Ok, so here is what I understand from what you guys are saying.
We know A=21
We therefore know B+C+D=69
The question never states how they could have voted.
Consequently, we could have seen a multitude of possibilities for B+C+D
1+1+67=69 ---> one is higher than 21, thus A is not highest
22+1+46=69 --->one is higher than 21, thus A is not highest
23+23+23=69 ---> all three are higher than 21, thus A is not higest.
I think the key to the answer is knowing
1) Disporportioned possibilities always leads to at least one higher than A
2) Evenly porportioned possibilities always leads to all three higher than A
I think one of the explanations someone gave explained the disporportioned possibility as a reason to choose B, but not taking into account the porportioned possibility could have still gotten the question wrong.
But since the porportioned result happens to correlate with what a disporportioned result is, we can still choose B as the answer.
This was a good question to think through. Thanks for question and dialogue guys. Amongst your conversation, I was able to conclude with what I just wrote. Very helpful. Thanks!
But this last option is the rea