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C

Let total employees = x
Taking the worst case.

St1: Mean = (40x/2 + 20x/2)/x = 30: INSUFF

St2: Mean = (60x/4 + 20*3x/4)/x = 30: INSUFF

Combined:
Mean = (40x/2+60x/4 + 20x/4) = 20+15+5 = 40. : SUFF
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I am getting C also

50% > 40 and 25% > 60

My question is can the 50%>40 include the above 60 years too ?

Let us say just 4 employees are there

1 > 60 => 61
2 > 40 => 41 + (61 from above)

The other 2 employees are 21 & 21

Hence average (21 + 21 + 41 + 61) /4 = 144/4 = 36

Hence C
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The answer is (C)


The mean would be te smallest if we pick ages that are the smallest possible, i.e. 21, 41, and 61

ST1. (41n+21n)/(2n) = 31 (Clearly NS)
ST2. (61n+21*3n)/(4n) = 31 (Clearly NS)

ST1 & ST2

The min mean is achieved at

(61n+41n+21*2n)/(4n) = 36 > 35 (So, SUFF)



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