The median of a set of 5 different numbers is x

Incorrectly assumed it to be a "mean" instead of "median".

The median of a set of 5 different numbers is x. If we add another set of 5 numbers to the existing set, will the new median of the new set be greater than x?

(1): Exactly 3 of the newly added numbers are greater than x

(2): Exactly 2 of the newly added numbers are equal to x

Assumption: data set is in ascending order.

Since there are 5 numbers in set at present, median will be at third position.

Statement 1: It translates into 2 new numbers of the 5 will be lesser than x. New position of x -> fifth.

Now median will be average of fifth and sixth position values. That is, x and any value greater than x
Hence, new median will always be greater than x.

Sufficient for a Y/N answer.

Statement 2: Since we only know two values are equal to x we do not know what will be the new position of x in the data set.

Case 1. If other three values are smaller, then x will be at sixth position. Which will decrease the value of median.
Case 2. If other three values are greater, then x will be at fifth position. Which will increase the value of median.

Not Sufficient for a Y/N answer.

Answer: Option A

Question:

The median of a set of 5 different numbers is X. >>>> _, _ ,X _, _
If we add another set of 5 numbers to the existing set, will the new median of the new set be greater than x? >>>> _, _, _, _, _, _,_ ,_, _, _ Is the median say Y of 10 nos.>X?

1) Exactly 3 of the newly added numbers are greater than x. >>>>> Exactly three nos. are greater than X but we also know from the question that there are definitely two more nos. (say A, B) greater than X. So,

_, _, _, _, _, X, >X, >X, >X, A, B or,
D,E _ _ _ X, A, B, >X, >X, >X or,
D,E, X, A, B, >X, >X, >X, _, _ or
D,E, X, >X, >X, >X, A, B,_, _

Note. Assuming D & E are less than X. Any further combinations will put nos. with median >X in the middle.

Now, it's clear that (X+>X)/2 will be >X. Sufficient.

2. Exactly 2 of the newly added numbers are equal to x >>>>> Exactly two nos. are equal to X but we also know from the question that there are definitely two nos. (say D & E) less than X. So,

D,E _, X, X, X, _, _,_, _ or, >>> Median will be equal to X.
_, D,E, X, X, X,_, _,_, _ or, >>> Median will be equal to X.
D,E, X, X, X,_, _,_, _,_ >>> Median will be > X as any number to the right of X will be >X.

Note. D & E are less than X.

So, the median can be =X or >X clearly not sufficient.

Answer: A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Clues: the original set contains two numbers less than X and two numbers greater than X.

1) Given that we are adding 5 numbers, if only 3 newly added numbers are greater than X, then the median will be (X + a number bigger than X)/2 which is greater than X. Sufficient.

2) Utilizing statement 1 since it also satisfies statement 2, the median is greater than X. Now find a case where the median is not greater than X. If only one added number is smaller than X, then the median is X. Not sufficient.

Correct Answer is A

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Answer is (A)

Since 5 different numbers in the set, it can be concluded that two of them less than x and two of them greater than x. New set contains ten numbers. In order to identify the new median be greater than x or not, we should compare the mean of 5th and 6th numbers of new set with x.

(1): Exactly 3 of the newly added numbers are greater than x
We can infer that two of newly added numbers are not greater than x. It means that the new set contains exactly 5 numbers which are greater than x and fifth number in the set (in ascending order) is equal to x. Hence, we can conclude it is sufficient.

(2): Exactly 2 of the newly added numbers are equal to x
If at least one of the newly added numbers is less than x, the new median will not be greater than x. If exactly three of the newly added numbers are greater than x, the new median will be greater than x.

Taking 1 statement : (1): Exactly 3 of the newly added numbers are greater than x
- Assuming set is in ascending order say; a b c d e
Median is middle value i.e. c in this case.
so c=x. So if we are adding 5 new numbers out of which 3 are > x, obviously the new median will shift to the right side.
say the new set -a -b a b c d e X Y Z.
Now the new median is (c+d)/2 which would be > c=x.
Eg. Old set: 1 2 3 4 5 ;x =3
New set -2 -1 1 2 3 4 5 6 7 8 ; x= (3+4)/2=3.4>old median(3)
Is the set is completely negative or fractions then also the same thing will apply.
So now A can be the answer.

