A bag contain 20 balls which x are red balls and y are blue balls. When 5 red balls were added and 10 blue balls were removed, the probability of getting a blue ball is ⅓. Find x and y.

We are given a problem where a bag contains a total of 20 balls, consisting of \(x\) red balls and \(y\) blue balls. When 5 red balls are added and 10 blue balls are removed, the probability of selecting a blue ball becomes \( \frac{1}{3} \). We are tasked with finding \(x\) and \(y\).

### Step-by-step solution:

1. **Initial conditions**:
   - There are 20 balls in total: 
   \[
   x + y = 20 \tag{1}
   \]
   where \(x\) is the number of red balls and \(y\) is the number of blue balls.

2. **New configuration after changes**:
   - 5 red balls are added: The new number of red balls is \( x + 5 \).
   - 10 blue balls are removed: The new number of blue balls is \( y - 10 \).
   - The total number of balls after these changes is \( x + 5 + (y - 10) = x + y - 5 = 20 - 5 = 15 \).

3. **Probability condition**:
   - The probability of selecting a blue ball from the new set of balls is given to be \( \frac{1}{3} \). This means the ratio of blue balls to the total number of balls is \( \frac{y - 10}{15} \):
   \[
   \frac{y - 10}{15} = \frac{1}{3} \tag{2}
   \]

4. **Solving the system of equations**:
   From equation \((2)\), solve for \(y\):
   \[
   \frac{y - 10}{15} = \frac{1}{3}
   \]
   Cross-multiply:
   \[
   3(y - 10) = 15
   \]
   \[
   3y - 30 = 15
   \]
   \[
   3y = 45
   \]
   \[
   y = 15
   \]

   Now substitute \(y = 15\) into equation \((1)\):
   \[
   x + 15 = 20
   \]
   \[
   x = 5
   \]

Thus, the number of red balls is \( x = 5 \) and the number of blue balls is \( y = 15 \).

### Final Answer:
- \( x = 5 \) (red balls)
- \( y = 15 \) (blue balls)

Would you like further details or have any questions? Here are 5 related questions to think about:

1. What if the probability of drawing a blue ball was \( \frac{2}{5} \) instead of \( \frac{1}{3} \)?
2. How would the problem change if 7 red balls were added instead of 5?
3. What if 12 blue balls were removed instead of 10?
4. Can you generalize the approach for different total numbers of balls in the bag?
5. How does the problem change if we draw two balls instead of one?

**Tip**: Always double-check the changes in total numbers when working with probabilities in similar problems.

A bag contains 20 balls, x of which are red and y are blue. After adding 5 red balls and removing 10 blue balls, the probability of drawing a blue ball is 1/3. Find x and y.

This problem involves finding the number of red and blue balls in a bag with a total of 20 balls. After adding and removing balls, the probability of drawing a blue ball becomes 1/3. Learn how to solve this problem step-by-step using algebraic and probability concepts. Suitable for Grades 9-10 students.

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