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Senior Manager  Joined: 10 Jul 2013
Posts: 294
The incomplete table above shows a distribution of scores fo  [#permalink]

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9 00:00

Difficulty:   55% (hard)

Question Stats: 67% (02:28) correct 33% (02:29) wrong based on 190 sessions

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Attachment: incomplete.png [ 3.28 KiB | Viewed 3051 times ]
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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Asif vai.....

Originally posted by Asifpirlo on 21 Aug 2013, 18:19.
Last edited by Bunuel on 22 Aug 2013, 04:08, edited 1 time in total.
Edited the question.
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Re: The incomplete table above shows a distribution of scores fo  [#permalink]

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6
3
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 84-60=24 pts below distributed among 3 students, so 8 pts per student. 83-8 = 75pts.
##### General Discussion
Intern  Joined: 21 Aug 2013
Posts: 6
Re: The incomplete table above shows a distribution of scores fo  [#permalink]

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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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Joined: 03 Oct 2013
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GMAT 1: 780 Q51 V46 Re: The incomplete table above shows a distribution of scores fo  [#permalink]

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1
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(84-60points) 24 points. Since there are 3 students. 24/3 is 8 . So the value will 8 less than the average which is 83-8= 75.
So 75 is the answer.
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Re: The incomplete table above shows a distribution of scores for a class  [#permalink]

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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:
2016-08-28_2152.png

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  [#permalink]

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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:
2016-08-28_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 ?
Manager  Joined: 04 Jan 2014
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GMAT 1: 660 Q48 V32 GMAT 2: 630 Q48 V28 GMAT 3: 680 Q48 V35 Re: The incomplete table above shows a distribution of scores for a class  [#permalink]

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5
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:
2016-08-28_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.

Senior Manager  S
Joined: 07 Sep 2014
Posts: 343
Concentration: Finance, Marketing
Re: The incomplete table above shows a distribution of scores for a class  [#permalink]

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4
1
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:
2016-08-28_2152.png

can ignore 83.

92-83 = 9 so +36
91-83 = 8 so +48

83-71 =12 so -60

total effect = 24
need to be distriubted in 3.
s0 83-8 =75
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Re: The incomplete table above shows a distribution of scores fo  [#permalink]

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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  [#permalink]

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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:
2016-08-28_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  [#permalink]

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1
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 2016-08-28_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

Attachments 24169342_10156971885658289_370232763_o.jpg [ 115.35 KiB | Viewed 1297 times ]

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Re: The incomplete table above shows a distribution of scores for a class  [#permalink]

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(1850+3x)/25=83
x=75

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Re: The incomplete table above shows a distribution of scores for a class  [#permalink]

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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:
2016-08-28_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

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_________________ Re: The incomplete table above shows a distribution of scores fo   [#permalink] 30 Dec 2018, 10:00
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