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

Question

During a recent semester at University X, 25 students enrolled in an economics class. Each student was enrolled in the university’s 4-year business program and took the course either as a traditional student (attending class and sitting for exams in person) or as an online student (listening to lectures and taking exams via computer), but not both. For each student, the table indicates whether he or she took the course online, along with his or her year in the program and scores on Exam 1, Exam 2, and the final exam. The final score was computed as a weighted mean of the scores on Exam 1, Exam 2, and the final exam, using the same weights for each student.

Student surname  
  Online
student?
(Y/N)  
  Year
in
program  
  Exam 1
score  
  Exam 2
score  
  Final
exam
score  
  Final
score  
 
  Abusuba 
 Y 
 2 
 89 
 87 
 85 
 86.5 
 
  Ardanin 
 N 
 1 
 85 
 83 
 84 
 84 
 
  Bar-Yaacov 
 Y 
 1 
 65 
 70 
 68 
 67.75 
 
  Benson 
 Y 
 1 
 77 
 80 
 75 
 76.75 
 
  Dedeoglu 
 N 
 2 
 90 
 96 
 95 
 94 
 
  Derezinski 
 Y 
 3 
 85 
 84 
 81 
 83.25 
 
  Garcia 
 Y 
 2 
 90 
 87 
 86 
 87.25 
 
  Hernandez 
 N 
 2 
 72 
 74 
 75 
 74 
 
  Jeyaretnam 
 Y 
 2 
 77 
 76 
 78 
 77.25 
 
  Lindt 
 Y 
 3 
 87 
 81 
 81 
 82.75 
 
  Mladek 
 N 
 4 
 64 
 75 
 76 
 72.75 
 
  Nguyen 
 N 
 3 
 70 
 74 
 72 
 72 
 
  Orlando 
 N 
 2 
 81 
 84 
 80 
 81.5 
 
  Pai 
 N 
 2 
 75 
 78 
 72 
 74.25 
 
  Parasarathy 
 N 
 2 
 88 
 91 
 95 
 92.25 
 
  Radzinsky 
 Y 
 3 
 91 
 95 
 100 
 96.5 
 
  Russell 
 N 
 4 
 51 
 69 
 72 
 66 
 
  Sweets 
 N 
 2 
 66 
 76 
 74 
 72.5 
 
  Sykes 
 N 
 3 
 51 
 69 
 73 
 66.5 
 
  Tachau 
 N 
 2 
 91 
 93 
 92 
 92 
 
  Tsosie 
 N 
 2 
 84 
 87 
 85 
 85.25 
 
  Underhill 
 N 
 1 
 77 
 75 
 71 
 73.5 
 
  Vladimirov 
 Y 
 3 
 69 
 75 
 74 
 73 
 
  Washburn 
 N 
 2 
 85 
 83 
 82 
 83 
 
  Zervos 
 N 
 2 
 95 
 97 
 98 
 97

For each of the following statements, select  Yes  if the statement is true based on the information provided; otherwise, select  No .

BLTN · Accepted Answer

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.

Sajjad1994 · Answer

Official Explanation

The score on the final exam had equal weight with the score on Exam 2 in computing the final score.

Let \(E_{1}\), \(E_{2}\), F, and S denote a student's Exam 1 score, Exam 2 score, final exam score, and final score, respectively. If F and \(E_{2}\) 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, \(S = yE_{1} + xE_{2} + xF.\) Since \(y = 1 − 2x,\) it follows that \(S = (1 − 2x)E_{1} + xE_{2} + xF.\) Solving this last equation for x in terms of \(E_{1}, E_{2}, F,\) and \(S, x = \frac{S−E_{1}}{E_{2}+F−2E_{1}}.\) In particular, it follows that this fractional expression must be constant.

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

The correct answer is No.

The median final score for all 25 students was 81.50.

Attachment:

2.jpg [ 160.18 KiB | Viewed 40929 times ]

For Exam 1 scores for students in year 3 of the program, the range was 40.

