monirjewel wrote:
A certain list consists of 21 different numbers. If n is in the list and n is 4 times the average(arithmetic mean) of the other 20 numbers in the list, then n is what fraction of the sum of the 21 numbers in the list?
(A) 1/20
(B) 1/6
(C) 1/5
(D) 4/21
(E) 5/21
A quick solution here is to
plug in some values that meet the given criteria.
Aside: I'm going to ignore the part about the numbers being different, since I have a feeling that this is the author's way of eliminating the possibility that all of the values equal zero (which would ruin the question).
So, the first 20 values (excluding n) could all equal 1, in which case their average (mean) would be 1.
Since n is 4 times the average, n would equal
4.
So, the sum of all 21 values is
24.
Question: n is what fraction of the sum of the 21 numbers in the list?
Answer:
4/
24= 1/6
Answer: B
IMPORTANT: Keep in mind that I broke the rule about all of the numbers being different. What's important here is that we COULD replace the 1's with 20 different numbers that still have a mean of 1, in which case the sum of the first 20 numbers would still be 20, which means the answer will remain
B.
Cheers,
Brent
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