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The incomplete table above shows a distribution of scores fo
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Updated on: 22 Aug 2013, 03:08
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The incomplete table above shows a distribution of scores for a class of 25 students. If the average (arithmetic mean) score for the class is 83, what score is missing from the table? A. 75 B. 77 C. 81 D. 84 E. 86
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Originally posted by Asifpirlo on 21 Aug 2013, 17:19.
Last edited by Bunuel on 22 Aug 2013, 03:08, edited 1 time in total.
Edited the question.




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Re: The incomplete table above shows a distribution of scores fo
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21 Aug 2013, 18:53
madhurs254 wrote: Asifpirlo wrote: The incomplete table above shows a distribution of scores for a class of 25 students. If the average (arithmetic mean) score for the class is 83, what score is missing from the table? A. 75 B. 77 C. 81 D. 84 E. 86 we can write it like 83= (92*4 + 91*6 + x*3 + 83*7 + 71*5)/ 25 Solving this will give x=75 Ans. Yeah, this is the long way. For the short way, simply calculate the deviations from the mean for each score to figure out what the other score is. 92 is 9 away from the mean, so 4x9=36 pts above 91 is 8 away from the mean, so 6x8=48 pts above 83 is equal, no worries 71 is 12 away from the mean, so 5x12=60 pts below. We have 84 pts above and 60 below, so the unknown score's deviation must be 8460=24 pts below distributed among 3 students, so 8 pts per student. 838 = 75pts.




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Re: The incomplete table above shows a distribution of scores fo
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21 Aug 2013, 18:45
Asifpirlo wrote: The incomplete table above shows a distribution of scores for a class of 25 students. If the average (arithmetic mean) score for the class is 83, what score is missing from the table? A. 75 B. 77 C. 81 D. 84 E. 86 we can write it like 83= (92*4 + 91*6 + x*3 + 83*7 + 71*5)/ 25 Solving this will give x=75 Ans.



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Re: The incomplete table above shows a distribution of scores fo
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20 Apr 2015, 18:02
Hi Easiest way of solving this questions finding the deviation from the mean. Average given is 83. 92 is 9 more than 83 and we have 4 students. So total points will be (9*4)36. 91 is 8 more than 83 and we have 6 students. So total points will be (8*6)48. 83 is more than the average. So no need of counting this. 71 is 12 less than 83 and we have 5 students. So total points will(12*5) 60. So total points more is 84 and less is 60. So the value will be less than 83(8460points) 24 points. Since there are 3 students. 24/3 is 8 . So the value will 8 less than the average which is 838= 75. So 75 is the answer.
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Re: The incomplete table above shows a distribution of scores for a class
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28 Aug 2016, 10:36
Bunuel wrote: The incomplete table above shows a distribution of scores for a class of 25 students. If the average (arithmetic mean) score for the class is 83, what score is missing from the table? A. 75 B. 77 C. 81 D. 84 E. 86 Attachment: 20160828_2152.png Answer is A. Sum of all = Average * 25 = 83*25=2075.  (1) From the table we can say, sum of all numbers = 92*4 + 91 *6 +83*7 + 71*5 + 3*X [X is the Value of Unknown score)  (2) Equating (1) and  (2), we will get X= 75.
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Re: The incomplete table above shows a distribution of scores for a class
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28 Aug 2016, 10:43
Bunuel wrote: The incomplete table above shows a distribution of scores for a class of 25 students. If the average (arithmetic mean) score for the class is 83, what score is missing from the table? A. 75 B. 77 C. 81 D. 84 E. 86 Attachment: 20160828_2152.png 92*4 + 91*6 + 3x+83*7+71*5 / 25 = 83 1850 + 3x = 2075 3x = 225 => x =75 Option A. Is there any short cut method ?



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Re: The incomplete table above shows a distribution of scores for a class
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28 Aug 2016, 16:08
msk0657 wrote: Bunuel wrote: The incomplete table above shows a distribution of scores for a class of 25 students. If the average (arithmetic mean) score for the class is 83, what score is missing from the table? A. 75 B. 77 C. 81 D. 84 E. 86 Attachment: 20160828_2152.png 92*4 + 91*6 + 3x+83*7+71*5 / 25 = 83 1850 + 3x = 2075 3x = 225 => x =75 Option A. Is there any short cut method ? Possible shortcut: First, find the unit digit of the total sum. Total sum = Avg * number of students = 83 * 25 > unit digit is 5. Next, find the unit digit of each pair from the table and add them up. 92 * 4 > xx8 91 * 6 > xx6 83 * 7 > xx1 71 * 5 > xx5 8+6+1+5 > 0 So, the remaining pair must end with 5. The number of student is 3 so the score must have 5 as the units digit. Only 75 from the given options satisfies. Answer (A).



