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If male students passed in Class A and Class B are in ratio 1:3 respectively and failed male students of class A and Class B are equal, then total male students who got passed from both classes together are what percent of total students of both the classes?

Question

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If male students passed in Class A and Class B are in ratio 1:3 respectively and failed male students of class A and Class B are equal, then total male students who got passed from both classes together are what percent of total students of both the classes?

Accepted Answer

**Solution:**

1. Let the number of male students who passed in Class A be $y$. Then, the number of male students who passed in Class B is $3y$ (given ratio $1:3$).

2. Let the total male students in Class A be $70x$ and in Class B be $90x$ (assuming total students proportional to $x$ for calculation).

3. Since the number of failed male students in Class A and Class B are equal, we have:
   $$70x - y = 90x - 3y$$

4. Simplify the equation:
   $$70x - y = 90x - 3y$$
   $$-y + 3y = 90x - 70x$$
   $$2y = 20x$$
   $$y = 10x$$

5. Total male students who passed from both classes:
   $$y + 3y = 4y = 4 	imes 10x = 40x$$

6. Total students in both classes:
   $$70x + 90x = 160x$$

7. Required percentage:
   $$\frac{40x}{160x} 	imes 100 = 25\%$$

**Note:** The original problem's total students were $280x$, so assuming total students as $280x$ (including females), the percentage becomes:

$$\frac{4y}{280x} 	imes 100 = \frac{4 	imes 10x}{280x} 	imes 100 = \frac{40x}{280x} 	imes 100 = \frac{40}{280} 	imes 100 = \frac{2}{14} 	imes 100 = 14 \frac{2}{7} \%$$

**Answer:** $14 \frac{2}{7} \%$

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