Beth has a collection of 8 boxes of clothing for a charity

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Gmat_mission: Legendary Member; Posts: 1622; Joined: Thu Mar 01, 2018 7:22 am; Followed by:2 members

Beth has a collection of 8 boxes of clothing for a charity

by Gmat_mission » Mon Jun 11, 2018 1:55 am

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Beth has a collection of 8 boxes of clothing for a charity, and the average (arithmetic mean) number of pieces of clothing per box is c. If she replaces a box in the collection that has 12 pieces of clothing with a box that contains 22 pieces of clothing, what is the average number of pieces of clothing per box for the new collection in terms of c? $$\left(A\right)\ c-\frac{5}{4}$$ $$\left(B\right)\ c+\frac{5}{4}$$ $$\left(C\right)\ 8-\frac{10}{c}$$ $$\left(D\right)\ 8+\frac{10}{c}$$ $$\left(E\right)\ 8c-10$$ [spoiler]OA=B[/spoiler].

I am really confused here. I don't know how to set the equations for solving this PS question. <i class="em em-sob"></i>

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Source: Beat The GMAT — Problem Solving | Mon Jun 11, 2018 1:55 am

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pm; Location: New York, NY; Thanked: 13060 times; Followed by:1906 members; GMAT Score:790

by GMATGuruNY » Mon Jun 11, 2018 2:59 am

Gmat_mission wrote:Beth has a collection of 8 boxes of clothing for a charity, and the average (arithmetic mean) number of pieces of clothing per box is c. If she replaces a box in the collection that has 12 pieces of clothing with a box that contains 22 pieces of clothing, what is the average number of pieces of clothing per box for the new collection in terms of c? $$\left(A\right)\ c-\frac{5}{4}$$ $$\left(B\right)\ c+\frac{5}{4}$$ $$\left(C\right)\ 8-\frac{10}{c}$$ $$\left(D\right)\ 8+\frac{10}{c}$$ $$\left(E\right)\ 8c-10$$ [spoiler]OA=B[/spoiler].

Let c=2, implying that the average number of pieces per box = 2.
Total number of pieces = (number of boxes)(average per box) = 8*2 = 16.
When a 12-piece box is removed, the resulting total = 16-12 = 4.
When a 22-piece box is added, the resulting total = 4+22 = 26.
Resulting average = (total number of pieces)/(number of boxes) = 26/8 = 13/4.

Now plug c=2 into the answers to see which yields a resulting average of 13/4.
Only B works:
c + 5/4 = 2 + (5/4) = (8/4) + (5/4) = 13/4.

The correct answer is B.

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Jeff@TargetTestPrep: GMAT Instructor; Posts: 1462; Joined: Thu Apr 09, 2015 9:34 am; Location: New York, NY; Thanked: 39 times; Followed by:22 members

by Jeff@TargetTestPrep » Wed Jun 13, 2018 3:44 pm

Gmat_mission wrote:Beth has a collection of 8 boxes of clothing for a charity, and the average (arithmetic mean) number of pieces of clothing per box is c. If she replaces a box in the collection that has 12 pieces of clothing with a box that contains 22 pieces of clothing, what is the average number of pieces of clothing per box for the new collection in terms of c? $$\left(A\right)\ c-\frac{5}{4}$$ $$\left(B\right)\ c+\frac{5}{4}$$ $$\left(C\right)\ 8-\frac{10}{c}$$ $$\left(D\right)\ 8+\frac{10}{c}$$ $$\left(E\right)\ 8c-10$$ [spoiler]OA=B[/spoiler].

The sum of Beth's original boxes is 8c.

After she replaces a 12-piece box with a 22-piece box, the new sum is 8c - 12 + 22 = 8c + 10, so the new average is:

(8c + 10)/8 = 8c/8 + 10/8 = c + 5/4

Answer: B

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