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

Question

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. What is the greatest number of votes that Ernie could have received?

A. 12
B. 13
C. 14
D. 15
E. 16

A. 12
B. 13
C. 14
D. 15
E. 16

johnalexander96 · Answer

I used the answer choices to approach that question.

Trying the maximum amount for Ernie, with the minimum amounts for the others while respecting the results order and total number of votes.

Starting with middle answer, if Ernie gets 14 votes, the others will get at min. 15, 16 and 17 votes besides the 40 already won by Alexa => adding these up gives 102 votes. 2 too many, so Ernie cannot get that many votes.

We can conclude that 13 is the right answer, B.

If any other faster method exists, happy to read about it.

Bunuel · Answer

Veritas Prep Official Explanation

To maximize Ernie, the thought process is similar – you want as even a distribution of B, C, D, and E as possible. Since Alexa is fixed at 40, then you have 60 votes left, and the most even distribution would be 15 each. But each value needs to be different, so you should look to allocate the votes around 15. B = 16, C = 15, D = 14 and E = 13 spreads the values out around the 15 average that you want and leaves you with 2 votes remaining. This isn’t enough to bump Ernie to 14 (you can’t do that without creating a tie among the other three), but you can give those votes to Bill and Charlie to keep them “ahead” and leave Ernie with a max of 13 votes.

asgopala · Answer

I tried plugging in the values and I started from from the choice E.

B+ C +D + E = 60

B + C+ D = 60 - 16 -> B + C +D = 44 ( now we have to test for value of B,C,D greater than 16 - say 17 + 18 + 19 =54 which is greater than 44). out

Basically this would elminate choices D and C too as the sum of the values that would be greater than 60 - value(E )

Choice B would result in a sum value ( B + C + D ) < 60 - Value(E) . Hence it would give us options to increase the values of B C D if needed to make them unique.

chetan2u · Answer

Using options is the safest
Let us start with the option C. 
If E is 16, the minimum possible values of D, C and B are 17, 18 and 19 =>  16+17+18+19+40\leq 100......110\leq 100 ..No
If you reduce E by 1, then total will become 110-4=106....No
If you reduce E further, then total will be 106-4=102.
Reduce E further by 1, =>  13+14+15+16+40\leq{100}.....98\leq 100 ...yes

Or

A+B+C+D+E=100
40+B+C+D+E=100
B+C+D+E=100-40=60
If we want to maximise the smallest, E, we have to take all values next to each other. 
E+3+E+2+E+1+E=60......4E+6=60.....4E=54.....E=13.5
As E is an integer, E has to be 13.

B

Regor60 · Answer

To maximize Ernie, minimize the others, with the sum of these 4 to equal 60

E = Ernie
E+ x= Dan
E+ y= Charles
E+ z = Bill

So the sum of these

4E+x+y+z= 60

So 60-x-y-z must be divisible by 4 and x,y,z must be minimized

The first number below 60 divisible by 4 is 56

Can x+y+z = 4 with no repeats of scores ?

Can't use 0 because that's 
 tied with Ernie

Minimum is 1+2+3 to avoid 
 ties but > 4

So 56 is out

Try 52

Can x+y+z = 8 with no 
 repeats ?

There are two ways: 1,3,4 
 and 1,2,5

So 52 works. 52/4 = 13

Posted from my mobile device

nazii · Answer

I assigned X to E, X+1 to D and X+2 to c and X+3 to B
therefore, 4x+6+40=100
4x=54
x=13.;;;
the least is 13

bumpbot · Answer

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

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