Let's denote the number of dimes Wendell has as \( d \). According to the problem, Wendell has 3 times as many nickels as dimes. Therefore, the number of nickels can be represented as \( 3d \).
Now, we know the values of dimes and nickels:
- Each dime is worth $0.10, so the total value of the dimes is \( 0.10d \) dollars.
- Each nickel is worth $0.05, so the total value of the nickels is \( 0.05(3d) = 0.15d \) dollars.
The total amount of money Wendell has in nickels and dimes can be expressed as: \[ 0.10d + 0.15d = 0.25d \]
According to the problem, Wendell has a total of $12 (or 1200 cents) in nickels and dimes: \[ 0.25d = 12 \]
To find \( d \), we will solve the equation: \[ d = \frac{12}{0.25} = 12 \div 0.25 = 12 \times 4 = 48 \]
Thus, Wendell has \( 48 \) dimes.
To confirm, we can check the number of nickels he has: \[ 3d = 3 \times 48 = 144 \]
Now, let's find the total value:
- The value of dimes: \( 48 \times 0.10 = 4.80 \) dollars.
- The value of nickels: \( 144 \times 0.05 = 7.20 \) dollars.
Adding these amounts gives: \[ 4.80 + 7.20 = 12.00 \text{ dollars} \]
So the calculations check out.
Therefore, Wendell has \( \boxed{48} \) dimes.
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