Question:
The combined mean age of a group of boys and girls in a school is 12. If the mean of the boys in that group is 14 and the mean of the girls is 9, then the percentage of boys in that group is
The combined mean age of a group of boys and girls in a school is 12. If the mean of the boys in that group is 14 and the mean of the girls is 9, then the percentage of boys in that group is
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Logic Tip: Use the Alligation Method for a much faster solution. Write the boys' mean (14) and girls' mean (9) on the top, and the combined mean (12) in the center. The ratio of boys to girls is the diagonal differences: $(12 - 9) : (14 - 12) = 3 : 2$.
Updated On: Apr 27, 2026
- 24%
- 36%
- 40%
- 60%
- 85%
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The Correct Option is D
Solution and Explanation
Concept:
The combined mean (weighted average) of two groups is given by the formula: $$\bar{x} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2}$$ where $n_1, n_2$ are the number of elements in each group and $\bar{x}_1, \bar{x}_2$ are their respective means.
Step 1: Set up the equation using the given means.
Let $n_b$ be the number of boys and $n_g$ be the number of girls. We are given: Combined mean $\bar{x} = 12$ Mean of boys $\bar{x}_b = 14$ Mean of girls $\bar{x}_g = 9$ Substitute these into the combined mean formula: $$12 = \frac{14n_b + 9n_g}{n_b + n_g}$$
Step 2: Solve for the ratio of boys to girls.
Multiply both sides by $(n_b + n_g)$ to eliminate the denominator: $$12(n_b + n_g) = 14n_b + 9n_g$$ $$12n_b + 12n_g = 14n_b + 9n_g$$ Rearrange the terms to group the boys and girls: $$12n_g - 9n_g = 14n_b - 12n_b$$ $$3n_g = 2n_b$$ From this, we find the ratio of boys to girls: $$\frac{n_b}{n_g} = \frac{3}{2}$$
Step 3: Calculate the percentage of boys.
The ratio $n_b : n_g = 3 : 2$ means out of every 5 students ($3 + 2$), 3 are boys. To find the percentage of boys: $$\text{Percentage} = \left(\frac{3}{5}\right) \times 100$$ $$\text{Percentage} = 3 \times 20 = 60$$
The combined mean (weighted average) of two groups is given by the formula: $$\bar{x} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2}$$ where $n_1, n_2$ are the number of elements in each group and $\bar{x}_1, \bar{x}_2$ are their respective means.
Step 1: Set up the equation using the given means.
Let $n_b$ be the number of boys and $n_g$ be the number of girls. We are given: Combined mean $\bar{x} = 12$ Mean of boys $\bar{x}_b = 14$ Mean of girls $\bar{x}_g = 9$ Substitute these into the combined mean formula: $$12 = \frac{14n_b + 9n_g}{n_b + n_g}$$
Step 2: Solve for the ratio of boys to girls.
Multiply both sides by $(n_b + n_g)$ to eliminate the denominator: $$12(n_b + n_g) = 14n_b + 9n_g$$ $$12n_b + 12n_g = 14n_b + 9n_g$$ Rearrange the terms to group the boys and girls: $$12n_g - 9n_g = 14n_b - 12n_b$$ $$3n_g = 2n_b$$ From this, we find the ratio of boys to girls: $$\frac{n_b}{n_g} = \frac{3}{2}$$
Step 3: Calculate the percentage of boys.
The ratio $n_b : n_g = 3 : 2$ means out of every 5 students ($3 + 2$), 3 are boys. To find the percentage of boys: $$\text{Percentage} = \left(\frac{3}{5}\right) \times 100$$ $$\text{Percentage} = 3 \times 20 = 60$$
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