Mary has $6.00 in nickels, dimes, and quarters. If she has twice as many dimes as quarters and ninety-eight more nickels than dimes, how many coins of each type does she have?

Let's define variables to represent the number of coins Mary has:

- Let $ q $ represent the number of quarters.
- Since Mary has twice as many dimes as quarters, the number of dimes is $ 2q $.
- Since she has 98 more nickels than dimes, the number of nickels is $ 2q + 98 $.

Now, we can write an equation to represent the total value of the coins. Each type of coin has a specific value:

- A quarter is worth $0.25, so the total value of the quarters is $ 0.25q $.
- A dime is worth $0.10, so the total value of the dimes is $ 0.10(2q) = 0.20q $.
- A nickel is worth $0.05, so the total value of the nickels is $ 0.05(2q + 98) = 0.10q + 4.90 $.

The total value of all the coins is $6.00, so we can set up the following equation:

\[
0.25q + 0.20q + (0.10q + 4.90) = 6.00
\]

Now, let's simplify and solve for $ q $:

\[
0.25q + 0.20q + 0.10q + 4.90 = 6.00
\]

\[
0.55q + 4.90 = 6.00
\]

Subtract 4.90 from both sides:

\[
0.55q = 1.10
\]

Divide by 0.55:

\[
q = \frac{1.10}{0.55} = 2
\]

Now that we know Mary has 2 quarters, we can calculate the number of dimes and nickels:

- Number of dimes: $ 2q = 2(2) = 4 $
- Number of nickels: $ 2q + 98 = 2(2) + 98 = 4 + 98 = 102 $

### Final Answer:
- Mary has **2 quarters**.
- Mary has **4 dimes**.
- Mary has **102 nickels**.

Would you like further clarification or details on any part of this solution?

### Related Questions:
1. What if Mary had $8.00 instead of $6.00 in the same ratio of coins?
2. How would the solution change if the number of dimes was triple the number of quarters?
3. How many coins would Mary have in total based on this problem?
4. How would the problem change if Mary had 98 more quarters than dimes instead of nickels?
5. What if Mary had $6.00 in pennies, nickels, and dimes with similar conditions?

### Tip:
To solve word problems efficiently, always define variables and write equations based on the relationships and values given.

Learn how to solve a coin value problem where Mary has $6.00 in quarters, dimes, and nickels. Explore the relationships between the number of coins and their values with a step-by-step solution. Suitable for students in Grades 7-9.

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