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

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.

Let's answer this question with the help of second sentence.
If two boys leave and three girls are added, the ratio of boys is still greater than girl.
so we have: X-2/y+3>1
==》 X>y+5,If we have at least 1 girl, so x>6 and the least possible amount for x = 7

Posted from my mobile device

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.