Attachment:

3.jpg [ 133.18 KiB | Viewed 40425 times ]

Sorting on Year in program reveals that there are 6 students in year 3—Derezinski, Lindt, Radzinsky, Vladimirov, Nguyen, and Sykes—with Exam 1 scores of 85, 87, 91, 69, 70, and 51, respectively. The lowest of these scores is 51 (Sykes) and the highest is 91 (Radzinsky), so the range of the scores is 91 − 51, or 40.

The correct answer is Yes.

bM22 · Answer

1.The score on the final exam had equal weight with the score on Exam 2 in computing the final score.
Firstly we don't have the weight that exam has on the exam score, so we cannot actually say that score on the final exam had equal weight with the score on Exam 2.
 Answer No.

2.The median final score for all 25 students was 81.50.
The final scores for all the students in ascending order: 
66.00, 66.50, 67.75, 72.00, 72.50, 72.75, 73, 73.50, 74, 74.25, 76.75, 77.25,  81.50 , 82.75, 83, 83.25, 84.00, 85.25, 86.50, 87.25, 92.00, 92.25, 94.00, 96.50, 97.
thus, we can see that the median is : 81.50.
 Answer Yes.

3. For Exam 1 scores for students in year 3 of the program, the range was 40.
Range = diff of highest and lowest score.
So Highest score in exam 1 = 91, lowest score in exam 1 = 41.
Thus the difference = 40, => range = 40. 
 So yes.

Apt0810 · Answer

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

Posted from my mobile device

AlphaCentauri4ly · Answer

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?

AndrewN · Answer

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

jabhatta2 · Answer

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.

avigutman · Answer

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.

jabhatta2 · Answer

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

avigutman · Answer

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.

zoezhuyan · Answer

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.

avigutman · Answer

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.

Vanshikakataruka · Answer

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?

avigutman · Answer

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 )

Abhishek2k7 · Answer

Assign weights a,b,c to E1, E2 and E3. So, that aE1+bE2+cE3=Final score. Now, to check relation between b and c we need to eliminate "a" from equations. For that, sort E1 values (1st column) and seek data where E1 values are same. Fortunately, in first 2 rows, when we subtract equations formed aftr applying weights from each other, we directly get c=1/2. For next similar set of E1 values i.e. 77, on subtracting rows from each other, we get equation b+7c=3.75, i.e. b=1/4.

swati.singh · Answer

I saw that there is no proper and convincing response given for statement 1, so I wanted to share what I did.

The approach I followed was to assume the weight of Exam 2 and the final exam  as 1 , and the weight of Exam 1  as x . Using a student's data like Zervos, solve for x, then apply these weights to another student, such as Ardanin, to check if the calculated final score matches the score in the table exactly.

Let the weight of E2 and Final Exam be 1 and 1 respectively, and let the weight of E1 be x.

Then, considering any data row which has easy numbers, say-

Zervos :

(95x + 97 + 98) / (1+1+x) = 97

=> x=1/2

So, then ratio of weights becomes:

E1 : E2 : Final Exam = 1⁄2 : 1 : 1

Or 1:2:2.

If this is true, this should fit other records too.

Checking for any one or two rows randomly:

Adranin :

85*1 + 83*2 + 84*2 / 5 = 419/5 = 83 point something ≠ 84 as mentioned.

Therefore, the answer is No.

Araevin · Answer

For Statement 1 since it's the most time consuming even with OE, here's the most efficient approach with minimal math. and using the datasets available
Step 1: Since we have to compare weights of Exam 2 (E2) and Final Exam Score (F), sort the table by Exam 1 (E1)
Step 2: Look for datasets where E1 scores are equal to eliminate the impact of E1 on Final Score (FS)
Step 3: First two rows in the sorted table (Russel and Sykes) have equal E1 scores and equal E2 scores. This lets us check that a 1 point difference in F has a 0.5 difference in FS. Clearly the weight of F is 50%.

Now we need to check whether E2 also has a 50% weight, which means E1 has a weight of 0%. Since sum of all weights needs to be a 100%.

Step 4: Sort the table by E2 and look for datasets where E2 scores are equal but E1 are not. There are multiple such sets, but 18th and 19th row (Tsosie and Abusuba) also have equal F scores. In these rows, if E1 has 0% weight then a change in E1 shouldn't change FS, but it does. This implies E1 has more than 0% weight and therefore E2 has less than 50% weight.

