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Can you provide a quick solution for question 1?
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Weightage of Judge 1, Judge 2: x
Weightage of Super Judge = y
Weightage of Audience: 2(2x+y)

Take celebrity 1 and celebrity 6:
for celebrity 1 you get equation: 7(2x) + 5y = 3(2x+y)(7.27)
for celebrity 6 you get: 9x + 7(5x+3y) = 3(2x+y)(7.13)
On subtracting and simplifying these 2 eqns: 59x = 21y
So, answer to 1st is Yes.
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Why didn't u consider the audience's weighted score in the 1st eqn?

arushi118
Weightage of Judge 1, Judge 2: x
Weightage of Super Judge = y
Weightage of Audience: 2(2x+y)

Take celebrity 1 and celebrity 6:
for celebrity 1 you get equation: 7(2x) + 5y = 3(2x+y)(7.27)
for celebrity 6 you get: 9x + 7(5x+3y) = 3(2x+y)(7.13)
On subtracting and simplifying these 2 eqns: 59x = 21y
So, answer to 1st is Yes.
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Hi sarkarsarbani,

Good catch, and your instinct is exactly right. The audience score must be in that equation. The final grade is a weighted average of all four inputs - Judge 1, Judge 2, Super Judge, and audience - so leaving the audience out of the numerator makes the equation incomplete.

Here's the full setup with the audience put back in.

Assign the weights

- Judge 1 weight = Judge 2 weight = x
- Super Judge weight = y
- Audience weight = twice the three judges' combined weight = 2(x + x + y) = 4x + 2y
- Total weight = 2x + y + (4x + 2y) = 6x + 3y

Write Celebrity 1's equation correctly

Celebrity 1 scores: J1 = 7, J2 = 7, SJ = 5, Audience = 8, Final = 7.27.

Numerator (every term, including audience):

7x + 7x + 5y + 8(4x + 2y) = 14x + 5y + 32x + 16y = 46x + 21y

So:

(46x + 21y) / (6x + 3y) = 7.27

Solving: 46x + 21y = 43.62x + 21.81y - 2.38x = 0.81y - y ≈ 2.94x.

That's why the answer is Yes: the Super Judge's weight comes out close to three times Judge 1's. The post you were reading dropped the 8(4x + 2y) audience piece - that's the term you sensed was missing.

Quick rehearsal to lock it in

Weighted averages always include every component. If two items score 6 and 9 with weights 1 and 2, the average is (6·1 + 9·2) / (1 + 2) = 24/3 = 8 - not (6 + 9)/2. Drop a weight or an item and the whole thing breaks. Same here: the audience term belongs in the sum.

Answer: Yes

sarkarsarbani
Why didn't u consider the audience's weighted score in the 1st eqn?


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Hi,
Where from is the -2.38x and -y coming in the equation?

egmat
Hi sarkarsarbani,

Good catch, and your instinct is exactly right. The audience score must be in that equation. The final grade is a weighted average of all four inputs - Judge 1, Judge 2, Super Judge, and audience - so leaving the audience out of the numerator makes the equation incomplete.

Here's the full setup with the audience put back in.

Assign the weights

- Judge 1 weight = Judge 2 weight = x
- Super Judge weight = y
- Audience weight = twice the three judges' combined weight = 2(x + x + y) = 4x + 2y
- Total weight = 2x + y + (4x + 2y) = 6x + 3y

Write Celebrity 1's equation correctly

Celebrity 1 scores: J1 = 7, J2 = 7, SJ = 5, Audience = 8, Final = 7.27.

Numerator (every term, including audience):

7x + 7x + 5y + 8(4x + 2y) = 14x + 5y + 32x + 16y = 46x + 21y

So:

(46x + 21y) / (6x + 3y) = 7.27

Solving: 46x + 21y = 43.62x + 21.81y - 2.38x = 0.81y - y ≈ 2.94x.

That's why the answer is Yes: the Super Judge's weight comes out close to three times Judge 1's. The post you were reading dropped the 8(4x + 2y) audience piece - that's the term you sensed was missing.

Quick rehearsal to lock it in

Weighted averages always include every component. If two items score 6 and 9 with weights 1 and 2, the average is (6·1 + 9·2) / (1 + 2) = 24/3 = 8 - not (6 + 9)/2. Drop a weight or an item and the whole thing breaks. Same here: the audience term belongs in the sum.

Answer: Yes


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