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

Question

The ratio of boys to girls in a certain coed school is greater than 1. When 2 boys leave and 3 girls are added to the school, the ratio still favors boys. What is the least number of boys that could have been originally enrolled in the school, assuming there was originally at least one girl?

(A) 1
(B) 4
(C) 6
(D) 7
(E) 13

Hi, i didnt really understand what the question is asking here. Is the minimum number of boys that are enrolled before the 2 left?

(A) 1
(B) 4
(C) 6
(D) 7
(E) 13

Mohammadmo · Answer

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 astill 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

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Mohammadmo · Answer

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

CareerGeek · Answer

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

RajatVerma1392 · Answer

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.

MegB07 · Answer

Hi, i didnt really understand what the question is asking here. Is the minimum number of boys that are enrolled before the 2 left?

Bunuel · Answer

The question is asking about the number of boys  before  the change (before 2 boys left and 3 girls were added).

The condition is: even after 2 boys leave and 3 girls join, the ratio still favors boys. That translates to:

(B - 2)/(G + 3) > 1
B > G + 5

Since there must have been at least 1 girl originally (G >= 1), the smallest possible value of B is greater than 6.

So the minimum number of boys originally enrolled is 7.

Answer: D.

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

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

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