Taking statement 2
Exactly 2 of the newly added numbers are equal to x
Eliminating by taking the set:
Set : 1 2 3 4 5; x=3
Now 5 new numbers are added out of which 2 are = to x. But we dont know about the rest of the 3 numbers
taking cases: 1: -3 -2 -1 1 2 3 3 3 4 5 (3 added are < x)
so now the new median is (2+3)/2 = 2.5 < old median (3)
but we add the 3 numbers such that those are > x
then the new median will be > 3.

So Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Hence Ans is A

The correct answer is option A.

Here is why:

The question is " Is X..." type question where you are looking for either a clear Yes or No. In this case the question asks if New_Median>X.
So if, New_Median>X then Yes
If New_Median< or = X then No.

Now let's review options. Assume the ascending order is _ _ X _ _

1) _ _ X _ _ _ _ _ - 3 newly added are greater than X, hence placed at the end.
Now the two remaining can be -
a. Equal to X, in which case X is <New_Median or =New_Median if #6 =X
b. Less than X, in which case X is < New_Median or =New_Median if #6 =X
Hence, first statement gives a clear answer.

2) _ _ X X X _ _ - 2 are exactly equal to X, hence place next to X
Now the rest 3 can be:
a. Greater than X, in which case the New_Median>X
b. Less than X, in which case New_Median<X
So no clear answer from the second statement.

The correct option in this case is option A.

Since there are 5 numbers in set at present, median will be at third position.

1: 3 new nos will be greater than x which means in total out of 10 nos x will be at 5 position and median will be avg of 5th and 6th no.
Hence, new median will always be greater than x.

Sufficient

2: Since we only know two values are equal to x we do not know what will be the new position of x in the data set.

Not Sufficient

Ans: A

PPhoenix77 - congrats on getting this one totally right! PM me to claim your reward!

If 5 nos. are added, new median will be average of 5th and 6th term.

Option 1: if 3 nos. added are greater than X, in that case exactly 5 nos. will be greater than x. Therefore x will be 5th term and 6th term will be a no. greater than x. This implies median i.e. average of T5 and T6 will be average of x and a no. greater than x i.e. a no. greater than x.
Therefore, A is sufficient

Option 2: If 2 nos. added are exactly x.
Let consider median of 7 nos. only i.e. median of original 5 nos. and two added x
in this case median will will be the 4th term i.e. x.

Now depending on value of 3 nos. median can be greater than or equal to x.
Ex: if 3 nos. added are greater than x: median of 10 nos. will be greater than x.
if 3 nos. added are less than x: median of 10 nos. will be less than x
if 2 added no. greater than x and 3rd less than x: median is x
if 2 no less than x and another greater: median is less than x.
so insufficent

Hence, A is answer

Hey Everyone!

Please find the official solution:

Solution

Steps 1 and 2: Understand the question stem and draw inferences:

Given:

• It is given that x is the median of a set of 5 numbers. These can be expressed as follows in the ascending order:
{p, q, x, r, s}
• Another set of 5 numbers are inserted. Let us express them as follows in the ascending order:
{a, b, c, d, e}

To Find:

• We have to find whether, after including the above mentioned new set of 5 numbers in the set {p, q, x, r, s}, the new median of the combined set will be greater than x.

Step 3: Analyse Statement 1: Exactly 3 numbers of the new set are greater than x

• Let us look at the new set:
o According to the new information c, d and e are greater than x
o That also means that a and b might be either lesser than x or equal to it

 So, when we arrange the numbers in the ascending order, the remaining 2 numbers i.e. a and b will definitely be on the left side of x
 That makes 4 numbers to the left of x (i.e. x is the 5th number) and all the other 5 numbers greater than x (2 from the original set and 3 from the new set of numbers which were added to the existing set)
 The new median will be the average of the 5th and the 6th numbers of the combined set and we know that the 6th number is greater than x
 Hence, (x + 6th number)/2 which is again greater than x as the 6th number is greater than x
• Hence the new median will always be greater than x.

