List P contains m numbers; list Q contains n numbers. If the two lists

Question

List P contains m numbers; list Q contains n numbers. If the two lists are combined to produce list R, containing m + n numbers, is the median of list R greater than the median of list P ?

(1) The smallest number in list Q is greater than the largest number in list P.

(2) m = n

Archit3110 · Answer

target is median of R > median of P
where P has m numbers and Q has n numbers and R has m+n numbers
#1
The smallest number in list Q is greater than the largest number in list P.
P = 1,1 & q = 1.5,1.5 ; R ; 1,1,1.5,1.5
median of P 1 median of R 1.5
sufficient
P = 1,1,1 ; Q = 2 ; R ; 1,1,1,2
median of P ; 1 and R 1
we get no insufficient
#2
m = n
P = 1,1,1 ; Q = 2,2,2 ; R = 1,1,1,2,2,2
median of P ; 1 and R ; 1.5
P = 1,1,1 ; Q = 1,1,1 ; R = 1,1,1,1,1,1
medain of P & Q ; 1 insufficient
from 1 &2
P = 1,1, and Q =2,2 R ; 1,1,2,2
Median of P ; 1 and R ; 1.5
sufficient
option C

alwaysHP · Answer

List P contains m numbers; list Q contains n numbers. If the two lists are combined to produce list R, containing m + n numbers, is the median of list R greater than the median of list P ?

(1) The smallest number in list Q is greater than the largest number in list P.
List Q = {2,2,2,2,2} P={1,1,1,1,1} : R= {1,1,1,1,1,2,2,2,2,2} median of P = 1; R= 1.5.
If Q= {2} P= {1,1,1,1} median of P = 1; R= 1.
So A is incorrect 2 cases 
(2) m = n
 List Q = {2,2,2,2,2} P={1,1,1,1,1} : R= {1,1,1,1,1,2,2,2,2,2} median of P = 1; R= 1.5. 
 C is the answer

Bunuel · Answer

MGMAT OFFICIAL EXPLANATION

Before looking at the statements, it is important to remember that nothing in the question assumes that the numbers in either list must all be different. For instance, if list P contains 3 numbers, {1, 2, 3} is one possible set, as is {1, 1, 1}.

(1): INSUFFICIENT. Under this condition, it is possible for the median of the combined list R to be greater than the median of set P. For instance, if list P contains the numbers 1, 2, and 3, and list Q contains the numbers 4, 5, and 6, then the median of list R (which contains all six numbers) is 3.5, which is greater than 2 (the median of list P).

It is also possible for the median of the combined set to equal the median of list P. For instance, if list P contains the numbers 1, 1, 1, and 1, and list Q contains the numbers 2, 2, and 2, then the median of list P is 1, and the median of list R is also 1.

(2): INSUFFICIENT. This information does not tell us whether the median of list R is greater than the median of list P. For instance, if list P contains the numbers 1, 2, and 3, and list Q contains the numbers 4, 5, and 6, then the median of list R (which contains all six numbers) is 3.5, which is greater than 2 (the median of list P).

It is also possible for the median of the combined set to equal the median of list P, as is perhaps most easily seen in cases in which all of the numbers are the same: e.g., if list P contains 1, 1, and 1, as does list Q, then the median of all three lists P, Q, and R is 1.

(1) AND (2): SUFFICIENT. If both conditions are true, then the set will have two middle values: the largest value in list P and the smallest value in list Q. According to statement (1), these values are different, so the median, found by averaging them, will be distinct from both; the median will be greater than the value from list P, and less than the one from list Q. Since the median is greater than the largest value in list P, it must be greater than every value in list P, and so greater than the median of P; therefore, this statement is sufficient.

The correct answer is C.

RamTee · Answer

This question is easier to solve if you are able to visualise the medians in relation with List P, Q and List R.

Since we want to determine the values of medians for List R and List P, it is important we understand some values in these lists and overall numbers in this list. The numbers will change the median spots hence the need to know something about 'm' and 'n'.

Option 1:  List Q's  smallest value > The largest value in List  P . Now if you are able to visualise this as a train, the first boogie of List Q only is behind Train (list) P. This would have been fine if List P has distinct numbers and #of values in List Q are more than list P, to ensure median of List R is definitely bigger than median of List P.

Hence insufficient

Option 2: Mentions only that m=n. This suggests that they have same number of values but nothing about List P w.r.t. List Q.

Hence insufficient.

If we check Option 1+ 2, we know that List Q's largest value will come after List P ends. Since m=n, then the median of List R will be at  (m+n)/2   which will always come right after 'm' numbers, and from List Q. Hence Median of List R > Median of List P .

sahilvermani · Answer

I mis-read median as mean and chose statement (1) to be sufficient:(.

SatvikVedala · Answer

Statement 1)
Conditions
1. What if set P has all distinctive numbers & Set Q has distinct single number
P has 1 to 16, Q has 17 => median increases
2. What if set P contains same numbers & Set Q has distinct single number
P contains 16 no's of 16, Q has 17 => median remains same
Insufficient

Statement 2)
Conditions
1. What if set P has same numbers as set Q
P has 16,16 Q has 16,16 => median remains same
2. What if set P has distinct numbers from set Q
P has 16,16, Q has 17,17 => median increases
Insufficient

Combining 1 + 2
P has distinct numbers that of Q
so median increases => Sufficient

bumpbot · Answer

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