Let's denote the number of boys in the school as \( b \). According to the problem, the number of girls \( g \) can be expressed as:
\[ g = b + 250 \]
The total number of pupils in the school is given as 3000:
\[ b + g = 3000 \]
Substituting the expression for \( g \) from the first equation into the second equation:
\[ b + (b + 250) = 3000 \]
This simplifies to:
\[ 2b + 250 = 3000 \]
Subtracting 250 from both sides gives:
\[ 2b = 2750 \]
Dividing both sides by 2 to solve for \( b \):
\[ b = 1375 \]
Now, substituting back to find the number of girls:
\[ g = b + 250 = 1375 + 250 = 1625 \]
To find the percentage of pupils who are boys, we calculate:
\[ \text{Percentage of boys} = \frac{b}{\text{Total pupils}} \times 100 = \frac{1375}{3000} \times 100 \]
To express the fraction \( \frac{1375}{3000} \) in simplest form, we can simplify it:
Finding the greatest common divisor (GCD) of 1375 and 3000:
- Factorizing:
- \( 1375 = 5^2 \times 11 \)
- \( 3000 = 3 \times 10^3 = 3 \times (2 \times 5^3) = 2^3 \times 3^1 \times 5^3 \)
The GCD is \( 5^2 = 25 \).
Now, simplifying:
\[ \frac{1375 \div 25}{3000 \div 25} = \frac{55}{120} \]
Thus, the fraction of boys in the school is \( \frac{55}{120} \).
To convert this to a percentage, we find:
\[ \frac{55}{120} \times 100 \approx 45.83% \]
So, in fraction form, the percentage of boys in the school is:
\[ \frac{55}{120} \text{ (or approximately 45.83%)} \]
Thus, the answer is:
\[ \frac{55}{120} \text{ of the pupils were boys.} \]
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