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Let's break this down step by step.

1. **Initial ratio**: The ratio of snails to newts is given as \( 3:4 \). This means for every 3 snails, there are 4 newts. Let the number of snails be \( 3x \) and the number of newts be \( 4x \), where \( x \) is a common factor.

2. **After newts enter**: After 12 more newts enter the pond, the new ratio of snails to newts becomes \( 3:5 \). The number of snails remains \( 3x \), but the number of newts becomes \( 4x + 12 \).

3. **Setting up the equation**: 
   \[
   \frac{\text{snails}}{\text{newts}} = \frac{3x}{4x + 12} = \frac{3}{5}
   \]
   Cross-multiply to eliminate the fractions:
   \[
   3(4x + 12) = 5(3x)
   \]
   Expanding both sides:
   \[
   12x + 36 = 15x
   \]
   Solving for \( x \):
   \[
   36 = 3x
   \]
   \[
   x = 12
   \]

4. **Calculating the number of snails**: 
   The number of snails is \( 3x \), so:
   \[
   3 \times 12 = 36
   \]

Thus, there are **36 snails** in the pond.

Would you like more details on any step?

Here are some related questions you might find helpful:

1. What is a ratio, and how is it applied in real-life problems?
2. How do you solve word problems involving ratios and proportions?
3. Can you explain more about cross-multiplication and its role in solving ratio problems?
4. What if the ratio of newts to snails were given instead? How would that affect the solution?
5. How do changes in numbers affect the ratio, and how is this used in algebraic equations?

**Tip:** When working with ratios, always express the values in terms of a common factor (like \( x \)) to simplify solving equations.

In a pond, the ratio of snails to newts was 3:4. 12 more newts then entered the pond, and the ratio of snails to newts became 3:5. Work out how many snails are in the pond.

Work out how many snails are in the pond by solving a ratio problem where the number of snails and newts is given, and 12 more newts enter the pond. Learn how to use cross-multiplication and proportions to find the solution. Suitable for students in Grades 7-9.

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