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Re: The incomplete table above shows a distribution of scores for a class
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28 Aug 2016, 21:00
Bunuel wrote: The incomplete table above shows a distribution of scores for a class of 25 students. If the average (arithmetic mean) score for the class is 83, what score is missing from the table? A. 75 B. 77 C. 81 D. 84 E. 86 Attachment: 20160828_2152.png can ignore 83. 9283 = 9 so +36 9183 = 8 so +48 8371 =12 so 60 total effect = 24 need to be distriubted in 3. s0 838 =75



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Re: The incomplete table above shows a distribution of scores fo
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23 Oct 2016, 10:31
Asifpirlo wrote: Attachment: incomplete.png The incomplete table above shows a distribution of scores for a class of 25 students. If the average (arithmetic mean) score for the class is 83, what score is missing from the table? A. 75 B. 77 C. 81 D. 84 E. 86 I tried to avoid working through big chunks of numbers, and wanted to find the last digit of the numbers only... 25*83  smth smth in the end 5. good. now 92*4  smth smth end is 8. 91*6  smth smth end is 6. 83*7  smth smth end is 1. 71*5  smth smth end is 5. so, last digit of the sum of these numbers is 8+6+1+5 = zero. smth smth 5  smth smth 0 = smth smth 5. since we need an integer number, and since the smth smth 5 number would be divided by 3, it is logically to assume that the last digit would be 5 as well... only A works...



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Re: The incomplete table above shows a distribution of scores for a class
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15 Nov 2017, 07:24
Bunuel wrote: The incomplete table above shows a distribution of scores for a class of 25 students. If the average (arithmetic mean) score for the class is 83, what score is missing from the table? A. 75 B. 77 C. 81 D. 84 E. 86 Attachment: 20160828_2152.png \(\frac{(92*4)+(91*6)+(x*3)+(83*7)+(71*5)}{25} = 83\) Or, \(368 + 546 +3x + 581 + 355 = 2075\) Or, \(3x = 225\) Or, \(x = 75\), answer will be (A) 75
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Re: The incomplete table above shows a distribution of scores for a class
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28 Nov 2017, 05:58
Hi I have a different approach to this question, Please confirm that this approach is right. See attachment. Abhishek009 wrote: Bunuel wrote: The incomplete table above shows a distribution of scores for a class of 25 students. If the average (arithmetic mean) score for the class is 83, what score is missing from the table? A. 75 B. 77 C. 81 D. 84 E. 86 Attachment: The attachment 20160828_2152.png is no longer available \(\frac{(92*4)+(91*6)+(x*3)+(83*7)+(71*5)}{25} = 83\) Or, \(368 + 546 +3x + 581 + 355 = 2075\) Or, \(3x = 225\) Or, \(x = 75\), answer will be (A) 75
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Re: The incomplete table above shows a distribution of scores for a class
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28 Nov 2017, 07:36
(1850+3x)/25=83 x=75 Sent from my ONEPLUS A3003 using GMAT Club Forum mobile app



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Re: The incomplete table above shows a distribution of scores for a class
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29 Nov 2017, 17:06
Bunuel wrote: The incomplete table above shows a distribution of scores for a class of 25 students. If the average (arithmetic mean) score for the class is 83, what score is missing from the table? A. 75 B. 77 C. 81 D. 84 E. 86 Attachment: 20160828_2152.png We are given that the average (arithmetic mean) score for the class is 83. Since we are given the average, we can find the difference between the average and each score and multiply their respective frequencies, and the sum of these products should be 0. That is, if we let n = the missing score, we can create the following equation: (92  83) x 4 + (91  83) x 6 + (n  83) x 3 + (83  83) x 7 + (71  83) x 5 = 0 36 + 48 + 3n  249 + 0  60 = 0 3n  225 = 0 3n = 225 n = 75 Alternate Solution: We can calculate a weighted mean for the 25 scores, using the formula average = sum/number: 83 = [4(92) + 6(91) + 3x + 7(83) + 5(71)]/25 2075 = 368 + 546 + 3x + 581 +355 2075 = 1850 + 3x 225 = 3x 75 = x Answer: A
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