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# Division R of Company Q has 1,000 employees. What is the average (arit

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Re: Division R of Company Q has 1,000 employees. What is the average (arit [#permalink]
cant we just add both averages together and then divide over 2? i know it´s a stupid question but pls guys be so kind and tell me logically why not

thank you
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Re: Division R of Company Q has 1,000 employees. What is the average (arit [#permalink]
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domleon wrote:
cant we just add both averages together and then divide over 2? i know it´s a stupid question but pls guys be so kind and tell me logically why not

thank you

No. That cannot be done. Check the following.

if total employees of division A is x1 and total salary is s1 then average = s1/x1.
If rest of employees is x2 and total salary of these is s2 then average = s2/x2.

Sum of the two averages = (s1/x1) + (s2/x2).

However, average at the company level = (s1+s2)/(x1+x2) and this is not the same as (s1/x1) + (s2/x2).

I hope, this helps.
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Re: Division R of Company Q has 1,000 employees. What is the average (arit [#permalink]
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We can express the average annual salary of Company Q as:

$$S = \frac{S_r * N_r+ S_{not-r}*N_{not-r}}{N_r+N_{not-r}}$$

where
- $$S_r$$ - The average annual salary of the employees in Division R
- $$S_{not-r}$$ - The average annual salary of the employees who are not in Division R
- $$N_r$$ - The number of employees in Division R
- $$N_{not-r}$$ - The number of employees who are not in Division R

So, we don't know the number of employees who are not in Division R.

You can also use huge/tiny numbers to check your answer. If $$N_{not-r}=0$$ or only 1 employee, S = 30K or will be very close to 30K. At the same time, if $$N_{not-r}>>1000$$, S will be very close to 35K. So we need to know at least $$\frac{N_{not-r}}{N_r}$$ ratio.
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Re: Division R of Company Q has 1,000 employees. What is the average (arit [#permalink]
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Re: Division R of Company Q has 1,000 employees. What is the average (arit [#permalink]
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