Weights of E2 and FS are not equal.

This approach could've been complicated if we didn't find the two perfect data sets that we did, but in table analysis using the data available and looking for patterns is the quickest way to solve them. Since it's an OG question, pretty sure the dataset was designed to be solved with above approach.

Even finding a part of the pattern instead of the full pattern can help save time. I originally only found the pattern up to Step 3 and used that info to validate a couple of rows (4th and 5th rows) in Step 4 (Nguyen and Hernandez) instead of looking for more data. Here, a 3 point change in F should have changed FS by 1.5 but it changed by 2 implying E1 has a weight more than 0%.

JROYENG97 · Answer

Here is how I answered Q1 instinctively and quickly.

If you find two entries (rows) where two of the 3 exam scores are the same between the two rows, you can easily find the weight of the exam that doesn't have the same score between the two rows. By filtering the table by final exam score, you see that the first two rows are an instance of this situation. We have:
Russel: E1: 51, E2: 69, FE: 72, FS: 66
Sykes: E1: 51, E2: 69, FE: 73, FS: 66.5

Since the only difference between the two rows is that Sykes has 1 more point on the FE, and it gives him 0.5 more points in his Final score, then we know the weight of the final exam has to be 50%. Therefore, we know exam 2 doesn't have the same weight as the final exam, as this would require for exam 1 to have a weight of 0%, but the last sentence of the introductory text implies this is not the case.

I like this approach as it only requires glancing at the data and virtually no calculation.

Hope this helps.

KarishmaB · Answer

The score on the final exam had equal weight with the score on Exam 2 in computing the final score

If this were true: 
 Abusuda: 89, 87, 85 WAvg = 86.5 
 Avg of Exam 2 and Final exam would be 86.  
 Avg of 86 and 89 is 86.5 which means weights given are 5:1

Ardanin: 85, 83, 84 WAvg = 84 
 Avg of Exam 2 and Final exam would be 83.5.  
 Avg of 83.5 and 85 is 84 which means weights given are 2:1

Since the weights are not consistent, this is not true.

Select  No

You should observe here that Exams 1 and 2 are likely to have equal weights and Final Exam is likely to have more weight. A quick check of a few values shows that this is true. The weights of the 3 are in the ratio 1:1:2. I did not do the calculations shown above to solve the question. I focused on observing the first few values because that made sense. GMAT normally wouldn't use un-intuitive logic.

The median final score for all 25 students was 81.50.

Sort the table using final score and count to the 13th value.

Select  Yes

For Exam 1 scores for students in year 3 of the program, the range was 40.

Sort the table using the Year column. Lowest value is 51 and highest is 91. Range = 40

Select  Yes

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Student surname	Online student? (Y/N)	Year in program	Exam 1 score	Exam 2 score	Final exam score	Final score
Abusuba	Y	2	89	87	85	86.5
Ardanin	N	1	85	83	84	84
Bar-Yaacov	Y	1	65	70	68	67.75
Benson	Y	1	77	80	75	76.75
Dedeoglu	N	2	90	96	95	94
Derezinski	Y	3	85	84	81	83.25
Garcia	Y	2	90	87	86	87.25
Hernandez	N	2	72	74	75	74
Jeyaretnam	Y	2	77	76	78	77.25
Lindt	Y	3	87	81	81	82.75
Mladek	N	4	64	75	76	72.75
Nguyen	N	3	70	74	72	72
Orlando	N	2	81	84	80	81.5
Pai	N	2	75	78	72	74.25
Parasarathy	N	2	88	91	95	92.25
Radzinsky	Y	3	91	95	100	96.5
Russell	N	4	51	69	72	66
Sweets	N	2	66	76	74	72.5
Sykes	N	3	51	69	73	66.5
Tachau	N	2	91	93	92	92
Tsosie	N	2	84	87	85	85.25
Underhill	N	1	77	75	71	73.5
Vladimirov	Y	3	69	75	74	73
Washburn	N	2	85	83	82	83
Zervos	N	2	95	97	98	97

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