Is there another way to solve this problem?
You can solve for x and y in terms of z, for example, then input various non-negative integer values of z to see if x and y come out as non-negative integers also. I get 16 distinct solutions this way.
 
  • #6
.Scott
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In Nickel Units,
the coins are: nickels(n):1, dimes(d):2, quarters(q):5
the total is: 317
Assuming no nickels, d+q=100, 2d+5q=317, so 3q=117, q=39

There are twenty solutions. You should should be able to work the rest yourself.
 
  • #7
Obviously, t is not any real number. t is going to come from the set of positive integers. You can't have half a nickle, or Sqrt(2) of a quarter.

Instead of introducing yet another variable (t) into your problem, make your parameter one of the coin denominations, either the nickle, the dime, or the quarter. By choosing the number of one of these coins, the original two equations you wrote will have two unknowns. These equations can then be solved, and the solutions inspected to see if they both fall into the set of positive integers.

If there are two unknowns now, will it take away one of the coins. So that either dimes, nickels, or quarter would not be found?
 
  • #8
Since there are more variables than equations I assume that there are many solutions, and try using a parametric representation.


x = 100 - y - z
y = 217 - 5t

where t = any real number


If I set t to be a number will it be correct?

For instance,

z = t ; t = 15
 
  • #9
SteamKing
Staff Emeritus
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All I am saying is that adding an arbitrary variable 't' to the analysis does nothing to advance a solution.

You have 3 variables already, but only two equations involving these three unknowns. Adding a fourth unknown does not help.

Post #6 showed how you could find the number of coins assuming there were no nickles. I think this is outside the scope of the problem, since the problem statement said the coins consisted of nickels, dimes, and quarters. However, if you assume, for instance, that the bowl contains 20 nickels, then you can modify the two equations to determine the number of dimes and quarters. Now, this is not saying that the solutions you get will make sense. You may get a negative number of quarters. If this is the case, then assume a greater number of nickels and repeat the solution.

Hint: This type of problem is easy to analyze with a spreadsheet.
 
  • #10
Ray Vickson
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In Nickel Units,
the coins are: nickels(n):1, dimes(d):2, quarters(q):5
the total is: 317
Assuming no nickels, d+q=100, 2d+5q=317, so 3q=117, q=39

There are twenty solutions. You should should be able to work the rest yourself.
I get a total of 16 solutions, not 20.
 
  • #11
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I get 16 solutions, too.
 
  • #12
.Scott
Homework Helper
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888
I get a total of 16 solutions, not 20.
You're right:
Nickels 0 to 45
Dimes 61 to 1
Quarters 39 to 54

The big clue is that:
3 nickels + 1 quarter = 4 coins (40 cents)
4 dimes = 4 coins (40 cents)
 

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