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If set S consists of the numbers 1, 5, -2, 8, and n, is 0 < n < 7 ?
(1) The median of the numbers in S is less than 5.
(2) The median of the numbers in S is greater than 1.
(1) The median of the numbers in S is less than 5.
(2) The median of the numbers in S is greater than 1.
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Stiv
If set S consists of the numbers 1, 5,-2, 8 and n, is 0 < n < 7?
Note that:
If a set has odd number of terms the median of a set is the middle number when arranged in ascending or descending order;
If a set has even number of terms the median of a set is the average of the two middle terms when arranged in ascending or descending order.
So the median of our set of 5 numbers: {-2, 1, 5, 8, n} must be the middle number, so it can be:
1 if \(n\leq{1}\);
5 if \(n\geq{5}\);
\(n\) itself if \(1\leq{n}\leq{5}\).
(1) The median of the numbers in S is less than 5 --> so either the median=1 and in this case \(n\leq{1}\) so not necessarily in the range \(0<n<7\) or median=n and in this case \(1\leq{n}<{5}\) and in this case \(0<n<7\) is always true. Not sufficient.
(2) The median of the numbers in S is greater than 1 --> again either the median=5 and in this case \(n\geq{5}\) so not necessarily in the range \(0<n<7\) or median=n and in this case \(1<{n}\leq{5}\) and in this case \(0<n<7\) is always true. Not sufficient.
(1)+(2) \(1<median<5\) --> \(median=n\) --> \(1<n<5\), so \(0<n<7\) is true. Sufficient.
Answer: C.
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General Discussion
Thank you! It's clear now.
Dear expert
Can the answer be D .
I understand the question asking whether 0<n<7
From Statement 1 , we get 1=<n<5, hence n is within 0<n<7 , thus it is sufficient
From statement 2, we get 1<n<=5, n is within 0<n<7, thus it is sufficient ,
Am I understand it wrongly
Appreciate could shed some lights
Can the answer be D .
I understand the question asking whether 0<n<7
From Statement 1 , we get 1=<n<5, hence n is within 0<n<7 , thus it is sufficient
From statement 2, we get 1<n<=5, n is within 0<n<7, thus it is sufficient ,
Am I understand it wrongly
Appreciate could shed some lights
apple08
Hi,
you do not get what you are assuming above.
i'll try to explain where you are going wrong..
If set S consists of the numbers 1, 5, -2, 8, and n, is 0 < n < 7 ?
(1) The median of the numbers in S is less than 5.
median means the central value..
lets put these numbers in ascending order..
-2,1,5,8 and n ; we are given median is less than 5....
so n can take any value less than 5..
if its 4... -2,1,4,5,8.... median is 4
if its 0... -2,0,1,5,8.... median is 1
so if n is between 1 and 5, n is the median and if n < =1, median will be 1...
so n <5..... insuff, as we are asked if 0 < n < 7..
n can be anything -100,-50
(2) The median of the numbers in S is greater than 1..
now lets put these numbers in ascending order..
-2,1,5,8 and n ; we are given median >1....
so n can take any value >1..
if its 2... -2,1,2,5,8.... median is 2
if its 6 ... -2,0,5,6,8.... median is 5
so if n is between 1 and 5, n is the median and if n > =5, median will be 5...
so n >1..... insuff, as we are asked if 0 < n < 7..
n can be anything 70,6 etc..
combined we know the median is between 1 and 5..
this means n will be the meadian..
so n is between 1 and 5, suff
C
hope helps
_________________
GMAT 1: 760 Q51 V42
Posts: 10,057
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.
If set S consists of the numbers 1, 5, -2, 8, and n, is 0 < n < 7 ?
(1) The median of the numbers in S is less than 5.
(2) The median of the numbers in S is greater than 1.
When it comes to inequality for DS questions, if range of que include range of con, it means the con is sufficient.
In the original condition, there is 1 variable(n), which should match with the number of equations. So you need 1 equation. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer.
For 1), if n<5, the range of que doesn’t include the range of con, which is not sufficient.
For 2), if 1<n, the range of que doesn’t include the range of con, which is not sufficient.
When 1) & 2), they become 1<n<5. The range of que includes the range of con, which is sufficient. Therefore, the answer is C.
For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
If set S consists of the numbers 1, 5, -2, 8, and n, is 0 < n < 7 ?
(1) The median of the numbers in S is less than 5.
(2) The median of the numbers in S is greater than 1.
When it comes to inequality for DS questions, if range of que include range of con, it means the con is sufficient.
In the original condition, there is 1 variable(n), which should match with the number of equations. So you need 1 equation. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer.
For 1), if n<5, the range of que doesn’t include the range of con, which is not sufficient.
For 2), if 1<n, the range of que doesn’t include the range of con, which is not sufficient.
When 1) & 2), they become 1<n<5. The range of que includes the range of con, which is sufficient. Therefore, the answer is C.
For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
chetan2u
Hi Chetan,
I also tried the same approach and marked E when I deduced 1<n<5 as question is stating if 0<n<7
ISN'T 1<n<5 subset of 0<n<7 , so how can we say n exists between 0 and 7
Please resolve the confusion
kanigmat011
Hi kani,
any value of n is between 1 and 5, this has to be true for whatever this set is a subset of...
say n is 2 or 3, it will always be between the main set..
so if you get your answer as 1<n<5 and the main Q asks you if n is a positive integer, that would be true since any value of n will be true
the vice versa will not be correct..
that is we get 0<n<7 and answer asks if 1<n<5,.. it is not suff, as it doe snot contain 6,5 and 1...
