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The average (arithmetic mean) cost per book for the 12 books on a cert
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28 Sep 2016, 13:30
28 Sep 2016, 13:30
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The average (arithmetic mean) cost per book for the 12 books on a certain table is k dollars. If a book that costs 18 dollars is removed from the table and replaced by a book that costs 42 dollars, then in terms of k, what will be the average cost per book, in dollars, for the books in the table?
A) k+2
B) k - 2
C) 12+ 24/k
D) 12- 24/k
E) 12k-6
A) k+2
B) k - 2
C) 12+ 24/k
D) 12- 24/k
E) 12k-6
Re: The average (arithmetic mean) cost per book for the 12 books on a cert
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28 Sep 2016, 18:39
28 Sep 2016, 18:39
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\(Average = \frac{Total Cost}{number of books}\)
Let S be the sum of costs of all the books.
\(K = \frac{S}{12}\)
One book costing 18 is replaced with one costing 42.
Sum of costs = (S-18) + 42 = S+24
Average = \(\frac{(S+24)}{12}\)
= \(\frac{S}{12}\) +2
= \(K+2\)
Let S be the sum of costs of all the books.
\(K = \frac{S}{12}\)
One book costing 18 is replaced with one costing 42.
Sum of costs = (S-18) + 42 = S+24
Average = \(\frac{(S+24)}{12}\)
= \(\frac{S}{12}\) +2
= \(K+2\)
Re: The average (arithmetic mean) cost per book for the 12 books on a cert
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15 Oct 2016, 09:18
15 Oct 2016, 09:18
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Logically, if we replace a book that was $18 with a book that was $42 --> the difference between the two is 24; this number is "absorbed" by the total number of books (i.e. 12), so each book's average goes up by 2
