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 least number of votes that Bill could have received?

A. 12
B. 15
C. 16
D. 17
E. 18

A. 12
B. 15
C. 16
D. 17
E. 18

CrackverbalGMAT · Answer

Since Alexa won with 40 Votes then the remaining 60 votes were cast by the other 4.

Also since Bill came second, he will have the second largest number of votes.

Since we have to get least number of votes for Bill, the number of votes for B,C, D will be in consecutive order. e might be a consecutive number or 2 or 3 numbers after Charlie.

Assuming all are consecutive, Let E, D and C and B get x, (x + 1), (x + 2) and (x + 3) votes respectively.

Therefore x + (x + 1) + (x + 2) + (x + 3) ≤ 60

4x + 6 ≤ 60

4x ≤ 54

x ≤ 13.5 = 13

So the other 3 other than Bill get 13, 14 and 15 votes.

Bill gets 60 - 13 - 14 - 15 = 18 votes

We can also have that the other 3 get 13, 14 and 16 votes

Then Bill gets 60 - 13 - 14 - 16 = 17 Votes

Option D

Arun Kumar

MaryKaaviya · Answer

Answer is  D

We have 60 votes to be shared by B,C,D and E

60/4=15

So the mid range for the votes is 15.

If E is 13, D is 14, C is 15, B is 18.

But we see that there are 3 values difference between B and C. So compensating the 1 from B to C we get

E =13, D=14, C=16,   B=17.

TheBunk · Answer

Thanks for your answer !

Once you know that the mid range is 15, how do you find "E is 13, D is 14, C is 15, B is 18" ?

MaryKaaviya · Answer

Since all the values must be distinct we cannot assign 15 to all. Since even number of values mid range can be C or D.

If C is 15, max values of D and E => D=C-1, E=C-2
then the Deficit of 3 will be added to B.

So C can only be 15. Only then the E can be higher than C.
And by Adding one to C from E, we can still maintain the order of values.
Thus C is 16.

MathRevolution · Answer

A = 40 and A > B > C > D > E.

Total votes: 100

Reamining votes: 100 - 40 = 60

=>  \frac{60}{ 4}  = 15 votes but no one received the same number of votes, therefore,

If Bill gets 16, then C can get 15, D can get 14 and E can get 13. But their sum is 58.

If Bill gets 17, C gets 16, D gets 14 and E gets 13

B + C + D + E = 17 + 16 + 14 + 13 = 60

Answer D

Bunuel · Answer

Veritas Prep Official Explanation

If you know Alexa received 40, then you need to figure out the least number of votes that Bill could have received in 2nd and still account for 60 votes with the other three candidates. To find that number quickly, divide 60 by 4 and know that, if they could each have the same number of votes, the answer would be 15. Since they cannot, you want to arrange the candidates around 15 so that it remains the average but each candidate has a different number. Imagine, for instance, if Bill had 16, Charlie 15, Dan 14, and Ernie 13. That combination would not quite work, as the sum would be 58 (average is below 15). The smallest number Bill could have is 17 (the others could have 16, 15, and 12).

GMATGuruNY · Answer

Since Alexa wins 40 of the 100 votes, the remaining 60 votes must be won by B, C, D and E.

We can PLUG IN THE ANSWERS, which represent the least number of votes that B could have received.
To MINIMIZE the votes for B, we must MAXIMIZE the votes for C, D and E, bearing in mind that no two candidates may receive the same number of votes.
Since the correct answer must be the least viable option, start with the smallest answer choice.
When the correct answer is plugged in, the four vote tallies for B, C, D and E will sum to 60 or more.

A: 12+11+10+9 = 42
B: 15+14+13+12 = 54
C: 16+15+14+13 = 58

Since only two more votes are needed to reach the threshold of 60, the next largest answer choice is viable.

D

bumpbot · Answer

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

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