# Math 20 voting question?

At the student council elections, 3 candidates ran for president. There were 584 students eligible to vote. The first candidate had 40 votes more than the second candidate and the third candidate had 20 fewer votes than the second candidate. What is the maximum number of votes the winning candidate could have received?

### 10 Answers

- 1 decade agoBest Answer
x = number of votes for candidate 1

y = number of votes for candidate 2

z = number of votes for candidate 3

total number of students eligible to vote = 584

x = (y + 40)

z = (y - 20)

y = ?

584 = x + y + z

584 = (y + 40) + y + (y - 20)

584 = 20 + 3y

584 - 20 = 3y

564 = 3y

divide both sides by 3

188 = y

x = 188 + 40

x = 228

z = 188 - 20

z = 168

*The maximum number of votes the winning candidate could have received is 228.

- miggitymaggzLv 51 decade ago
Since you have 3 unknowns (3 candidates), you need 3 equations to determine those unknowns. Let A = 1st candidate, B = 2nd candidate, C = 3rd candidate:

A = 40 + B

B = 20 + C

A + B + C = 584

Substituting 2nd eqn in to 1st:

A = 40 + (20 + C) = 60 + C

Plug new A and B in to 3rd eqn:

(60 + C) + (20 + C) + C = 584

80 + 3C = 584

3C = 504

C = 168

therefore, B = 188 and A = 228

check: 168 + 188 + 228 = 584 .. bango

A = 1st Candidate = 228 votes.

- 1 decade ago
584 = x + x-40 + x -60 where x is the winning candidates number of votes

584 = 3x -100

684 = 3x

(divide by 3)

228 = x

so the winning candidate got 228 votes the others got 188 and 168

- MAHAANIM07Lv 41 decade ago
Let 2nd candidate=x votes

(x+40)+x+(x-20)=584

or3x+20=584

or 3x=584-20=564

or x=564/3=188

so x+40=188+40=228 ans

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- carlstromLv 43 years ago
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- Anonymous1 decade ago
Most votes possible for the winner is 228.

Now what do I win??

- 1 decade ago
figure it out ure self dum-dum...

really seriously figure it out yourself ya frikin DA