In a school 3/8 of the number of students are girls and the rest are boys. One-third of the number of boys are below 10 years and 2/3 the number if girls are also below 10 years. If the number of students of age 10 or more years is 260. then the number of boys in the school is:
300
The problem describes a school with a certain number of students, divided into boys and girls. We are given the fraction of students who are girls and, consequently, the fraction who are boys. The students are also categorized by age: those below 10 years and those aged 10 or more years. We are given the fraction of boys below 10 and the fraction of girls below 10. Finally, we know the total number of students aged 10 or more, and we need to find the total number of boys in the school.
Let's break down the information given:
Let the total number of students in the school be $T$.
Based on the given fractions, we can express the number of boys and girls in terms of $T$:
Now, let's find the fraction of students in different age groups for both boys and girls:
We know that the total number of students aged 10 or more years is 260. This total is the sum of boys aged 10 or more and girls aged 10 or more.
Number of students aged 10 or more = (Fraction of boys aged 10 or more) + (Fraction of girls aged 10 or more)
$260 = \frac{5}{12} T + \frac{1}{8} T$
To add the fractions on the right side, we find a common denominator for 12 and 8, which is 24.
So, $260 = \frac{10}{24} T + \frac{3}{24} T = \frac{13}{24} T$
Now, we can solve for $T$:
$T = 260 \times \frac{24}{13}$
Since $260 \div 13 = 20$, we have:
$T = 20 \times 24 = 480$
The total number of students in the school is 480.
The problem asks for the number of boys in the school. We found earlier that the number of boys is $\frac{5}{8}$ of the total number of students ($T$).
Number of boys = $\frac{5}{8} T$
Substitute the value of $T = 480$:
Number of boys = $\frac{5}{8} \times 480$
Since $480 \div 8 = 60$, we have:
Number of boys = $5 \times 60 = 300$
Thus, the number of boys in the school is 300.
| Category | Fraction of Total Students | Number of Students (if T=480) |
|---|---|---|
| Total Students | $T$ | 480 |
| Girls | $\frac{3}{8} T$ | $\frac{3}{8} \times 480 = 180$ |
| Boys | $\frac{5}{8} T$ | $\frac{5}{8} \times 480 = 300$ |
| Boys below 10 | $\frac{5}{24} T$ | $\frac{5}{24} \times 480 = 100$ |
| Boys ≥ 10 | $\frac{5}{12} T$ | $\frac{5}{12} \times 480 = 200$ |
| Girls below 10 | $\frac{1}{4} T$ | $\frac{1}{4} \times 480 = 120$ |
| Girls ≥ 10 | $\frac{1}{8} T$ | $\frac{1}{8} \times 480 = 60$ |
| Total ≥ 10 | $\frac{13}{24} T$ | $200 + 60 = 260$ (Matches given information) |
| Step | Description | Applied in this Problem |
|---|---|---|
| 1 | Identify total unknown quantity (e.g., Total Students). Represent it with a variable (e.g., $T$). | Let Total Students = $T$. |
| 2 | Express different categories (e.g., boys/girls) as fractions of the total. | Girls = $\frac{3}{8} T$, Boys = $\frac{5}{8} T$. |
| 3 | Express sub-categories (e.g., age groups within boys/girls) as fractions of the total. | Boys < 10 = $\frac{1}{3} \times \frac{5}{8} T = \frac{5}{24} T$, Boys ≥ 10 = $\frac{5}{8} T - \frac{5}{24} T = \frac{5}{12} T$. Girls < 10 = $\frac{2}{3} \times \frac{3}{8} T = \frac{1}{4} T$, Girls ≥ 10 = $\frac{3}{8} T - \frac{1}{4} T = \frac{1}{8} T$. |
| 4 | Use the given numerical information to form an equation involving the variable $T$. | Students ≥ 10 = (Boys ≥ 10) + (Girls ≥ 10). $260 = \frac{5}{12} T + \frac{1}{8} T = \frac{13}{24} T$. |
| 5 | Solve the equation to find the value of the total quantity $T$. | $\frac{13}{24} T = 260 \implies T = 260 \times \frac{24}{13} = 480$. |
| 6 | Use the value of $T$ to find the specific number asked for in the question. | Number of boys = $\frac{5}{8} T = \frac{5}{8} \times 480 = 300$. |
Word problems involving fractions require careful reading to understand what each fraction refers to (e.g., fraction of the total, or fraction of a subgroup). Here are some key points:
Practice with different types of fraction problems helps in quickly identifying how the fractions relate to the overall total and setting up the correct calculations.
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