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

For example,

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

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Davefor MBA