The average value of all the pennies, nickels, dimes, and quarters in Paula's purse is 20 cents. If she had one more quarter, the average value would be 21 cents. How many dimes does she have in her purse?

To solve this problem, let's define the variables and set up equations based on the given conditions.

Let:
- \( p \) be the number of pennies.
- \( n \) be the number of nickels.
- \( d \) be the number of dimes.
- \( q \) be the number of quarters.

The value of the coins in cents is:
- Pennies: \( p \) cents
- Nickels: \( 5n \) cents
- Dimes: \( 10d \) cents
- Quarters: \( 25q \) cents

The total number of coins is \( p + n + d + q \).

The total value of the coins is \( p + 5n + 10d + 25q \) cents.

### Step 1: Average value of 20 cents
The average value of all the coins is 20 cents. Thus,
\[
\frac{p + 5n + 10d + 25q}{p + n + d + q} = 20
\]

Multiplying both sides by \( p + n + d + q \) gives:
\[
p + 5n + 10d + 25q = 20(p + n + d + q) \quad \text{(1)}
\]

### Step 2: Average value of 21 cents after adding one quarter
If Paula had one more quarter, the total value and the total number of coins would be:
- Total value: \( p + 5n + 10d + 25q + 25 \) cents
- Total number of coins: \( p + n + d + q + 1 \)

The new average value is 21 cents. Thus,
\[
\frac{p + 5n + 10d + 25q + 25}{p + n + d + q + 1} = 21
\]

Multiplying both sides by \( p + n + d + q + 1 \) gives:
\[
p + 5n + 10d + 25q + 25 = 21(p + n + d + q + 1)
\]

Expanding and simplifying:
\[
p + 5n + 10d + 25q + 25 = 21p + 21n + 21d + 21q + 21
\]

### Step 3: Simplifying the second equation
Rearranging terms gives:
\[
p + 5n + 10d + 25q + 25 = 21p + 21n + 21d + 21q + 21
\]

\[
4 = 20p + 16n + 11d - 4q
\]

So we have:
\[
4 = 20p + 16n + 11d - 4q
\]

### Step 4: Solve the system of equations
We have two equations:
1. \( p + 5n + 10d + 25q = 20(p + n + d + q) \)
2. \( 4 = 20p + 16n + 11d - 4q \)

From equation (1):
\[
p + 5n + 10d + 25q = 20p + 20n + 20d + 20q
\]
\[
19p + 15n + 10d - 5q = 0
\]
\[
19p + 15n + 10d = 5q
\]
\[
19p + 15n + 2d = 25
\]

Now solve this system:
1. \( 19p + 15n + 10d = 5q \)
2. \( 4 = 20p + 16n + 11d - 4q \)

By solving these equations, we get:
### Solve for 4
\( 20p + 16n + 11d - 4q = 4 \)

Thus,
\[
25 - 4q = 4
\]
So \( q = 4 \).

Now,
\[
19p + 15n + 10d = 25

So
d = 3.

Therefore, the number of dimes is 3.

Would you like details on any part of this solution or have any questions?

1. How to solve linear equations?
2. What's the value of 25 dimes?
3. How to set up algebraic equations from word problems?
4. How to find the average of numbers?
5. What's a penny worth?

**Tip:** When solving word problems, carefully define and interpret variables to set up accurate equations.

Learn how to solve Paula's purse problem involving pennies, nickels, dimes, and quarters. Discover how to calculate averages and set up algebraic equations to find the number of dimes. Suitable for students in Grades 7-9.

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