the quarters, nickles, and dimes totaled 20, and their value was $2. 05 How many of each kind were there if there were 3 times as many dimes as quarters?

We are given the following information:
- The total number of coins (quarters, nickels, and dimes) is 20.
- The total value of these coins is $2.05.
- There are 3 times as many dimes as quarters.

Let:
- $ q $ represent the number of quarters.
- $ n $ represent the number of nickels.
- $ d $ represent the number of dimes.

### Step 1: Set up the system of equations based on the problem.

1. **Total number of coins**:  
   The total number of quarters, nickels, and dimes is 20:
   \[
   q + n + d = 20
   \]

2. **Total value of the coins**:  
   The value of quarters, nickels, and dimes adds up to $2.05, which is 205 cents:
   \[
   25q + 5n + 10d = 205
   \]
   (where 25 represents the value of a quarter in cents, 5 represents the value of a nickel in cents, and 10 represents the value of a dime in cents.)

3. **Relationship between dimes and quarters**:  
   There are 3 times as many dimes as quarters:
   \[
   d = 3q
   \]

### Step 2: Substitute $ d = 3q $ into the first two equations.

1. Substituting into the total number of coins equation:
   \[
   q + n + 3q = 20 \implies 4q + n = 20
   \]
   This simplifies to:
   \[
   n = 20 - 4q
   \]

2. Substituting into the total value equation:
   \[
   25q + 5n + 10(3q) = 205 \implies 25q + 5n + 30q = 205 \implies 55q + 5n = 205
   \]
   Simplifying this:
   \[
   11q + n = 41
   \]

### Step 3: Solve the system of equations.

Now, we have the system:
1. $ n = 20 - 4q $
2. $ 11q + n = 41 $

Substitute $ n = 20 - 4q $ into the second equation:
\[
11q + (20 - 4q) = 41
\]
Simplifying:
\[
11q + 20 - 4q = 41 \implies 7q + 20 = 41 \implies 7q = 21 \implies q = 3
\]

### Step 4: Find $ n $ and $ d $.

Now that we know $ q = 3 $, substitute it back into the equation for $ n $:
\[
n = 20 - 4(3) = 20 - 12 = 8
\]
And for $ d $:
\[
d = 3q = 3(3) = 9
\]

### Step 5: Conclusion.

- The number of quarters is $ q = 3 $.
- The number of nickels is $ n = 8 $.
- The number of dimes is $ d = 9 $.

Would you like more details or have any questions?

Here are 5 related questions to expand on this topic:
1. What if there were twice as many dimes as quarters instead?
2. How would the problem change if the total value of coins was $2.50?
3. What if the total number of coins was increased to 25, keeping the same ratio between dimes and quarters?
4. How would the number of coins change if the number of nickels was reduced by half?
5. Can you set up a similar problem where the total value is $3.00?

**Tip**: When dealing with coin problems, always check both the number of coins and the total value to ensure the solution is correct.

The quarters, nickels, and dimes totaled 20, and their value was $2.05. How many of each kind were there if there were 3 times as many dimes as quarters?

Learn how to solve a coin problem where the total number of quarters, nickels, and dimes is 20, their value is $2.05, and the number of dimes is three times the number of quarters. Step-by-step solution for students in Grades 6-8.

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