The ratio of boys to girls in a certain coed school is greater than 1.

Given, (b - 2)/(g + 3) > 1
—> b - 2 > g + 3
—> b > g + 5

number is girls, g = 1 at least
—> b > 1 + 5
—> b > 6
—> Minimum value of boys, b = 7

IMO Option D

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As per question, let boys = b and girls = g
1. \(\frac{(b)}{(g)}\) >1 => b > g
2.From the second condition, we can get that there should be at least 2 boys in the school. = > b >2
now, if 3 girls are added even then the ratio favors to the boys that means:

\(\frac{(b-2)}{(g+3)}\) > 1
Given, there was at least one girl in the school => g>= 1


Using options:
1. We can eliminate 1, as this will give -1 boys which is not possible.
2. If b=4 => \(\frac{(4-2)}{(1+3)}\) => \(\frac{(2)}{(4)}\) => \(\frac{(1)}{(2)}\)
that is violating the condition.
3. If b=6 => \(\frac{(6-2)}{(1+3)}\) => \(\frac{(4)}{(4)}\) => 1
4. If b=7 => \(\frac{(7-2)}{(1+3)}\) => \(\frac{(5)}{(4)}\) =>\(\frac{(b)}{(g)}\) >1
This holds true.
We don`t need to check other option, as that is large number than 7.
So, least number of boys were 7 in the school.

IMO the answer is D.

Please hit kudos if you like the solution.

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