ratios + weighted average

The y key stopped responding on my key board so reposting the same question!



At a certain company, the average (arithmetic mean) number of years of experience is 9.8 years for the male employees and 9.1 years for the female employees. What is the ratio of the number of the company's male employees to the number of the company's female employee?

(1) There are 52 male employees at the company
(2) The average number of years of experience for the company's male and female employees combined is 9.3 years.



[spoiler]OA: B [/spoiler]
My question is how is the number of years of service related to the number of males/females. Lets sa the average ears of experience of females
is 10 (there could be 2 females wiht 5 earsof experiecne each or 5 females with 2 ears of experience each) The ratio fo males to females changes.

Why is not E)? Can someone please clear my confusion?

Sorry for the confusion

I figured it out!

Thanks Ian!!

The average of averages =
Sum of experiences of both groups / total employees

9.3 = (M)(9.8) +(F)(9.1) / (M + F)

from there - you can find M/F
Sufficient.