Lucky found $8.25 in pennies, nickels, dimes, and quarters while walking home from school one week. When she depostied this money in the bank, she noticed that she had twice as many nickels as pennies, 1 fewer dime than nickels, and 1 more quarter than nickels. How many quarters did Lucky find that week?

Options

A. 3
B. 9
C. 16
D. 21

Answer

To solve the problem, we will define variables for each type of coin Lucky found and set up equations based on the information given.

### Step-by-Step Solution

1. **Define Variables:**
Let:
- \( p \) = number of pennies
- \( n \) = number of nickels
- \( d \) = number of dimes
- \( q \) = number of quarters

2. **Set Up Relationships:**
According to the problem:
- Lucky has twice as many nickels as pennies:
\[ n = 2p \]
- She has one fewer dime than nickels:
\[ d = n - 1 \]
- She has one more quarter than nickels:
\[ q = n + 1 \]

3. **Total Value of Coins:**
The total amount of money is $8.25, which is equal to 825 cents. The value of the coins can be expressed as:
\[ p + 5n + 10d + 25q = 825 \]

4. **Substitute Relationships into the Total Value Equation:**
Substitute \( n = 2p \), \( d = n - 1 = 2p - 1 \), and \( q = n + 1 = 2p + 1 \) into the total value equation:
\[ p + 5(2p) + 10(2p - 1) + 25(2p + 1) = 825 \]

5. **Simplify the Equation:**
Expanding the equation:
\[ p + 10p + 20p - 10 + 50p + 25 = 825 \]
Combine like terms:
\[ 81p + 15 = 825 \]

6. **Solve for \( p \):**
Subtract 15 from both sides:
\[ 81p = 810 \]
Divide by 81:
\[ p = 10 \]

7. **Find the Number of Nickels, Dimes, and Quarters:**
- Number of nickels:
\[ n = 2p = 2 \times 10 = 20 \]
- Number of dimes:
\[ d = n - 1 = 20 - 1 = 19 \]
- Number of quarters:
\[ q = n + 1 = 20 + 1 = 21 \]

### Final Answer:
Lucky found **21 quarters** that week.