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

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Re: The average (arithmetic mean) cost per book for the 12 books on a cert [#permalink]
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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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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]
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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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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Hope this is a correct way to solve it as well:

The difference in sum: $$42-18=24$$

$$k+\frac{24}{12}=k+2$$

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