There were 5 candidates (Alexa, Bill, Charlie, Dan, and Ernie) vying

We have
40>B>C>D>E>=1 and B+C+D+E=60
To maximize D, B and C should be as close to D as possible. E should be 1.
How about B=D+2, C=D+1, and E=1
(D+2)+(D+1)+(D) + (1)=60 => 3D = 56 => D = 18.666666666...
18.66666... is the maximum D, but D must be an integer.
D has to be 18
40,21,20,18,1
OR
40,20,19,18,3

The answer is (C).

Given: There were 5 candidates (Alexa, Bill, Charlie, Dan, and Ernie) vying for student council, and 100 total votes were cast. Everyone received at least one vote, and no two candidates received the same number of votes. Alexa won the election with 40 votes, Bill came in 2nd, Charlie in 3rd, Dan in 4th, and Ernie in last.

Asked: What is the greatest number of votes that Dan could have received?

To maximize Dan's votes, we have to maximize votes received by Bill, Charlie and Dan and reduce their differences to minimum 1 and to minimise votes received by Ernie to 1 vote.

Votes received by Alexa = 40
Let votes received by Dan be x.
Votes received by Bill = x+2
Votes received by Charlie = x+1
Votes recieved by Dan = x
Votes received by Ernie = d

Total votes = 40 + (x+2) + (x+1) + x + d = 100
43 + 3x + d = 100
3x = 57 - d
x = 18; Since d can not be 0 and should be minimum 3.

IMO C

We have A +B +C +D +E = 100
or, 40 +B +C +D +E = 100 or, B +C +D +E = 60 for maximum value of D, E = 1 and C = D+1, B = D+2

Or, (D+2) +(D+1)+D+1 = 60 or, D = 56/3 = 18.6

We cannot take D = 19, then Sum will exceed 100, so D (max) = 18.

I think C.

A>B>C>D>E

A=40

For max D, B,C,D must be as close as possible. Best case scenario is they are all consecutive numbers.
So, if E gets 0, remaining 60 votes can be divided into 21,20,19 votes.
But E has to get at least one vote, so the only way to do this is if we take one vote from 19 and give it to E.

So D will have 18 votes
Ans C

Veritas Prep Official Explanation

To maximize Dan, you’ll want to minimize the others. Alexa is fixed at 40, and you can minimize Ernie by putting him at 1. That leaves 59 votes to split between Bill, Charlie, and Dan, with B > C > D. Your goal, then? As even a distribution as possible, so you should look for a number below 59 that is divisible by 3. 57 = 3 • 19, but then you cannot give each of the three candidates a different total (there are only two votes left from the 59). So choose 54: That allows for each candidate to receive 18, and now you simply need to allocate the other 5 votes so that the candidates have different vote totals (either D = 18, B = 20, and C = 21; or D = 18, B = 19, and C = 23).
_________________

a =40
b=x+2
c=x+1
d=x
e=1


x+2+x+1+x+1=60
x=18(2/3)

so 18 is max for d