A bag contains some green balls, some blue balls, and 7 red balls. The probability of picking one green ball is $\frac{1}{7}$ more than the probability of picking one red ball. The probability of picking one blue ball is $\frac{9}{35}$ more than the probability of picking one red ball. Find the total number of blue balls in the bag.

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**Solution:**

1. Let the total number of balls in the bag be $X$.

2. Number of red balls = 7 (given).

3. Let the number of green balls be $a$ and the number of blue balls be $b$.

4. Probability of picking one red ball = $\frac{7}{X}$.

5. Probability of picking one green ball is $\frac{1}{7}$ more than that of red ball:
   $$
   \frac{a}{X} = \frac{7}{X} + \frac{1}{7}
   $$

6. Probability of picking one blue ball is $\frac{9}{35}$ more than that of red ball:
   $$
   \frac{b}{X} = \frac{7}{X} + \frac{9}{35}
   $$

7. From the first equation:
   $$
   \frac{a}{X} - \frac{7}{X} = \frac{1}{7} \implies \frac{a - 7}{X} = \frac{1}{7} \implies a - 7 = \frac{X}{7} \implies a = 7 + \frac{X}{7}
   $$

8. From the second equation:
   $$
   \frac{b}{X} - \frac{7}{X} = \frac{9}{35} \implies \frac{b - 7}{X} = \frac{9}{35} \implies b - 7 = \frac{9X}{35} \implies b = 7 + \frac{9X}{35}
   $$

9. Since $a$ and $b$ must be integers, $X$ must be a multiple of 35.

10. Let $X = 35$.

11. Calculate $a$ and $b$:
    - $a = 7 + \frac{35}{7} = 7 + 5 = 12$
    - $b = 7 + \frac{9 \times 35}{35} = 7 + 9 = 16$

12. Total balls check:
    $$
    a + b + 7 = 12 + 16 + 7 = 35 = X
    $$

13. Therefore, the total number of blue balls is $16$.

**Answer:** 16

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