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23 Feb 2025, 00:50
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Q1 - The score on the final exam had equal weight with the score on Exam 2 in computing the final score (Yes/No)?
Just thought I'll add my solution here for the first question (yes/no) - since I didn't see anyone attempt something similar.
Step1: Let's assume x, y and z as the weights for each exam that's used to calculate the final score.
By definition x + y + z, i.e., sum of the weights should be equal to 1.
Eq.1) x + y + z = 1
Step 2: Next let's take 2 samples from the table given. Feel free to take any sample here, but for ease of calculation I'll be taking samples where the final score is a whole-number. I'll be multiplying the intermediate scores (exam1, exam2, etc) by weights to express the final score.
Sample 1: Russel - Intermediate Scores: 51, 69, 72; Final Score: 66
Eq. 2) 51x + 69y + 72z = 66
Sample 2: Nguyen - Intermediate Scores: 70, 74, 72; Final Score: 72
Eq. 3) 70x + 74y + 72z = 72
Step 3: Since, we're only interested in weights for exam 2 and final exam, let's simplify the equations 2 and 3 using equation 1 to remove the scores for Exam 1. To simplify equation 2) - multiply equation 1 by (x's co-efficient; 51) and subtract from equation 2. Perform same for equation 3.
Subtracting 51*Eq. 1 from Eq. 2
Eq. 4) 18y + 21z = 15 (Divide by 3 further)
6y + 7z = 5
Subtracting 70*Eq. 1 from Eq. 3
Eq. 5) 4y + 2z = 2
Step 4: Since, we're asked to determine if the weights y and z are same, we can use proof by contradiction, i.e., assume they are the same and prove by contradiction that they are not. This is the easiest step and can be done much earlier. But, I've included it here to run through the reasoning more clearly. Applying the logic we get the following -
Eq. 4.1) 13y = 5
Eq. 5.1) 6y = 2
We can see that we're getting two values of y, which isn't possible. Hence, we can conclude that the weights for exam 2 and final exam are not the same.
Just thought I'll add my solution here for the first question (yes/no) - since I didn't see anyone attempt something similar.
Step1: Let's assume x, y and z as the weights for each exam that's used to calculate the final score.
By definition x + y + z, i.e., sum of the weights should be equal to 1.
Eq.1) x + y + z = 1
Sample 1: Russel - Intermediate Scores: 51, 69, 72; Final Score: 66
Eq. 2) 51x + 69y + 72z = 66
Sample 2: Nguyen - Intermediate Scores: 70, 74, 72; Final Score: 72
Eq. 3) 70x + 74y + 72z = 72
Step 3: Since, we're only interested in weights for exam 2 and final exam, let's simplify the equations 2 and 3 using equation 1 to remove the scores for Exam 1. To simplify equation 2) - multiply equation 1 by (x's co-efficient; 51) and subtract from equation 2. Perform same for equation 3.
Subtracting 51*Eq. 1 from Eq. 2
Eq. 4) 18y + 21z = 15 (Divide by 3 further)
6y + 7z = 5
Subtracting 70*Eq. 1 from Eq. 3
Eq. 5) 4y + 2z = 2
Eq. 4.1) 13y = 5
Eq. 5.1) 6y = 2
13 Sep 2025, 08:30
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For the first part, we see that there're two students for whom the final score is same as Final Exam, and E1 and E2 are at same distance (Nguyen). Which confirms that weights of E1 and E2 are same. But at the same time, for Zervos the final score is same as E2 but the E1 and final scores are not at same distance, which confirms that the weights are different. Hence, weights for E2 and Final exams are not the same.
PocusFocus
IESE School Moderator
21 Sep 2025, 15:58
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The first statement is a good opprtunity to use the weighted average theory:
'The score on the final exam had equal weight with the score on Exam 2 in computing the final score.'
Took Ardanin:
Exam 1=85; Exam 2 = 83; Final Exam = 84 --> Final Score = 84 (weighthed score)
If you look at the distances of the points to the weighted score you'll realize that "weight if the exam 1 = weight of exam 2"
Now took another student let's say "Hernandez", then apply again the the theory to this set:
Exam 1 is 2twice distant from the average as the final exam, so final Exam weight double than exam 1, notice that exam 2 is nuled due that is located in the average.
As weight Exam 1 = weight Exam 2 = 0.5 weight of Final Exam, the answer is no.
Notice that almost all this could be performed mentally, once you have clarity in the theory.
'The score on the final exam had equal weight with the score on Exam 2 in computing the final score.'
Took Ardanin:
Exam 1=85; Exam 2 = 83; Final Exam = 84 --> Final Score = 84 (weighthed score)
If you look at the distances of the points to the weighted score you'll realize that "weight if the exam 1 = weight of exam 2"
Now took another student let's say "Hernandez", then apply again the the theory to this set:
Exam 1 is 2twice distant from the average as the final exam, so final Exam weight double than exam 1, notice that exam 2 is nuled due that is located in the average.
As weight Exam 1 = weight Exam 2 = 0.5 weight of Final Exam, the answer is no.
Notice that almost all this could be performed mentally, once you have clarity in the theory.
29 Nov 2025, 05:57
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parkhydel
I may have the quick way of checking for the first sub-question.
Know that if two terms remain constant, and the third term changes, then the mean’s change would be that term’s change times its weight.
Ex: With 50, 51, 52, the average is 51. Assume that for the 52 term, its weight is 40%. If 52 increases to 53, then average changes to 51 + (53-52) * 40% = 51.4
Now, with this in mind, a clever trick that we can use here is to sort by the Exam 1 score!
Then, look for cases where either only the second term (Exam 2 score) changes or the third term (Final exam score) changes.
Look at Russell and Skyes:
Russel: 51, 69, 72 -> Final: 66
Sykes: 51, 69, 73 -> Final: 66.5
So, when the Final Exam Score moves up by 1, the Final Score moves up by 0.5
Therefore, the Final Exam Score's weight is: 0.5/1 = 0.5
We also know that all 3 scores both contribute to the Final Score in some way. Therefore, each of their percentages is non-zero.
Given this, if the Final Exam Score's weight is 0.5, then the Exam 2 Score's weight has to definitely be lower than 0.5
=> The Final Exam Score's weight and the Exam 2 Score's weight are definitely not equal!
This all looks like a lot, but if you understand the nature of weighted averages, then this line of thinking becomes easier to conjure up.
