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For a collection of measurements, the average value was found to be m.

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For a collection of measurements, the average value was found to be m. [#permalink]

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New post 02 Jan 2018, 23:54
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For a collection of measurements, the average value was found to be m. If every value in the collection was squared, what is the average value of the resulting collection?

(1) m = 5
(2) The sum of all measurements in the original collection is 100.
[Reveal] Spoiler: OA

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Re: For a collection of measurements, the average value was found to be m. [#permalink]

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New post 03 Jan 2018, 03:30
Bunuel wrote:
For a collection of measurements, the average value was found to be m. If every value in the collection was squared, what is the average value of the resulting collection?

(1) m = 5
(2) The sum of all measurements in the original collection is 100.


E.

There a few ways to solve this:

Using logic: knowing an average is enough to find the average of a new series in which a uniform change has been made to all values (say, multiplying by 2 or adding 1). But since squaring is basically multiplying each value in a collection by a different number (itself) - we can't know the average of the new series without knowing more about the values. Knowing the original sum (statement 2) isn't enough, since there are many different ways to get to this sum. The information is insufficient.

A precise Solution:
This question describes an equation: average = sum/ n. Statements 1 and 2 only give us one of the three factors needed - so they are clearly insufficient. Combining them we have both average and sum - which is enough to find the the third factor (n) for the original equation. However, we are being asked here about a new series, one in which all values are squared - we have the n, but don't know the sum - and thus don't have enough information to find the average.


Solution with the Alternative method of using numbers:
1) this could be a collection of just one number -"5" - in which case the average value of the squared collection is 25. But it could also be a collection of 10 and 0, in which case the average of the squared collection is (0 + 100)/2 = 50. Two different answers - Insufficient!
2) Once again, we'll take the easiest possible example - this could be a collection of only the number 100 > average of the squared collection = 10,000. But if it's a collection of (0, 200) > average of squared = (0 + 40,000)/2 = 20,000. Two different answers - Insufficient!
Combined: here the original collection could be, say, 20 times 5, but it could also be 100 and nineteen zeros. The average value for the first series squared would of course be 25, and for the second series (10,000 + 0 + 0... = 10,000)/20= 500. Two different answers - Insufficient!
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Re: For a collection of measurements, the average value was found to be m. [#permalink]

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New post 03 Jan 2018, 09:58
Bunuel wrote:
For a collection of measurements, the average value was found to be m. If every value in the collection was squared, what is the average value of the resulting collection?

(1) m = 5
(2) The sum of all measurements in the original collection is 100.


S1 - m=5
There can be any set of nos. with average 5 whose average of the squares will be different.
For example consider, nos. 4,5,6. Average =5
Average of squares = (16+25+36)/3 = 25.6
If we take different set of nos. the average will be different.
Insufficient.

S2- Sum of the measurements = 100.
Again, there can be any set of nos.
Insufficient.

Combining, also insufficient since there will be many numbers having sum 100 and average 5.
Answer E.
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Re: For a collection of measurements, the average value was found to be m.   [#permalink] 03 Jan 2018, 09:58
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For a collection of measurements, the average value was found to be m.

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