Given below is the number of male and female students in classes A, B, and C. Some data are missing which you need to find based on the instructions provided.

| | A | B | C |
|--------|----|----|----|
| Boys | 50 | -- | -- |
| Girls | -- | 80 | 60 |

Note:

1. The probability of selecting a boy from class A is $\frac{5}{12}$.
2. The probability of selecting a boy from all the boys of all classes is $\frac{14}{19}$ such that the boy selected is either from class B or class C.
3. The probability of selecting a boy from class B is equal to the probability of selecting a boy from class C.

**Question:** How many boys are there in class C?

Question

Given below is the number of male and female students in classes A, B, and C. Some data are missing which you need to find based on the instructions provided.

|        | A  | B  | C  |
|--------|----|----|----|
| Boys   | 50 | -- | -- |
| Girls  | -- | 80 | 60 |

Note:

1. The probability of selecting a boy from class A is $\frac{5}{12}$.
2. The probability of selecting a boy from all the boys of all classes is $\frac{14}{19}$ such that the boy selected is either from class B or class C.
3. The probability of selecting a boy from class B is equal to the probability of selecting a boy from class C.

**Question:** How many boys are there in class C?

Accepted Answer

**Solution:**

1. Let the number of boys in class B be $x$ and in class C be $y$.

2. From the problem, boys in class A = 50.

3. Probability of selecting a boy from class A is given as $\frac{5}{12}$. This probability is:
   $$\frac{	ext{boys in A}}{	ext{total students in A}} = \frac{50}{50 + 	ext{girls in A}} = \frac{5}{12}$$

4. Let the number of girls in class A be $g_A$. Then:
   $$\frac{50}{50 + g_A} = \frac{5}{12}$$

5. Cross-multiplying:
   $$12 	imes 50 = 5 	imes (50 + g_A)$$
   $$600 = 250 + 5 g_A$$
   $$5 g_A = 350$$
   $$g_A = 70$$

6. So, girls in class A = 70.

7. Total boys in all classes = $50 + x + y$.

8. Total girls in all classes = $70 + 80 + 60 = 210$.

9. Total students in all classes = boys + girls = $(50 + x + y) + 210$.

10. Probability of selecting a boy from all boys such that the boy is from class B or C is $\frac{14}{19}$.

This means:
   $$\frac{x + y}{50 + x + y} = \frac{14}{19}$$

11. Cross-multiplying:
   $$19(x + y) = 14(50 + x + y)$$
   $$19x + 19y = 700 + 14x + 14y$$
   $$19x - 14x + 19y - 14y = 700$$
   $$5x + 5y = 700$$
   $$x + y = 140$$

12. Given that the probability of selecting a boy from class B equals that from class C:

$$\frac{x}{x + 80} = \frac{y}{y + 60}$$

13. Cross-multiplying:
   $$x(y + 60) = y(x + 80)$$
   $$xy + 60x = xy + 80y$$
   $$60x = 80y$$
   $$\frac{x}{y} = \frac{80}{60} = \frac{4}{3}$$

14. Let $x = 4k$ and $y = 3k$.

15. From step 11:
   $$x + y = 140$$
   $$4k + 3k = 140$$
   $$7k = 140$$
   $$k = 20$$

16. Therefore:
   $$x = 4 	imes 20 = 80$$
   $$y = 3 	imes 20 = 60$$

**Answer:** Number of boys in class C is $\boxed{60}$.

Answer

Answer

Answer