SOLUTION: Suppose the average of 999 numbers is also 999. Suppose further that from these numbers you chose 729 of them and their computed average is coincidentally also 729. What would be

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Question 1148996: Suppose the average of 999 numbers is also 999. Suppose further that from these numbers you chose 729 of them and their computed average is coincidentally also 729. What would be the average of the remaining numbers?
Found 2 solutions by josmiceli, MathTherapy:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +T+ = the total value of the 999 numbers
+999+=+T%2F999+
+T+=+999%5E2+
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Let +A+ = the average of remaining +270+ numbers
Tje total value of the +729+ number is +729%2A729+
+999+=+%28+729%5E2+%2B+A%2A270+%29+%2F+999+
+999%5E2+=+729%5E2+%2B+270A+
+998001+=+531441+%2B+270A+
+466560+=+270A+
+A+=+1728+ answer
Get a 2nd opinion if needed


Answer by MathTherapy(10553) About Me  (Show Source):
You can put this solution on YOUR website!
Suppose the average of 999 numbers is also 999. Suppose further that from these numbers you chose 729 of them and their computed average is coincidentally also 729. What would be the average of the remaining numbers?
If average of 999 numbers is 999, then sum of those 999 numbers = 999(999), or 9992

If average of 729 numbers is 729, then sum of those 729 numbers = 729(729), or 7292

Therefore, sum of the other 270 (999  -  729) numbers is: 9992 - 7292

Further, the average of these 270 remaining numbers would be: 

Although somewhat irrelevant, the answer can also be expressed as: 

What to learn from this: The average of a LIST of 999 numbers is 999. Of these 999 numbers, 729 have an average of 729, and the remaining 270 

                         have an average of 999 + 729. Quite UNUSUAL, isn't it?