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The average (arithmetic mean) cost per book for the 12 books on a cert
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Coruja
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The average (arithmetic mean) cost per book for the 12 books on a cert [#permalink] 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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rudra316
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Re: The average (arithmetic mean) cost per book for the 12 books on a cert [#permalink] 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] 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] 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] New post  15 Dec 2016, 13:39
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Great Official Question.
Here is my solution to this one=>
Mean (12)=k

\(Mean = \frac{Sum}{#}\)

Hence Sum(12)=12k

Now new Sum(12)=12k-18+42=12k+24

New mean =\(\frac{12k+24}{12}=k+2\)

Hence A
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Re: The average (arithmetic mean) cost per book for the 12 books on a cert [#permalink] 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] 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] New post  02 Jan 2020, 15:29
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Hope this is a correct way to solve it as well:

The difference in sum: \(42-18=24\)

\(k+\frac{24}{12}=k+2\)

Answer: A
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Re: The average (arithmetic mean) cost per book for the 12 books on a cert [#permalink] New post  09 Mar 2020, 12:43
rudra316 wrote:
\(Average = \frac{Total Cost}{number of books}\)
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
= \(K+2\)



Hi Guys I'm trying to figure out how you got from here

= \(\frac{S}{12}\) +2

to here

= \(K+2\)[/quote]
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Re: The average (arithmetic mean) cost per book for the 12 books on a cert [#permalink]
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