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Re: Beth has a collection of 8 boxes of clothing for a charity, [#permalink]
1
Carcass 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 collections 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?

A) c- \(\frac{5}{4}\)
B) c+ \(\frac{5}{4}\)
C) 8- \(\frac{10}{c}\)
D) 8+\(\frac{10}{c}\)
E) 8c-10


One box contains 12 items of clothing
Let t, u, v, w, x, y, z = the number of items in each of the other 7 boxes

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.
We can write: (t + u + v + w + x + y + z + 12)/8 = c

If she replaces a box in the collections 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?
If we replace the box with 12 items with a box with 22 items, the NEW average = (t + u + v + w + x + y + z + 22)/8
If we replace 22 with 12 + 10, then the NEW average = (t + u + v + w + x + y + z +12 + 10)/8

USEFUL PROPERTY: (a + b)/c = a/c + b/c
We apply the above property to write: NEW average = (t + u + v + w + x + y + z +12)/8 + 10/8
Replace first fraction with c to get: NEW average = c + 10/8
Simplify remaining fraction to get: NEW average = c + 5/4

Answer: B

Cheers,
Brent
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