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 ?

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?

_________________

SC Butler has resumed! Get

two SC questions to practice, whose links you can find by date,

here.Choose life.