Comments on: What the Heck is a “Weighted Average”?
https://www.knewton.com/resources/blog/test-prep/what-the-heck-is-a-weighted-average/
Tue, 14 Apr 2015 19:37:41 +0000hourly1http://wordpress.org/?v=4.3.1By: tina
https://www.knewton.com/resources/blog/test-prep/what-the-heck-is-a-weighted-average/#comment-6247
Sat, 26 Oct 2013 17:25:00 +0000http://www.knewton.com/stage/?p=18528#comment-6247Answer is B .. Use the concept of weighted average.. if the 2 averages and combined averages are known then the ratio of male to female can be found
]]>By: tina
https://www.knewton.com/resources/blog/test-prep/what-the-heck-is-a-weighted-average/#comment-6246
Sat, 26 Oct 2013 17:24:00 +0000http://www.knewton.com/stage/?p=18528#comment-6246no, it is always less
]]>By: Kris
https://www.knewton.com/resources/blog/test-prep/what-the-heck-is-a-weighted-average/#comment-3919
Fri, 06 Apr 2012 19:49:00 +0000http://www.knewton.com/stage/?p=18528#comment-3919Answer=C
•From the problem we understand,•Average experience of male employees=9.8•Average experience of female employees=9.1•Required info= male employees/female employees. Ratio••From 1)•Number of male employees= 52•Hence, total experience of male employees=52*9.8• but this gives us no information about females in the company. Their number is not dependent on the number of males and their experience is in no way related for us to calculate their number•Hence 1) is insufficient.•2) average years of experience for all employees=9.3Let F be the number of females and M the number of males•(F*9.1+m*9.1)/(F+M)=9.3•But 2) provides us no info on F and M•Consider 1) and 2),•M=52•Hence, ((F*9.1)+(52*9.8))/(F+52)=9.3•This can be used to solve for F. Hence, C, both 1) and 2) are together sufficient.
]]>By: R. Wayne Moorhead
https://www.knewton.com/resources/blog/test-prep/what-the-heck-is-a-weighted-average/#comment-3481
Fri, 23 Dec 2011 18:23:00 +0000http://www.knewton.com/stage/?p=18528#comment-3481Can the weighed average be greater than the average?
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