# How many coins are there?

• physics=world
In summary, there are twenty possible solutions for the number of nickels, dimes, and quarters on the table with a total value of $15.85. These solutions can be found by setting the number of nickels to 0 and solving for the number of dimes and quarters, then adjusting the number of nickels and repeating the process until all possible solutions have been found. physics=world 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?

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.

physics=world said:
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 physics=world said: 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.

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.

SteamKing said:
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?

physics=world said:
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

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.

.Scott said:
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.

I get 16 solutions, too.

Ray Vickson said:
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)

## 1. How do you determine the total number of coins?

The total number of coins can be determined by counting them one by one or by weighing them and using the average weight of a single coin to calculate the total number.

## 2. Are there any specific types of coins that are not included in the count?

Yes, some coins may not be included in the count such as commemorative coins, collector's coins, or foreign coins that are not in circulation in the country.

## 3. What is the estimated number of coins currently in circulation?

According to the Federal Reserve, there are approximately 140 billion coins in circulation in the United States as of 2021.

## 4. How many coins are minted each year?

The number of coins minted each year varies depending on the demand for currency and the need to replace old or damaged coins. In recent years, the US Mint has produced an average of 14 billion coins annually.

## 5. What is the most common denomination of coins?

The most common denomination of coins is the penny, with over 130 billion in circulation as of 2021. This is followed by the nickel, dime, and quarter, in that order.

• Precalculus Mathematics Homework Help
Replies
7
Views
676
• Precalculus Mathematics Homework Help
Replies
5
Views
811
• Precalculus Mathematics Homework Help
Replies
1
Views
2K
• Precalculus Mathematics Homework Help
Replies
3
Views
2K
• Calculus and Beyond Homework Help
Replies
5
Views
1K
• Precalculus Mathematics Homework Help
Replies
2
Views
981
• Precalculus Mathematics Homework Help
Replies
3
Views
1K
• Precalculus Mathematics Homework Help
Replies
2
Views
771
• Precalculus Mathematics Homework Help
Replies
3
Views
1K
• Precalculus Mathematics Homework Help
Replies
2
Views
833