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

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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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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

Spoiler: OA

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

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28 Sep 2016, 17:39

$Average = \frac{Total Cost}{number of books}$
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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29 Sep 2016, 10:17

Coruja wrote:

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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15 Oct 2016, 08:18

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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15 Dec 2016, 13:39

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]

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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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27 Feb 2017, 05:48

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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02 Jan 2020, 15:29

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] 02 Jan 2020, 15:29

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