_________________
Point Noted
Thanks Chetan
Thanks Chetan
Posts: 35
If set S consists of the numbers 1, 5, -2, 8, and n, is 0 < n < 7 ?
(1) The median of the numbers in S is less than 5.
(2) The median of the numbers in S is greater than 1.
Given numbers are
-2 1 5 8 and n
Statement 1:
if median is less than 5
it can be 1 or n
if it is 1 => n<1
if it is n => n<5
not sufficient
Statement 2:
median is greater than 1
=>median can be 5 or n
if 5 =>n>5
if n => 1<n<5
not sufficient
Together
median is >1 and <5
=> n is 1<n<5
thus answer option C
(1) The median of the numbers in S is less than 5.
(2) The median of the numbers in S is greater than 1.
Given numbers are
-2 1 5 8 and n
Statement 1:
if median is less than 5
it can be 1 or n
if it is 1 => n<1
if it is n => n<5
not sufficient
Statement 2:
median is greater than 1
=>median can be 5 or n
if 5 =>n>5
if n => 1<n<5
not sufficient
Together
median is >1 and <5
=> n is 1<n<5
thus answer option C
HarveyKlaus
Hi
Arrange the numbers in ascending order..
-2,1,5,8...
1) median is less than 5..
So n can be anything below 5 that is n<5..
If n is between 1 and 5, median is n,
And if n<1, median is 1
So n can be in the range 0 to 7 or less than 0
Insuff.
2) median is greater than 1.
So n >1..
If n is between 1 and 5, median is n, otherwise 5.
Insuff
Combined n<5 and n>1, so 1<n<5..
So ans is YES for 0<n<7
Suff
C
_________________
HarveyKlaus
Set \(S = \{-2, 1, 5, 8, n \}\)
(1) If median of the numbers in S is less than 5, we have n<5.
If \(n<-2\), set \(S = \{ n, -2, 1, 5, 8 \}\). Median of S is 1.
If \(-2<n<1\), set \(S = \{ -2, n, 1, 5, 8 \}\). Median of S is 1.
If \(1<n<5\), set \(S = \{ -2, 1, n, 5, 8 \}\). Median of S is n<5.
In all cases, which have median of S is less than 5, we have \(n<5\). This means n coud be in \((0, 7)\) or could be out of \((0, 7)\).
Insufficient.
(2) If median of the numbers in S is greater than 1, we have \(n>1\).
If \(n>8\), set \(S = \{ -2, 1, 5, 8, n \}\). Median of S is 5.
If \(5<n<8\), set \(S = \{ -2, 1, 5, n, 8 \}\). Median of S is 5.
If \(1<n<5\), set \(S = \{ -2, 1, n, 5, 8 \}\). Median of S is \(n>1\).
In all cases, which have median of S is greater than 1, we have \(n>1\). This means n coud be in \((0, 7)\) or could be out of \((0, 7)\).
Insufficient.
Combine (1) & (2)
(1) lead to \(n<5\)
(2) lead to \(n<1\)
(1) & (2) lead to \(1<n<5\), means that \(0<n<7\). Sufficient
Stiv
\(S = \left\{ { - 2,1,5,8} \right\} \cup \left\{ n \right\}\)
\(?\,\,\,:\,\,\,0 < n < 7\)
\(\left( 1 \right)\,\,\,{\rm{Med}}\left( S \right) < 5\,\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,n = 0\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\,\,\,\,\,\,\,\,\left[ {Med\left( S \right) = {\rm{Med}}\left( {\left\{ { - 2,0,1,5,8} \right\}} \right) = 1} \right]\,\, \hfill \cr \\
\,{\rm{Take}}\,\,n = 1\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\,\,\,\,\,\,\,\,\left[ {Med\left( S \right) = {\rm{Med}}\left( {\left\{ { - 2,1,1,5,8} \right\}} \right) = 1} \right]\,\,\, \hfill \cr} \right.\,\)
\(\left( 2 \right)\,\,\,{\rm{Med}}\left( S \right) > 1\,\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,n = 7\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\,\,\,\,\,\,\,\,\left[ {Med\left( S \right) = {\rm{Med}}\left( {\left\{ { - 2,1,5,7,8} \right\}} \right) = 5} \right]\,\, \hfill \cr \\
\,{\rm{Take}}\,\,n = 6\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\,\,\,\,\,\,\,\,\left[ {Med\left( S \right) = {\rm{Med}}\left( {\left\{ { - 2,1,5,6,8} \right\}} \right) = 5} \right]\,\,\, \hfill \cr} \right.\)
\(\left( {1 + 2} \right)\,\,\,\,1 < {\rm{Med}}\left( S \right) < 5\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle\)
\(\left( * \right)\,\,\left\{ \matrix{\\
\,n \le 0\,\,\, \Rightarrow \,\,\,\,Med\left( S \right) = 1\,\,,\,\,\,\,{\rm{impossible}} \hfill \cr \\
\,n \ge 7\,\,\, \Rightarrow \,\,\,\,Med\left( S \right) = 5\,\,,\,\,\,\,{\rm{impossible}} \hfill \cr} \right.\,\,\,\,\,\,\)
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
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