IMO A

Let Volume in 6 Tanks be V1, V2, V3, V4, V5, V6
Let average be of the Volumes Be A
Now SD = Sqrt{[(V1-A)^2 + (V2-A)^2 + (V3-A)^2 + (V4-A)^2 + (V5-A)^2 + (V6-A)^2]/6} = 10

Evaluating Statement 1

For each tank 30% of water (amount at beginning) was removed

Now new Volumes 0.7 V1, 0.7 V2, 0.7 V3, 0.7 V4, 0.7 V5, 0.7 V6
New average = 0.7 A
Now new SD = Sqrt{[(0.7V1-0.7A)^2 + (0.7V2-0.7A)^2 + (0.7V3-0.7A)^2 + (0.7V4-0.7A)^2 + (0.7V5-0.7A)^2 + (0.7V6-0.7A)^2]/6}

SD = Sqrt{0.49[(V1-A)^2 + (V2-A)^2 + (V3-A)^2 + (V4-A)^2 + (V5-A)^2 + (V6-A)^2]/6}
SD = Sqrt{0.49 x 100} as we know that the previous sd was 10
SD = 0.7 x 10
SD = 7 Gallons

Conceptually, one could easily avoid the calculations above if one realizes that a constant increase or decrease on all individual elements would result a similar increase or decrease on SD

Hence sufficient


Evaluating Statement 2
New Average was 63

Now this new average could be attained in infinite number of ways resulting in infinite SDs, therefore statement two is insufficient

For Example

{1,1,1,1,1,373}
{63,63,63,63,63,63}

Hi Akansha,

Great question, and good job with your hunch about the right answer. Standard deviation is basically a measure of how "deviant," on average, each element of a set is from the set's true average. In this problem we are told that there are six tanks, each of which may contain a different volume of water, but that on average, a given tank's water volume is 10 gallons deviant from the true average water content. Call their volumes a, b, c, d, e, and f, and call their average x. After they are each reduced by 30%, their new volumes will be .7a, .7b, .7c, .7d, and .7e, and their new average .7x. The average ("standard") deviation from that average .7x will have been scaled down by the exact same factor, so the new standard deviation will be .7 times the old one, and since we're given the old one, 10, we can compute that new one to be .7*10, i.e. 7 gallons.

This may be easier to see with cleaner numbers. Consider, for example, the set {12, 15, 18, 21, 24}. This set is characterized by a certain standard deviation from its average 18. If I now scale this set down by a constant -- let's say I make everything 1/3 of what it was -- I'll get {4, 5, 6, 7, 8}. This new set is characterized by a much smaller average ("standard") deviation -- specifically, it's 1/3 what it was before -- from the new average 6. Now, if I then go and double everything in that set and come out with {8, 10, 12, 14, 16}, I also wind up doubling my standard deviation from the new average 12.

To sum up, multiplying or dividing every element in a set by the same constant -- the same scale factor -- scales the standard deviation by that same constant... the same way if you hold a magnifying glass over a map and you know the magnification factor it is uniformly applying to everything, you know for sure that everything will become that many times as deviant from some fixed point.

(The GMAT will NEVER require you to grind out the calculation of a standard deviation using the mathematical formula, but if you're curious, the way to calculate it is to take the square root of the average of the squares of all the differences from a set's average. Yikes! So, in the example above with {12, 15, 18, 21, 24}, the average is 18, and the differences between each term and that average are -6, -3, 0, 3, and 6. The squares of those differences are then 36, 9, 0, 9, and 36, and the average of those squares is (36+9+0+9+36)/5 = 18. The square root of 18, i.e. 3root2, is your standard deviation. If you apply this algorithm to the scaled down and partway-back-up new sets, you can verify that their standard deviations are, as expected, root 2 and 2root2, respectively. But I repeat, it is NOT necessary to perform that calculation ever on the GMAT!)

Hope that helps!

Ashley Newman-Owens
GMAT Instructor
Veritas Prep

Thanks a lot.. yeah it very much clear now!!

@Akansha , Please use spoiler so that others can attempt without getting biased !

Thanks