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The average (arithmetic mean) cost per book for the 12 books on a cert

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The average (arithmetic mean) cost per book for the 12 books on a cert  [#permalink]

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New post 28 Sep 2016, 12:30
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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
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Re: The average (arithmetic mean) cost per book for the 12 books on a cert  [#permalink]

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New post 28 Sep 2016, 17:39
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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
= \(K+2\)
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Re: The average (arithmetic mean) cost per book for the 12 books on a cert  [#permalink]

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New post 29 Sep 2016, 10:17
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Coruja wrote:
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


Quote:
he average (arithmetic mean) cost per book for the 12 books on a certain table is k dollars.


Total cost of 12 books = 12k

Quote:
If a book that costs 18 dollars is removed from the table and replaced by a book that costs 42 dollars


Total cost of books now = 12k - 18 + 42 Or, 12k +24

Quote:
what will be the average cost per book, in dollars, for the books in the table


Average cost will be \(\frac{(12k + 24 )}{24}\) = k + 2

Hence answer will be (A) k + 2

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Re: The average (arithmetic mean) cost per book for the 12 books on a cert  [#permalink]

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New post 15 Oct 2016, 08:18
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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
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Re: The average (arithmetic mean) cost per book for the 12 books on a cert  [#permalink]

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New post 15 Dec 2016, 13:39
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Re: The average (arithmetic mean) cost per book for the 12 books on a cert  [#permalink]

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New post 27 Feb 2017, 03:29
Total cost = 12k
After replacement the total cost will be 12k -18 +42 = 12k +24
Average cost is (12k +24)/12 = k+2. Option A
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Re: The average (arithmetic mean) cost per book for the 12 books on a cert  [#permalink]

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New post 27 Feb 2017, 05:48
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we can see the book that has replaced cost more (42) and so we can calculate the diff which ultimately increased the avg cost of the books.
so 42-18=24 can be equally divided among 12 books that would be +2 and now as we know the previous avg that k so new avg=old avg(k)+(2)addition in each books cost =k+2
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Re: The average (arithmetic mean) cost per book for the 12 books on a cert  [#permalink]

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Re: The average (arithmetic mean) cost per book for the 12 books on a cert   [#permalink] 16 Oct 2018, 16:07
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