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Re: The incomplete table above shows a distribution of scores for a class [#permalink]
Could you please clarify in the problem why it is -5/5 the answer and not +5/5 in your method? i.e. we have 5 leftover at the end, why would the answer not be 79 instead of 77

sahasonjoy
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Re: The incomplete table above shows a distribution of scores for a class [#permalink]
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kelly_jacques wrote:
Could you please clarify in the problem why it is -5/5 the answer and not +5/5 in your method? i.e. we have 5 leftover at the end, why would the answer not be 79 instead of 77

sahasonjoy



Refer your PM

+5 here means that if the average of the four numbers were 78, the total would be give 5 spare.

So, the FIFTH number should cater for the extra 5, and how does it do it. It reduces 5 from total if it were 78.
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Re: The incomplete table above shows a distribution of scores for a class [#permalink]
Dear Friend
would you pls clarify the last part of your method? what is the concept behind this method? Still I was not able to figure out why did you reduce 5/5 from 78!
Thanks in advance for your help
chetan2u wrote:
kelly_jacques wrote:
Could you please clarify in the problem why it is -5/5 the answer and not +5/5 in your method? i.e. we have 5 leftover at the end, why would the answer not be 79 instead of 77

sahasonjoy



Refer your PM

+5 here means that if the average of the four numbers were 78, the total would be give 5 spare.

So, the FIFTH number should cater for the extra 5, and how does it do it. It reduces 5 from total if it were 78.
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Re: The incomplete table above shows a distribution of scores for a class [#permalink]
Wonder if there is a less math-intensive/logic based approach that can be used here?

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

The incomplete table above shows a distribution of scores for a class of 20 students. If the average (arithmetic mean) score for the class is 78, what score is missing from the table?

A. 73
B. 75
C. 77
D. 79
E. 81

Attachment:
Class_Incomplete.JPG


Responding to a pm: Use deviations from mean approach for quick calculations.

Mean is 78.
83 is 5 in excess so we have 5*5 = 25 excess (because 83 appears 5 times)
70 has 8 deficit from 78 and appears 6 times so 48 deficit
92 has 14 excess and appears 3 times so 42 excess.
64 has 14 deficit.
Currently excess is 67 and deficit is 62. So we need another 5 deficit to make them equal. Since the unknown number appears 5 times, it must have a deficit of 1 i.e. the number must be 77.

Answer (C)

The following video on Arithmetic Mean will help:

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