hard math question?

Question

Every year three students are nominated to be “Student of the Year.” After a review process, 10 teachers must rank each student first, second or third place (no ties), with each first-place vote worth three points, each second-place vote worth two points, and each third-place vote worth one point.
Under these conditions, what is the smallest number of first-place votes that it would take to ensure that a student would win the Student of the Year? What is the smallest number of first-place votes a student could receive and still win the Student of the Year?
Please show work and procedure.

Answer 1

It simplifies the problem a little if you shift the scoring down one point:
first place (f) = 2 pts
second pl (2) = 1 pt
third place (t) = 0 pts

Maximum possible for a student = 20 pts
Minimum possibe = 0
Total points available = 30
So 16 will ensure victory.

First part:
We want the maximum number of first places necessary to ensure victory. Since we are trying to ensure victory with first places alone, let's say I get x first places and the rest in thirds. My points = 2x

Now let's try to maximize the second guy's score. He gets 10-x first places, and x seconds. His score = 2*(10-x) + x = 20-x.

We require 2x > 20-x
3x > 20
x ≥ 7
ANS 7

*********
Second part:
Now we want to find the minimum no. of first places that might still win. So let's say I win x first places and the rest in second. My points = 2x + 10-x = 10+x.

The others must score less than this. We wish to get their scores as close together as possible.
Let the second guy get y firsts and 10-y thirds. His score = 2y.
Let the third guy get 10-x-y firsts, x seconds, y thirds. His score = 20-2x-2y+x = 20-x-2y

So 2y < 10+x
And 20-x-2y < 10+x or 2y > 10-2x
10 - 2x < 2y < 10 + x
Clearly x = 1 will satisfy that inequality with y=5.
x = 0 will at best give you a tie.

So it is possible to win with just 1 first place.
First: 1,9,0 = 11 pts
Second: 5,0,5 = 10 pts
Third: 4,1,5 = 9 pts


ANS.
First part: 7
Second part: 1

Answer 2

There are 60 points available in total and they are distributed among three students. The winner needs at least 21 (final score 21-20-19).

It is possible to get 21 with 1 first place (3) and 9 second places (18).

7 first-place ranks (21) would ensure a win.

A student could get 6 first places and 4 third places (18+4 = 22) while another student could get the remaining 4 first places plus 6 second places (12 + 12 = 24). Therefore, getting 6 first places is NOT sufficient.

Answer 3

1. 6 If you had 5 first-place votes and 5 seconds, you could still tie with another, the third getting all third-place votes.

2. 1. you have 1 first and 9 seconds for 21
one guy has 5 firsts and 5 thirds for 20
the other 4 firsts, 1 second, and 5 thirds for 19

Answer 4

q1
what is the smallest number of first-place votes that it would take to ensure that a student would win the Student of the Year?

let x be # of 3p votes, y be # of 2p votes and z be # of 1p votes.
so x + y + z = 10

you asked for ensured win # of 3p, so we'll assume the winner got all 3p and 1p, while the runner-up with all the left 3p and 2p.
so # of winner's 3p = # of runner-up's 2p

3x + 1*(10-x) > 2x + 3*(10-x)
2x + 10 > 30 - x
3x > 20
x > 20/3
x = 7

q2
What is the smallest number of first-place votes a student could receive and still win the Student of the Year?

so we assume winner win more on his 2p counts rather than his 3p. and he won the voting just by 1 more the mean.

3x + 2y = 21
x + y = 10
x = 1 and y = 9