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Beth has a collection of 8 boxes of clothing for a charity, and the av

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New post 17 Jul 2016, 10:11
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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 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
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Re: Beth has a collection of 8 boxes of clothing for a charity, and the av  [#permalink]

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New post 18 Jul 2016, 02:24
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(Total items in all boxes / total no of boxes) = average item in each box = c

Total items in all boxes = c*(total no of boxes) = 8c

she replaces a box in the collections that has 12 pieces of clothing with a box that contains 22 pieces of clothing
New total items in all boxes = 8c - 12 + 22 = 8c+10

New average = (New total items in all boxes / total no of boxes) = (8c+10)/8 = c+5/4

Hence, answer will be B.

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Re: Beth has a collection of 8 boxes of clothing for a charity, and the av  [#permalink]

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New post 18 Jul 2016, 07:01
That's the way to go about on this one, yessir!
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Re: Beth has a collection of 8 boxes of clothing for a charity, and the av  [#permalink]

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New post 23 May 2017, 00:10
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14101992 wrote:
(Total items in all boxes / total no of boxes) = average item in each box = c

Total items in all boxes = c*(total no of boxes) = 8c

she replaces a box in the collections that has 12 pieces of clothing with a box that contains 22 pieces of clothing
New total items in all boxes = 8c - 12 + 22 = 8c+10

New average = (New total items in all boxes / total no of boxes) = (8c+10)/8 = c+5/4

Hence, answer will be B.


Hi there!
Thank you for your answer.

If I may ask you how you canceled (8c+10)/8 and got c+5/4? Would you please explain?
Thank you!
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Re: Beth has a collection of 8 boxes of clothing for a charity, and the av  [#permalink]

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New post 24 May 2017, 10:10
michaelkalend wrote:
14101992 wrote:
(Total items in all boxes / total no of boxes) = average item in each box = c

Total items in all boxes = c*(total no of boxes) = 8c

she replaces a box in the collections that has 12 pieces of clothing with a box that contains 22 pieces of clothing
New total items in all boxes = 8c - 12 + 22 = 8c+10

New average = (New total items in all boxes / total no of boxes) = (8c+10)/8 = c+5/4

Hence, answer will be B.


Hi there!
Thank you for your answer.

If I may ask you how you canceled (8c+10)/8 and got c+5/4? Would you please explain?
Thank you!

michaelkalend , because the answer is a year old and because original poster might have unfollowed this topic, I will answer.

Sometimes it's the little things we miss. :)

14101992 split the numerator to simplify, thus:

\(\frac{(8c + 10)}{8}\)

= \(\frac{8c}{8}\) + \(\frac{10}{8}\)

Factor out 8 in the first term. That leaves c

Simplify second term, \(\frac{10}{8}\) = \(\frac{5}{4}\)

And you have c + \(\frac{5}{4}\)

Hope that helps.
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Re: Beth has a collection of 8 boxes of clothing for a charity, and the av  [#permalink]

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New post 24 May 2017, 11:33
genxer123 wrote:
michaelkalend wrote:
14101992 wrote:
(Total items in all boxes / total no of boxes) = average item in each box = c

Total items in all boxes = c*(total no of boxes) = 8c

she replaces a box in the collections that has 12 pieces of clothing with a box that contains 22 pieces of clothing
New total items in all boxes = 8c - 12 + 22 = 8c+10

New average = (New total items in all boxes / total no of boxes) = (8c+10)/8 = c+5/4

Hence, answer will be B.


Hi there!
Thank you for your answer.

If I may ask you how you canceled (8c+10)/8 and got c+5/4? Would you please explain?
Thank you!

michaelkalend , because the answer is a year old and because original poster might have unfollowed this topic, I will answer.

Sometimes it's the little things we miss. :)

14101992 split the numerator to simplify, thus:

\(\frac{(8c + 10)}{8}\)

= \(\frac{8c}{8}\) + \(\frac{10}{8}\)

Factor out 8 in the first term. That leaves c

Simplify second term, \(\frac{10}{8}\) = \(\frac{5}{4}\)

And you have c + \(\frac{5}{4}\)

Hope that helps.