Hence, Statement (1) ALONE is sufficient.

This eliminates answer choices B, C and E.

Step 4: Analyse Statement 2: Exactly 2 numbers of the new set are equal to x

• Let us consider 2 cases here:
o Case 1: All the other numbers (which are not equal to x) of the newly added set of 5 numbers are lesser than x

 As the numbers are less than x they will be to the left of x when the numbers of the combined set are arranged in the ascending order
 In total there will be 5 numbers to the left of x. Hence, x will be the 6th number
 Here the new median will be (5th number + x)/2 which is lesser than x
o Case 2: All the other numbers are greater than x

 As the numbers are greater than x they will be to the right of x when the numbers of the combined set are arranged in the ascending order
 In total there will be 5 numbers to the right of x
 Here the new median will be (x + 6th number)/2 which is more than x
o Hence, we can’t say anything concrete based on only the 2nd statement

Hence, Statement (2) ALONE is not sufficient.

This eliminates answer choice D.


Hence the Correct Answer: Option (A)

Alternate Solution

Steps 1 and 2: Understand the question stem and draw inferences:

Given:
• It is given that x is the median of a set of 5 numbers. These can be expressed as follows in the ascending order:

o {1, 2, 3,4 ,5}, Here the median (x) = 3

• Another set of 5 numbers are inserted. Let us express them as follows in the ascending order:

o {a, b, c, d, e}

To Find:
We have to find whether, after including the above mentioned new set of 5 numbers in the set {1, 2, 3,4, 5} the new median of the combined set will be greater than 3 or not.

Step 3: Analyse Statement 1:
Exactly 3 numbers of the new set are greater than x
• This means 3 elements in the new set is greater than 3, and the rest 2 are ≤3.

o Let us assume the set to be: {1, 2, 4, 5, 6}

• Now, combining both the sets we get:

o {1, 1, 2, 2, 3, 4, 4, 5, 5, 6}

• Hence the new median will always be greater than 3 (3.5 in this case) because we have a number greater than 3 to the right of 3.
Hence, Statement (1) ALONE is sufficient.

Step 4: Analyse Statement 2:
Exactly 2 numbers of the new set are equal to x
• Let us consider 2 cases here:

o Case 1: All the other numbers (which are not equal to x) of the newly added set of 5 numbers are lesser than x.

 Let us assume the set to be {0, 1, 2, 3, 3}

 In that case, the final set would be: {0, 1, 1, 2, 2, 3, 3, 3, 4, 5}

 Median for this set = 2.5, which is lesser than 3.

o Case 2: All the other numbers are greater than x

 Let us assume the set to be {3, 3, 4, 5, 6}

 In that case, the final set would be: {1, 2, 3, 3, 3, 4, 4, 5, 5, 6}

 Median for this set = 3.5, which is greater than 3.

Since this statement does not lead us to any unique answer, Statement (2) ALONE is not sufficient.

Step 5: Combine both Statements
As Statement 1 alone is sufficient to answer the question, we do not need to analyze further.

Hence the Correct Answer: Option: A

Well, I did all of the legwork and selected the wrong answer.

Not sure if my approach is right.

Set 1: {t,u,x,w,v}
Set 2: {a,b,c,d,e}

Let's assume in each set that the integers are in ascending order.

The median of a set of 5 different numbers is x. If we add another set of 5 numbers to the existing set, will the new median of the new set be greater than x?

(1): Exactly 3 of the newly added numbers are greater than x
d,e,t,u,x,a,b,c,w,v ---> new median is x + a/ 2 (a,b,c are greater than x)
t,u,x,d,e,a,b,c,w,v --> new median is e + a /2 (here d and e are equal to x)

In both cases the new median is larger (because a > x)

Sufficient
(2): Exactly 2 of the newly added numbers are equal to x

t,u,x,d,e,a,b,c,w,v ---> e + a / 2 is the new median and is greater than x
a,b,c,t,u,x,d,e,w,v ---> u + x /2 is the new median and is less than x

Insufficient.

A.