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In 1995 Division A of Company X had 4850 customers. If there

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Intern
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Joined: 12 Feb 2009
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In 1995 Division A of Company X had 4850 customers. If there [#permalink]

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New post 08 May 2009, 08:22
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In 1995 Division A of Company X had 4850 customers. If there were 86 service errors in Division A that year, what was the service error rate in number of service errors per 100 customers for Division B of Company X?

(1) In 1995 the overall service error rate for Divisions A and B combined was 1.5 service errors per 100 customers

(2) In 1995 Division B had 9350 customers, none of whom are customers of A.

Why isn't (1) alone sufficient?

((86/4850)*100))+(X*100)=1.5

I think i see now. You're taking a weighted avg, so can't solve for X without knowing the weight?

Really is
((86/4850)*100))+(X/Y*100)=1.5

Kudos [?]: 10 [0], given: 0

Senior Manager
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Joined: 08 Jan 2009
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Kudos [?]: 175 [0], given: 5

Re: Word Translations [#permalink]

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New post 08 May 2009, 18:09
What you did is Total Avg = (Avg) A + (Avg) B .

The above eqn is does not hold good. I took two sets of numbers
A = ( 2,4,7,8) b =( 2,3,6,7,5) where Sum A = 21 and Sum B = 23
21/4 + 23/5 /= 45/8
9.45 /= 5.5

So Total Avg = Sum of terms in A and B / No of terms.

i think both statements would be needed to solve. Correct me if i am wrong.

Kudos [?]: 175 [0], given: 5

Re: Word Translations   [#permalink] 08 May 2009, 18:09
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In 1995 Division A of Company X had 4850 customers. If there

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