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10 students took a chemistry exam that was graded on a scale of 0 to 100. Five of the students were in Dr. Adams’ class and the other five students were in Dr. Brown’s class. Is the median score for Dr. Adams’ students greater than the median score for Dr. Brown’s students?
(1) The range of scores for students in Dr. Adams’ class was 40 to 80, while the range of scores for students in Dr. Brown’s class was 50 to 90.
(2) If the students are paired in study teams such that each student from Dr. Adams’ class has a partner from Dr. Brown’s class, there is a way to pair the 10 students such that the higher scorer in each pair is one of Dr. Brown’s students.
(1) The range of scores for students in Dr. Adams’ class was 40 to 80, while the range of scores for students in Dr. Brown’s class was 50 to 90.
(2) If the students are paired in study teams such that each student from Dr. Adams’ class has a partner from Dr. Brown’s class, there is a way to pair the 10 students such that the higher scorer in each pair is one of Dr. Brown’s students.
30 Jun 2010, 22:39
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This is a Yes/No Data Sufficiency question that never requires you to identify the specific values of the medians-- just to answer a question about their relative values.
Remember that the median--as opposed to the mean--only depends on the value of the middle term (or 2 middle terms if you have an even number of elements in your list, but each class has an odd number of students here) when all the terms are listed in order. For example:
For the set {1, 1, 1, 2, 100} -- median = 1
For the set {1, 1, 1, 1, 1} -- median = 1
For the set {-1000, -100, 1, 20, 99999}-- yup, median still equals 1.
Statement 1 tells you about the RANGE of scores in class A(dams). Range only involves the endpoints of the list. The middle term could be ANY number within that range. For example:
{40, 40, 40, 40, 80} -- median = 40
{40, 50, 60, 70, 80} -- median = 60
{40, 80, 80, 80, 80} -- median = 80
Likewise, a range of 50 to 90 could give you any of the following for class B(rown):
{50, 50, 50, 50, 90} -- median = 50
{50, 51, 62, 84, 90} -- median = 62
{50, 90, 90, 90, 90} -- median = 90
In order for the statement to be sufficient to answer the question, you would have to be say whether all possible medians for all possible set As were always greater than (or always less than) the medians for all possible set Bs. Since we have a "sometimes it might be, sometimes it might not" situation, this statement is INSUFFICIENT.
Statement 2, however, tells you that it's possible to pair up each student so that each score in set A corresponds to a HIGHER score in set B.
So if you listed out the elements in set A in ascending order: { p, q, r, s, t}
then set B must contain:
{#bigger than p, #bigger than q, #bigger than r, #bigger than s, #bigger than t}
What is the median of set A? The middle term-- r. What's the median of set B? Something bigger than r. Even though I have NO IDEA what those numbers are, if I put them in order I know that the middle term of set B will have to be greater than the middle term of set A. SUFFICIENT.
Hope this helps.
Remember that the median--as opposed to the mean--only depends on the value of the middle term (or 2 middle terms if you have an even number of elements in your list, but each class has an odd number of students here) when all the terms are listed in order. For example:
For the set {1, 1, 1, 2, 100} -- median = 1
For the set {1, 1, 1, 1, 1} -- median = 1
For the set {-1000, -100, 1, 20, 99999}-- yup, median still equals 1.
Statement 1 tells you about the RANGE of scores in class A(dams). Range only involves the endpoints of the list. The middle term could be ANY number within that range. For example:
{40, 40, 40, 40, 80} -- median = 40
{40, 50, 60, 70, 80} -- median = 60
{40, 80, 80, 80, 80} -- median = 80
Likewise, a range of 50 to 90 could give you any of the following for class B(rown):
{50, 50, 50, 50, 90} -- median = 50
{50, 51, 62, 84, 90} -- median = 62
{50, 90, 90, 90, 90} -- median = 90
In order for the statement to be sufficient to answer the question, you would have to be say whether all possible medians for all possible set As were always greater than (or always less than) the medians for all possible set Bs. Since we have a "sometimes it might be, sometimes it might not" situation, this statement is INSUFFICIENT.
Statement 2, however, tells you that it's possible to pair up each student so that each score in set A corresponds to a HIGHER score in set B.
So if you listed out the elements in set A in ascending order: { p, q, r, s, t}
then set B must contain:
{#bigger than p, #bigger than q, #bigger than r, #bigger than s, #bigger than t}
What is the median of set A? The middle term-- r. What's the median of set B? Something bigger than r. Even though I have NO IDEA what those numbers are, if I put them in order I know that the middle term of set B will have to be greater than the middle term of set A. SUFFICIENT.
Hope this helps.
General Discussion
Please explain in detail...
10 students took a chemistry exam that was graded on a scale of 0 to 100. Five of the students were in Dr. Adams’ class and the other five students were in Dr. Brown’s class. Is the median score for Dr. Adams’ students greater than the median score for Dr. Brown’s students?
(1) The range of scores for students in Dr. Adams’ class was 40 to 80, while the range of scores for students in Dr. Brown’s class was 50 to 90.
(2) If the students are paired in study teams such that each student from Dr. Adams’ class has a partner from Dr. Brown’s class, there is a way to pair the 10 students such that the higher scorer in each pair is one of Dr. Brown’s students.
10 students took a chemistry exam that was graded on a scale of 0 to 100. Five of the students were in Dr. Adams’ class and the other five students were in Dr. Brown’s class. Is the median score for Dr. Adams’ students greater than the median score for Dr. Brown’s students?
(1) The range of scores for students in Dr. Adams’ class was 40 to 80, while the range of scores for students in Dr. Brown’s class was 50 to 90.
(2) If the students are paired in study teams such that each student from Dr. Adams’ class has a partner from Dr. Brown’s class, there is a way to pair the 10 students such that the higher scorer in each pair is one of Dr. Brown’s students.
