Set K contains every multiple of 6 from 18 to 306 inclusive.

Set K contains every multiple of 6 from 18 to 306 inclusive. If w is the median of set K, and x is the average (arithmetic mean) of set K, what is the value of w-x?

Quiz

Answer/Solution

0

Steps/Work

Set K ={18,24,30,....306}
Let the no. of terms be n.
so we know the AP formulae to calculate the the tn term:
tn= a+(n-1)*d where d = common difference of the terms.
306= 18 +(n-1)*6
n=49.
So the set K consist of n terms.
The median of the set having odd nos of elements is (n+1)/2., which in this case is 25.
Let's find the 25th term using the same formulae again:
t25= 18+(25-1)*6
t25= 162
So , the median of the set K is 162 i.e. w=162.
Now. lets find the average (arithmetic mean) of the set. For that we need to find the sum of all the elements first, lets call it S.
Since, set k is nothing but a Arithmetic Progression series having first element(a) as 18, common difference (d) as 6 and no. of terms(n) as 49.
Using the formulae to calculate sum of an AP series, which is
S= n/2[2a + (n-1)*d], we will calculate the sum.
so, S= 49/2 [2*18 + (49-1)*6]
This gives us S= 7938.
Now Arithmetic mean of Set K = 7938/no. of terms= 7938/49= 162.
So x= 162.
Now, (w-x) = (162-162)= 0.
Therefore Correct Answer = C

General Calculator / Auto Solver

Set k contains every multiple of from to inclusive. If w is the median of set k, and x is the average ( arithmetic mean ) of set k, what is the value of w - x?

Calculated Answer

6

Step by Step Solution

In order to solve problem, you must evaluate n0 + n1 + n0 * ((n2 - n1) / n0) / 2 - (n1 + n0 * ((n2 - n1) / n0) / 2).

n0 =6
n1 =18
n2 =306