think of the 6 volumes as a, b, c, d, e, f
mean = x = (a+b+c+d+e+f)/6
std. deviation = {[(x~a)^2+(x~b)^b+(x~c)^2....+(x~f)^2]/6}^1/2 = 10
now if you take out 30% of the vol. then you are left with 70% in each of them
a' = 0.7a ....and so on
new mean = y = (0.7a + 0.7b ....+0.7f)/6 = 0.7x
std. deviation = {[(y~a')^2+(y~b')^b+...(xy~f')^2]/6}^1/2
== {[0.7(x~a)^2 + .....0.7(x~f)^2]/6}^1/2
sub'ing value given as 10 gallon we get the new std. deviation
(B) gives you the mean but not anythine else.
So (A)