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The incomplete table above shows a distribution of scores for a class

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

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New post 28 Aug 2016, 09:54
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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

[Reveal] Spoiler:
Attachment:
2016-08-28_2152.png
2016-08-28_2152.png [ 5.7 KiB | Viewed 1401 times ]
[Reveal] Spoiler: OA

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

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New post 28 Aug 2016, 10:36
Bunuel wrote:
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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

[Reveal] Spoiler:
Attachment:
2016-08-28_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 [#permalink]

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New post 28 Aug 2016, 10:43
Bunuel wrote:
Image
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

[Reveal] Spoiler:
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 ?

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

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New post 28 Aug 2016, 16:08
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msk0657 wrote:
Bunuel wrote:
Image
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

[Reveal] Spoiler:
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.

Answer (A).

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

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New post 28 Aug 2016, 21:00
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Bunuel wrote:
Image
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

[Reveal] Spoiler:
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 for a class [#permalink]

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New post 15 Nov 2017, 07:24
Bunuel wrote:
Image
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

[Reveal] Spoiler:
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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New post 28 Nov 2017, 05:58
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Hi
I have a different approach to this question, Please confirm that this approach is right. See attachment.

Abhishek009 wrote:
Bunuel wrote:
Image
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

[Reveal] Spoiler:
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
24169342_10156971885658289_370232763_o.jpg [ 115.35 KiB | Viewed 194 times ]

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

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New post 28 Nov 2017, 07:36
(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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New post 29 Nov 2017, 17:06
Bunuel wrote:
Image
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

[Reveal] Spoiler:
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

Answer: A
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Re: The incomplete table above shows a distribution of scores for a class   [#permalink] 29 Nov 2017, 17:06
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