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To solve the problem of how many dimes John has in his pocket, let's define the variables and set up the equations based on the given information.

Let:
- $ n $ be the number of nickels.
- $ d $ be the number of dimes.

We have two pieces of information:
1. The total number of coins is 50.
2. The total value of the coins is $3.50.

From this, we can set up the following equations:

1. $ n + d = 50 $
2. $ 0.05n + 0.10d = 3.50 $

We can solve these equations step-by-step:

### Step 1: Solve for $ n $ in terms of $ d $
From the first equation:
\[ n = 50 - d \]

### Step 2: Substitute $ n $ into the second equation
\[ 0.05(50 - d) + 0.10d = 3.50 \]

### Step 3: Simplify and solve for $ d $
\[ 0.05 \times 50 - 0.05d + 0.10d = 3.50 \]
\[ 2.50 + 0.05d = 3.50 \]
\[ 0.05d = 1.00 \]
\[ d = 20 \]

So, John has 20 dimes in his pocket.

Thus, the correct answer is:

C) 20

Would you like more details or have any questions about this solution? Here are some related questions to consider:

1. How many nickels does John have in his pocket?
2. What is the total value of the nickels in John's pocket?
3. How would the equations change if the total value was $4.00 instead?
4. What if John had quarters instead of dimes; how would you set up the equations?
5. Can you set up and solve the equations using a matrix approach?
6. How can you check your solution to ensure it's correct?
7. What if John had a combination of nickels, dimes, and quarters? How would you solve that?
8. How can you graphically represent the solution to this problem?

**Tip:** When solving systems of equations, always double-check your substitutions and arithmetic to ensure accuracy.

Learn how to solve a problem involving the number of dimes John has in his pocket using systems of linear equations and algebra. This problem covers equations related to coin values and is suitable for students in Grades 7-9.

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