Ans should be A

stmt 1: For each tank,30%of the volumn of water that was in the tank at the begining of the experiment was removed during the experiment.

let volume of water in each of the six tanks be a,b,c,d,e,f respectivley

so mean M = (a+b+c+d+e+f)/6

after 30% water is removed the volume of water in each of the tanks is
7a/10 ,7b/10 ..........

& new mean = 7/10 * M

so std dev = sqrt ((7a/10 -7M/10)^2 +......)/5

= sqrt (7^2/10^2 ( (a-M)^2 +(b-M)^2 ....)/5) )

= 7/10 * sqrt((a-M)^2 +(b-M)^2 ....)/5)

=7/10 *10 =7

Here one can say that since all the values are decreased by the same percentage, the mean will decrease by the percentage, and since the SD is the mean distance of values from the mean, so it must decrease by the same percent (imagine this a miniature distribution of the original)

so we can directly say that the new sd = 70% of old i.e 7

SUFF

stmt 2: New mean is 63, but this does not convet the changes that took place i.e the amt of water removed from the tanks(could be same, more from 1 less from other etc). so INSUFF

So ans should be A