Seems so easy! Thank you very much!
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Beth has a collection of 8 boxes of clothing for a charity, and the av  [#permalink]

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New post 13 Jun 2018, 07:16
Avigano 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



hi pushpitkc generis,

my "approach" to this problem was similar to that of a cancer - sideways and not straightforward :) perhaps thats why i missed something :)

let total number of clothing be \(x\), now since one box containing \(12\) pieces was replaced by bigger box containing \(22\)--> \(22-12 =10\)

so number of boxes did not change. so we have \(\frac{x+10}{8} =c\) ---> \(\frac{x+5}{4} =c\) this is actually the new average number isnt it ? :?

so what`s my next step ? :) i did this \(x+5=4c\) now what ? :-) :hurt:
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Beth has a collection of 8 boxes of clothing for a charity, and the av  [#permalink]

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New post 13 Jun 2018, 08:44
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dave13 wrote:
Avigano 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

hi pushpitkc generis,

my "approach" to this problem was similar to that of a cancer - sideways and not straightforward :) perhaps thats why i missed something :)

let total number of clothing be \(x\), now since one box containing \(12\) pieces was replaced by bigger box containing \(22\)--> \(22-12 =10\)

so number of boxes did not change. so we have \(\frac{x+10}{8} =c\) ---> \(\frac{x+5}{4} =c\) this is actually the new average number isnt it ? :?

so what`s my next step ? :) i did this \(x+5=4c\) now what ? :-) :hurt:

dave13 , fairly close. Conceptual error:
Quote:
so we have \(\frac{x+10}{8} =c\) ---> \(\frac{x+5}{4} = c\) this is actually the new average number isnt it?
No, that last equation is not the new average.
Sum 2 = x + 10
\(n\) has not changed. You are correct.

Careful: (x+10) is after ADDING to original total

You're working in reverse. To get to original average \(c\), SUBTRACT what got added.

Then you could say instead (use MINUS 10)

(New SUM 2 = x) - (added #), divided by 8, equals old average, thus:
\(\frac{x-10}{8}=c\)
This logic gets a little confusing. Split the LHS numerators:

\(\frac{x}{8}-\frac{10}{8}=c\)

\(\frac{x}{8}=c+\frac{10}{8}\)

\(\frac{x}{8}=c+\frac{5}{4}\)

\(x\)= Sum 2
\(8 = n\)
So \(\frac{Sum_2}{n}=A_{new}\)
AND thus \(\frac{x}{8}=A_{new}=c+\frac{5}{4}\)

That works. You needed subtraction instead of addition on LHS. This approach is hard, but you were close. That state of affairs indicates that you mostly or entirely understand the concepts.

This approach is hard in part because it includes another variable that is not needed. But if you like the method, so be it! :)

Another approach without \(x\)? Let's use
A, Average (# of items per box)
n, number of boxes
S, Sum (total items)
\(A*n=S\) and hence \(A=\frac{Sum}{n}\)

Stage 1
\(A*n=Sum_1\)
\(A=c\)
\(n=8\)
\(Sum_1=?\)

Interim: \(Sum_1\) changes.
12 are removed (-12) and 20 are added (+20). That process = \(Sum_2\) at Stage 2

Stage 2
\(A_2= ?\)
\(n=8\)
\(Sum_2\): (Stage 1 Sum) - 12 + 20 = ?

We want new average. Use rearranged formula. If \(A*n=Sum\), then \(A=\frac{Sum}{n}\)

\(A_2=\frac{Sum_2}{n}\)

See whether that works for you? :-)
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Re: Beth has a collection of 8 boxes of clothing for a charity, and the av  [#permalink]

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New post 13 Jun 2018, 16:31
Avigano 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



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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Re: Beth has a collection of 8 boxes of clothing for a charity, and the av  [#permalink]

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New post 02 Feb 2019, 09:08
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Avigano 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

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Re: Beth has a collection of 8 boxes of clothing for a charity, and the av  [#permalink]

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New post 15 Feb 2019, 03:26
The average can also be thought of as the value each input takes IF all of the inputs are equal. For example, the average of 6,7,8 is 7. because 6+7+8=7+7+7...

Now 10 more pieces of clothing get added. These need to be distributed evenly. 10/8=5/4

Since each box averaged c pieces of clothing before, each box now averages c+5/4 pieces of clothing.
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Re: Beth has a collection of 8 boxes of clothing for a charity, and the av  [#permalink]

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New post 03 Mar 2019, 22:51
Avigano 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



Please verify if my method is sound

We can use Test It. Let's say C = 10

That means 10 = (SUM)/8

SUM = 80

Now if one of the boxes were replaced with a box that has 22 pieces of clothing, the new SUM will be 80 + (22-12) = 90
The new value of C should be 90/8 = 11.25

Now plug in 10 in each of the answer choices to see which answer gives 11.25.

A) c- \(\frac{5}{4}\)

10- \(\frac{5}{4}\) = Does not equal 11.25

B) c+ \(\frac{5}{4}\)

10 + \(\frac{5}{4}\) = 11.25 (MATCH)


C) 8- \(\frac{10}{c}\)

8- \(\frac{10}{10}\) = 7 No Match

D) 8+\(\frac{10}{c}\)

8+\(\frac{10}{10}\) = 9 NO match

E) 8c-10 Wayy to big!
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Re: Beth has a collection of 8 boxes of clothing for a charity, and the av   [#permalink] 03 Mar 2019, 22:51
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