Let the Set P be (10, 4, 7, 18) and Set Q be (x, 10, 4, 7, 18, 25).

Question

Let the Set P be (10, 4, 7, 18) and Set Q be (x, 10, 4, 7, 18, 25).
If the median of Set Q is exactly 4 higher than the median of Set P, what is the value of x? 
A. 10
B. 12
C. 13
D. 15
E. 16

abannore · Answer

Answer: D

P={4,7,10,18}
Median of P (Mp) = 17/2 = 8.5

Let median of set Q be Mq
Mq=4+Mp
Mq=4+8.5
Mq=12.5

If 10<x<18, 
Q={4,7,10,x,18,25}
Mq=(10+x)/2
(10+x)/2=12.5
x=15

Even if 7<x<10, same equation of median will form resulting in x=15 which fails the condition. In all other postions of x, value of x won't satisfy the median condition.

So x=15

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rajamech · Answer

First of all arrange elements of P and Q in ascending orders:
P (4, 7, 10, 18)
Q (4, 7, 10, 18, 25) and x can come anywhere in Q.
Median of P is (7+10)/2 or 8.5
Let's solve the possibility where x will come with 10 in the median equation:
Value of x is more/less than 10 then equation will be 
(10+x)/2 = 8.5+4
x= 15

Answer is D

CrackverbalGMAT · Answer

To calculate the median, the values in the set have to be arranged in ascending order.

Elements of set P can be arranged as 4,7,10,18. The median is the average of 7 and 10. Median of set P = 8.5

Elements of set Q can be arranged as 4,7,10,18, 25. Note that we need to decide the location of x depending on the median. 
We know that the median of set Q is 4 more than the median of set P. Therefore, median of set Q = 12.5.

Set Q has 6 elements, hence the median will be the average of the 3rd and 4th value in the set. 
If x<4 OR x>25, 3rd value = 10 and 4th value = 18. Median ≠ 12.5. Therefore, x is neither less than 4 nor greater than 25.
If 4<x<10 OR 18<x<25, a similar conclusion may be drawn.

Therefore, x has to be between 10 and 18 and hence Median =  \frac{10 + x }{ 2}  = 12.5.

Simplifying, we get x = 15.  The correct answer option is D .

Hope that helps!

bumpbot · Answer

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Let the Set P be (10, 4, 7, 18) and Set Q be (x, 10, 4, 7, 18, 25).

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Let the Set P be (10, 4, 7, 18) and Set Q be (x, 10, 4, 7, 18, 25).

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