During an experiment, some water was removed from each of the 6 water tanks. If the standard deviation of the volumes of water in the tanks at the beginning of the experiment was 10 gallons, what was the standard deviation of the volumes of water in the tanks at the end of the experiment?

1) For each tank, 30% of the volume of water that was in the tank at the beginning of the experiment was removed during the experiment.

2) The average (arithmetic mean) volume of water in the tanks at the end of the experiment was 63 gallons.

Please share your answers and explanations.

What is OA? Is it E?

The answer is (A).

I couldn't come up with a proper explanation. The only way I can think of is that if the same 30% is removed from each tank, the standard deviation may remain the same. I know that the standard deviation of the sample doesn't change if we add or subtract the same constant value to the sample values. Here as per Statement (1), even though the same 30% of water is removed, it need not be the same value. So I am confused.

Std. Dev. = sqrt[sum for all terms of (X - mean)^2/n], where X is each term and n is the number of terms.

This is probably not the best way to represent it (the limitations of posting on a forum, alas!) but for our purposes it is OK. On the GMAT it is more important to understand what std. dev. represents than to know how to calculate it.

Qualitatitively, std. dev. is a measure of the spread of the data.

Harish, the source of your confusion is your statement "I know that the standard deviation of the sample doesn't change if we add or subtract the same constant value to the sample values." That is only true if all of the samples have the same quantity to begin with (std. dev. = 0)!

The more accurate statement would have been "The standard deviation of the sample changes by a known factor if we add or subtract the same percentage to each of the sample values." If the samples each decrease by 30%, the mean decreases by 30%, and the (X - mean) decreases by 30% for each term. You don't really have to complete the calculation to see that the resulting std. dev. will be smaller than the original 10 by some factor (I believe the result would be 7, but you can check my math).

Great explanation. Thanks a lot.

Ron,

can you please explain me how can (X - mean) decreases by 30%?
If X and mean both are decreased by the same percentage then there difference of X and mean will still be same. Right?
for Ex: say
X = 10 and Mean = 5
so initially X- Mean = 5

Now lets say we reduce both by some n%age. ie. reduce both of a value of 2, then

X = 10 - 2 => 8
Mean = 5 - 2 => 3

Now again X - Mean = 5 (same as the initial one)

Please correct me if I'm wrong.

Lets say V = Avergage vol - (V1 + V2 + V3 + .. V6 )/6

Sqr(Old Std Dev ) = (V1-V)^2 + (V2-V)^2 + ... (V6-V)^2

New Mean = (.7*V1 + .7*V2 +.7*V3 ... )/6 = .7*V ; .7 times the old mean

say V1' = new volume
V' = new Mean volume

Sqr(New Std Dev) = (V1' - V')^2 + (V2'-V')^2 ... + (V6'-V')^2
= (.7)^2 { Old Std Dev}

whoa there.

you can PICK ONE change to make:
EITHER
* change the numbers by the same percentage,
OR
* change the numbers by the same numerical amount.

you cannot do both of these at once, unless the two numbers are the same to start with, because the same absolute change will constitute a different percentage of each of the two starting numbers.
in the example you've quoted, 2 is 20% of 10, so you're decreasing the original value of 10 by 20%. by contrast, 2 is 40% of 5, so you're decreasing the original value of 5 by 40%.
if you were to decrease by identical percentages, then you'd either do 10 - 4 and 5 - 2 (40% decrease each), or 10 - 2 and 5 - 1 (20% decrease each).