Assume b represents the number of boys and g represents the number of girls in a classroom. We know that there is at least one boy and one girl, and there are more girls than boys. Which expression would have a larger value?


(1) (gb)/2


(2) (g+b)/2


(3) There is not enough information.


(4) Both expressions are equal

GAMEMASTERX40  Jun 2, 2018
edited by GAMEMASTERX40  Jun 2, 2018


In expressions (1) and (2), there is only one difference: the signage. Since there is at least one boy, the minimum number that b/2 can equal is 1/2 or 0.5. There is no maximum. 


Do you think that subtracting or adding at least 0.5 to the number of girls in the classroom will result in the larger value? Certainly adding it would make it the larger value!


We still have to consider the possibility where they could be equal, though. This prompts the following equation:


\(g-\frac{b}{2}=g+\frac{b}{2}\) For the time being, we will assume that these values are equal. Doing this will allow us to discover how many boys or girls will result in an equilibrium. Notice how the g's cancel out.
\(-\frac{b}{2}=\frac{b}{2}\) Add b/2 to both sides. 
\(b=0\) According to the answer I just got, there would have to be 0 boys in order for \(g-\frac{b}{2}\) to be equal to \(g+\frac{b}{2}\) . This is impossible because the problem states that there is at least one boy. Therefore, (4) is not possible.


This means that the original idea stands: \(g+\frac{b}{2}\) always results in the larger value, in this case. 

TheXSquaredFactor  Jun 2, 2018

So... the answer is (2)? Then the expression (g+b)/2 would mean there are more girls than boys as the larger value.

GAMEMASTERX40  Jun 3, 2018

Yes, you are correct!


I have realized that you have added parentheses to (1) and (2), which were hitherto nonexistent. The logic, however, is the same, and the answer still remains the same, too.  

TheXSquaredFactor  Jun 3, 2018
edited by TheXSquaredFactor  Jun 3, 2018

Yeah, I just forgot to add the parentheses to the problem. Thanks for the help!

GAMEMASTERX40  Jun 3, 2018

6 Online Users


New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.