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We are given the following information:

1. The total value of Nancy's coins is 85 pence.
2. She has only 2 pence coins and 5 pence coins.
3. The ratio of the number of 2 pence coins to the number of 5 pence coins is 1:3.
4. We are asked to find how many more 5 pence coins she has than 2 pence coins.

### Step-by-Step Solution:

Let:
- \( x \) represent the number of 2 pence coins.
- \( 3x \) represent the number of 5 pence coins (since the ratio of 2 pence coins to 5 pence coins is 1:3).

The total value of the coins is 85 pence, so we can write the equation:

\[
2x + 5(3x) = 85
\]

Simplifying this equation:

\[
2x + 15x = 85
\]

\[
17x = 85
\]

Solving for \( x \):

\[
x = \frac{85}{17} = 5
\]

Thus, Nancy has:
- \( x = 5 \) 2 pence coins.
- \( 3x = 3 \times 5 = 15 \) 5 pence coins.

Finally, the number of additional 5 pence coins compared to 2 pence coins is:

\[
15 - 5 = 10
\]

### Conclusion:
Nancy has 10 more 5 pence coins than 2 pence coins.

Would you like further details or have any questions?

Here are 5 related questions to expand your understanding:
1. What would be the total number of coins if Nancy had the same number of 2 pence coins as 5 pence coins?
2. How would the solution change if the ratio were 2:3 instead of 1:3?
3. What would the total value be if Nancy had only 5 pence coins?
4. How would the problem change if the total value were 100 pence instead of 85 pence?
5. If Nancy had 7 more 5 pence coins than 2 pence coins, how many coins would she have in total?

**Tip**: When solving word problems involving ratios, always represent the unknown quantities in terms of a variable to set up equations effectively.

Nancy has some coins with a total value of 85 pence. She has only 2 pence coins and 5 pence coins. The ratio of the number of 2 pence coins to the number of 5 pence coins is 1:3. Nancy has more 5 pence coins than 2 pence coins. How many more?

Learn how to solve a problem involving the total value of coins in pence, given a ratio of 2 pence and 5 pence coins. This step-by-step guide uses algebraic techniques to find the difference in the number of coins. Suitable for students in Grades 6-8.

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