|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
26 May 2017, 04:43
Kudos
Bookmarks
There was an experiment on the increasing number of bacteria in the culture dish at the beginning and at the end of each month for 1 year. If the standard deviation of the number of bacteria at the beginning of each month was 2,000, what was the standard deviation of the number of bacteria at the end of each month?
(1) The number of bacteria at the beginning of each month increased by 50% when compared to the number of bacteria at the end of each month.
(2) The average (arithmetic mean) number of bacteria at the beginning of each month is 12,000,000.
(1) The number of bacteria at the beginning of each month increased by 50% when compared to the number of bacteria at the end of each month.
(2) The average (arithmetic mean) number of bacteria at the beginning of each month is 12,000,000.
26 May 2017, 04:45
Kudos
Bookmarks
Official Solution:
In the original condition, if you assume the standard deviation of the bacteria at the beginning of each month as \(S(B)\), and the standard deviation of the bacteria at the end of each month as \(S(E)\) (here, \(S(X)=\)standard deviation of \(X, B=\) the state at the Beginning, and \(E=\)the state at the End, \(S(B)=2,000\), and the question\(=S(E)=\)?. However, if the elements in a certain set X move in parallel, the standard deviation always keeps the same value, so you get \(S(aX + b) = |a|S(X)\). In other words, from, \(aX + b\), the \(b\) moves in parallel, and therefore it has no effect on the standard deviation. Also, for DS questions, when one condition is a number and the other condition is a ratio (percent), the ratio (percent) is likely to be the answer.
Thus, by solving con 1) and con 2), con 1) is a ratio, and con 2) is a number, so con 1) is most likely to be the answer.
In the case of con 1), in fact, the number of bacteria has increased by 505 since the beginning of each month, so you get \(S(E) = S(B + 50 \% B) = S(1.5B) = |1.5|S(B) = 1.5(2,000) = 3,000\), and hence it is unique and sufficient.
In the case of con 2), the average (arithmetic mean) does not affect the standard deviation, and hence it is not sufficient. The answer is A.
Answer: A
In the original condition, if you assume the standard deviation of the bacteria at the beginning of each month as \(S(B)\), and the standard deviation of the bacteria at the end of each month as \(S(E)\) (here, \(S(X)=\)standard deviation of \(X, B=\) the state at the Beginning, and \(E=\)the state at the End, \(S(B)=2,000\), and the question\(=S(E)=\)?. However, if the elements in a certain set X move in parallel, the standard deviation always keeps the same value, so you get \(S(aX + b) = |a|S(X)\). In other words, from, \(aX + b\), the \(b\) moves in parallel, and therefore it has no effect on the standard deviation. Also, for DS questions, when one condition is a number and the other condition is a ratio (percent), the ratio (percent) is likely to be the answer.
Thus, by solving con 1) and con 2), con 1) is a ratio, and con 2) is a number, so con 1) is most likely to be the answer.
In the case of con 1), in fact, the number of bacteria has increased by 505 since the beginning of each month, so you get \(S(E) = S(B + 50 \% B) = S(1.5B) = |1.5|S(B) = 1.5(2,000) = 3,000\), and hence it is unique and sufficient.
In the case of con 2), the average (arithmetic mean) does not affect the standard deviation, and hence it is not sufficient. The answer is A.
Answer: A
Moderator:
