|
|
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 is a set A of numbers which has m as its average (arithmetic mean) and R as its standard deviation. If a number x is added to this set and the standard deviation of the new set is r, is \(r \le R\)?
(1) \(x = m\)
(2) \(m < x < m + R\)
(1) \(x = m\)
(2) \(m < x < m + R\)
26 May 2017, 04:45
Kudos
Bookmarks
Official Solution:
If you take the 1st step of the variable approach and modify the original condition and the question, Set \(A=\{a_{1}, a_{2},\ldots \ldots , a_{n}\} mean=m\) and standard deviation=R.
New set \(= \{a_{1}, a_{2}, \ldots \ldots , a_{n}, x\}\) standard deviation=r, and since you have to know 2 sets of elements, there are many variables. Therefore, E is most likely to be the answer.
By solving con 1) and con 2), since the standard deviation shows the distance between each element and the average
In the case of con 1), it becomes the mean of x=set A, which is added to the new set, and since the average that is not distant from the mean is added, r, which is the standard deviation of the new set, agrees with \(r \le R\) , hence yes, it is sufficient.
In the case of con 2), the x added to the new set becomes the number smaller than the standard deviation of set S \((m < x < m + R)\), and therefore, r, which is the standard deviation of the new set, agrees with \(r \le R\), hence yes, it is sufficient. The answer is D. This question too is "CMT 4(B: if you get A or B too easily, consider D)". Con 1) is easy and con 2) is really hard, so by CMT 4(B), the answer is D.
Answer: D
If you take the 1st step of the variable approach and modify the original condition and the question, Set \(A=\{a_{1}, a_{2},\ldots \ldots , a_{n}\} mean=m\) and standard deviation=R.
New set \(= \{a_{1}, a_{2}, \ldots \ldots , a_{n}, x\}\) standard deviation=r, and since you have to know 2 sets of elements, there are many variables. Therefore, E is most likely to be the answer.
By solving con 1) and con 2), since the standard deviation shows the distance between each element and the average
In the case of con 1), it becomes the mean of x=set A, which is added to the new set, and since the average that is not distant from the mean is added, r, which is the standard deviation of the new set, agrees with \(r \le R\) , hence yes, it is sufficient.
In the case of con 2), the x added to the new set becomes the number smaller than the standard deviation of set S \((m < x < m + R)\), and therefore, r, which is the standard deviation of the new set, agrees with \(r \le R\), hence yes, it is sufficient. The answer is D. This question too is "CMT 4(B: if you get A or B too easily, consider D)". Con 1) is easy and con 2) is really hard, so by CMT 4(B), the answer is D.
Answer: D
Moderator:
