During a recent semester at University X, 25 students enrolled in an

1)No,since nothing is mentioned about the weights of each exam so we cannot confidently say that final exam and exam 2 have equal weights

2)Yes,when we arrange all the scores and see then the median is 81.5

3)Yes,For Exam 1 ,year 3 highest is 91 and lowest is 51 so the range is 91-51=40

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the first guess is right, but the explanation is wrong.

If last two exams: Ex1 and Final Ex were waited evenly, than Benson's final score ought to be >77
Due to reason that last 2 scores were 80 & 75, but in realty his score was downgraded = 76.75

Of. Explanation.

Let E1, E2, F, and S denote a student's Exam 1 score, Exam 2 score, final exam score, and final score, respectively. If F and E2 are equally weighted in computing the final score, then there must be constants x and y such that y + x + x = 1 and for each student, S = yE1 + xE2 + xF. Since y = 1 − 2x, it follows that S = (1 − 2x)E1 + xE2 + xF. Solving this last equation for x in terms of E1, E2, F, and S, x = S−E1/E2+F−2E1. In particular, it follows that this fractional expression must be constant.

For Abusuba—the first student listed in the table—E1 = 89, E2 = 87, F = 85, and S = 86.50, so S−E1/E2+F−2E1 = 86.50−89/87+85−2(89), which is equal to 512. For Ardanin—the second student listed in the table—E1 = 85, E2 = 83, F = 84, and S = 84.00, so S−E1/E2+F−2E1 = 84.00−85/83+84−2(85), which is equal to 13. Since 5/12 ≠ 1/3, the fractional expression is not constant. Therefore, F and E2 are not equally weighted in computing the final score.

I think the best way to quickly validate Q#1 is to consider an evenly spaced set of E1, E2 & F like that of Ardanin:

The Arithmetic mean of 83, 84 and 85(E2,F,E1) is clearly 84 which is equal to the final score. The only way it's possible is if either they are equally weighed or weighed more toward the average. For example: Average of the set 83,84,84,84,84,85 would still be 84 as adding average to a set would not change the average.

On the other hand, any number other than the value of mean is added to the set then the new mean will certainly deviate from the average. For example in case of E2&F weighing the same(Barring E1):

83, 83,84,84,85-> The Mean shifts from 84 to a number between 83 and 84.

Hence, this set alone proves that E2&F can't have equal weights without E1 carrying the same weight. Either way, we got both Yes and No, so we cannot determine the weights to compare.

AndrewN Was my thought process correct?

Watch those superlatives—here, best in best way. But yes, I agree that your thought process is sound. In keeping with my simple-is-often-better approach, I looked to the first few rows of data only to deduce that no, the exams in question cannot be given equal weight in computing the final score.

An easy way to work with averages is to set the average and then calculate the value above or below that average for each data point. In this manner, you can quickly work through a set to figure out the average, without working through lengthy calculations. For instance, say that I had the following set: {4, 9, 22, 28, 37}. Of course, I could add them up and divide by 5, or I could set the average to something that looked reasonable—say, 20—and then simply keep track of how far above or below my target average each number in the set was:

20 | {4, 9, 22, 28, 37}

20 | {-16, -11, +2, +8, +17}

Since the sum of the numbers in brackets is 0, I know the average is 20, right on the nose. What if I had selected 15?

15 | {4, 9, 22, 28, 37}

15 | {-11, -6, +7, +13, +22}

Note that the sum of the numbers in brackets is now +25. When I average this value by dividing by the five data points, I get +5. Thus, after correcting the estimate by adding 5, I get the true average: 15 + 5 = 20.

How does this relate to the question at hand? Well, we are given the final score, so we can set this value as our average. We can see quite clearly that the second row values will average to 84: {85, 83, 84} is {+1, -1, 0}. But if we check the top row, the picture comes out a bit different, relative to the final score of 86.5: {89, 87, 85}. That should be 87 (from {+2, 0, -2}) if everything carries equal weight. So, now the question becomes, how can the score be 86.5? If the latter two tests carry equal weight, they would average to 86, and then they would have to carry a lot more weight than Exam 1 to pull the final score down to 86.5. It seems more reasonable that that 86.5 comes from averaging the first two exam scores and then balancing that out with the final exam score: 89 and 87 average to 88; 88 and 85 average to 86.5 (1.5 from each value). This looked promising. Now, I just wanted proof by checking the third row:

67.75 | {65, 70, 68}

The average of 65 and 70 is 67.5 (2.5 from each value).

67.75 | {67.5, 68}

Yep, that one checks out. The average would be 67.75.

I know the process looks lengthy when I type everything out, but trust me, once the first domino fell, the answer followed soon after. Thank you for thinking to ask me about this one. I have to run to my next lesson!

- Andrew

Hi avigutman - was curious, how would you solve the 1st problem only. Solved the 2nd and 3rd . Feel free to include this riddle in the AMA if you think the class could perhaps benefit with this thinking

Different people have put in different strategies but i was curious how you would do it.

Not sure if my proof is 100 %






Not sure how to proove Weightage of E2 is unequal to Weightage of Final score however.

You're right, jabhatta2, your reasoning doesn't actually answer the question that was asked. I suspect that you didn't focus your reasoning on the task at hand, but rather tried to make random observations and inferences about the data and see if any of those are helpful.

The only way I could think of to verify whether the last two exams were equally weighted was to search for a student whose first score was at exactly the midpoint between the 2nd and final scores. If I can find a student like that, then I'll know for sure whether or not those last two scores are equally weighted. For such a student, the overall average will match the first score if and only if the last two scores are equally weighted.

Jeyaretnam was such a student, and the overall average doesn't match the first score, so I conclude that the answer is NO.

Edited to add: I just thought of another way to do this: find a student whose 1st score matches his overall average, and check whether the 2nd and 3rd scores are equidistant from the overall average. They'll be equidistant if and only if they're equally weighted. Unfortunately, such a student doesn't exist in the data, so this method isn't useful in this particular problem.

Thanks so much avigutman - In both methods, you are looking at uniform sets

Red - Exam 1 DOES NOT MATCH the overall Average
Green - Exam 1 DOES MATCH the overall Average.

I think i thought of another scenario.

E2 = 80
Final = 80
E1 = 90

Final average = 85

i think this prooves the weight of E2 = weight of Final.

^^ actually no -- my scenario doesnt proove that

Well, jabhatta2, I'm not sure what you mean by uniform sets, but to clarify, in Red I'm not just saying "Exam 1 DOES NOT MATCH the overall Average". This observation is only useful if E1 is at the midpoint of E2 and E3, i.e. E1 = (E2
+E3)/2



Yeah, as I think you realized on your own, any time E2 = E3 we have no way of knowing their weights. They can be anything.

dear experts: avigutman, AndrewN,MartyTargetTestPrep ,

I am little confused by the first question,
The score on the final exam had equal weight with the score on Exam 2,
so why we consider E1, we just check E2 and final exam, it is possible that E1 accounts 50%, E2 accounts 25%, E3 accounts 25%, because we have no more information , so we can not check whether the final exam had equal weight with the score on Exam 2,

can I move through 1st question like this.

No, zoezhuyan. I highlighted in the quote the reason we can't do that. The question wasn't asking whether we have enough information. Rather, the question implied that we do have enough information.

dear AndrewN,
I read the whole topic, I don't think it is reasonable to evaluate whether final score under assuming E2 FE have equal weight is same as final score on table. because you can have multiply possibilities that E2 FE have equal weight and get a different final score from that on table, while E2, FE have equal weight.

I know OA is always correct, while I have no idea where I missed.

­so in this ques part 1 - weighted average I used a technique but I am not sure if the approach is correct, can someone please confirm 

I considered the scoring of Hernandis coz his scores were integers
so I 
(72a+74b+75c)/ a+b+c=74 where Abc were asigned weights to Exam1,exam2 and final respectively 
74B got cancelled on both sides giving me c=2a 

then i took Nguyens score
(70a+74b+72c)/a+b+C=72
now 72c got cancelled giving me A=B 

so i made a ratio of A:B:C= 1:1:2 

Giving my final as a "no" 

Is this approach correct?

 

­Yes, Vanshikakataruka, it's mathematically correct, but I would argue that it's not appropriate for the GMAT (too time consuming and doesn't promote quantiative reasoning)