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# Set K contains every multiple of 6 from 18 to 306 inclusive. If w is

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Set K contains every multiple of 6 from 18 to 306 inclusive. If w is [#permalink]

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17 Sep 2015, 00:57
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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?

A) -6
B) -3
C) 0
D) 3
E) 6

Kudos for a correct solution.
[Reveal] Spoiler: OA

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Re: Set K contains every multiple of 6 from 18 to 306 inclusive. If w is [#permalink]

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17 Sep 2015, 01:44
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I would solve this problem in a calculative way, but with less chances of errors !
First find median of the Set K.

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.

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Re: Set K contains every multiple of 6 from 18 to 306 inclusive. If w is [#permalink]

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17 Sep 2015, 02:50
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Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.

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?

A) -6
B) -3
C) 0
D) 3
E) 6

Since 18/6 =3, 306/6=51, K= {6*3(=18), 6*4, ...., 6*50, 6*51(=306)}. The number of elements of K is, therefore, 49(=51-3+1). Since is arithmetic progress and the number of elements is odd, the average of sequence is equal to the median of the sequence. So w-x=0. The answer is C.

Well, if you insist a tedious computations :
The sum of all elements --> 6*(3+4+....+51)= 6*(54*49)/2 = 3*2*27*49=6*27*49
The average of the set ---> 6*27*49/49= 6*27=162

Since the number of elements is 49(odd), the median is the 25th element(=6*27=162).

As a result x=162, w=162, so w-x=o, as we predicted.
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Re: Set K contains every multiple of 6 from 18 to 306 inclusive. If w is [#permalink]

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17 Sep 2015, 16:03
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Bunuel wrote:
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?

A) -6
B) -3
C) 0
D) 3
E) 6

Kudos for a correct solution.

According to the properties of evenly spaced sets,
Mean = Median = (first+last)/2

Set K is evenly spaced as all the elements are multiples of 6
Since, mean = median
their difference w-x = 0

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Re: Set K contains every multiple of 6 from 18 to 306 inclusive. If w is [#permalink]

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17 Sep 2015, 21:56
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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?

A) -6
B) -3
C) 0
D) 3
E) 6

Kudos for a correct solution.

15 sec only need to answer or the time need to read the question .

Note that the question clearly says it a consecutive multiple of 6 from 6 to 306 inclusive .
so avg = median
so its 0 C ans .

Property of consecutive integer property ---- avg = median
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Re: Set K contains every multiple of 6 from 18 to 306 inclusive. If w is [#permalink]

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19 Sep 2015, 23:44
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In any set of numbers having the same common difference ie., consecutive multiples of a number,

Mean = Median = $$\frac{(First Term + Last Term)}{2}$$

Since we know mean = median, we don't even have to bother to calculate either of them as we only want the difference between the two, which will always be 0.
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Posts: 5769
Re: Set K contains every multiple of 6 from 18 to 306 inclusive. If w is [#permalink]

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20 Sep 2015, 00:04
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MathRevolution wrote:
Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.

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?

A) -6
B) -3
C) 0
D) 3
E) 6

Since 18/6 =3, 306/6=51, K= {6*3(=18), 6*4, ...., 6*50, 6*51(=306)}. The number of elements of K is, therefore, 49(=51-3+1). Since is arithmetic progress and the number of elements is odd, the average of sequence is equal to the median of the sequence. So w-x=0. The answer is C.

Well, if you insist a tedious computations :
The sum of all elements --> 6*(3+4+....+51)= 6*(54*49)/2 = 3*2*27*49=6*27*49
The average of the set ---> 6*27*49/49= 6*27=162

Since the number of elements is 49(odd), the median is the 25th element(=6*27=162).

As a result x=162, w=162, so w-x=o, as we predicted.

Hi,
even if the numbers are even but in arithmetic progression, the median and average should be the same.....
so its not important that the number of elements is odd...
only difference in odd and even elements will be that the median will be a part of the set in odd numbers
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Re: Set K contains every multiple of 6 from 18 to 306 inclusive. If w is [#permalink]

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20 Sep 2015, 00:06
Bunuel wrote:
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?

A) -6
B) -3
C) 0
D) 3
E) 6

Kudos for a correct solution.

median and mean are the same in sets having numbers evenly spaced out or in arithmetic progression..
C
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Re: Set K contains every multiple of 6 from 18 to 306 inclusive. If w is [#permalink]

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27 Sep 2015, 11:32
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Bunuel wrote:
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?

A) -6
B) -3
C) 0
D) 3
E) 6

Kudos for a correct solution.

Gmatprepnow Official Solution:

If set K contains every multiple of 6 from 18 to 306, then set K looks like this: {18, 24, 30, 36, 42, . . ., 294, 300, 306}. Since the numbers in set K are equally spaced, the median (w) and mean (x) of this set will be equal.

Do we need to calculate the median and mean of set K?

No. We can answer the question without making any calculations whatsoever.

If w and x are equal, then w-x = 0, so the answer to the question is C.

So, when it comes to statistics questions on then GMAT, be sure to watch out for equally spaced numbers in a set. If they’re equally spaced, you may be able to save yourself a lot of time.
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Re: Set K contains every multiple of 6 from 18 to 306 inclusive. If w is [#permalink]

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28 Feb 2017, 03:43
To find the total number of numbers in the set, we can say that
306 = 18 + (n-1)6 or n = 49.
Median term (w) will be 25th term which will be equal to 18 +(25-1)6 = 162
Average (x) = {n/2( first term +last term)}/n
= (first term +last term)/2
= (306+18)/2 = 162.
Therefore w - x = 162-162 = 0. Option C
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Re: Set K contains every multiple of 6 from 18 to 306 inclusive. If w is [#permalink]

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02 Mar 2017, 04:51
don't need to do much calculation as we can clearly see that its an A.P. so median and average will always be the same
so w-x=0
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Re: Set K contains every multiple of 6 from 18 to 306 inclusive. If w is [#permalink]

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02 Mar 2017, 07:49
Bunuel wrote:
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?

A) -6
B) -3
C) 0
D) 3
E) 6

Kudos for a correct solution.

The set of numbers is = 18, 24, 30 ....... 300 , 306

Or, we may write the set as 6 ( 3 , 4 , 5 ..... 50 , 51 )

There will be 51 - 3 + 1 = 49 numbers

Median will be 49+1/2 = 25th term which will be 27*6 = 162

Mean will be 1323/49 = 27 , so the value of mean will be 27*6 = 162

Thus , Mean - Median = 162 - 162 => 0

Hence, correct answer must be (C) 0
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Re: Set K contains every multiple of 6 from 18 to 306 inclusive. If w is [#permalink]

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02 Mar 2017, 18:12
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Bunuel wrote:
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?

A) -6
B) -3
C) 0
D) 3
E) 6

Since we have a set of evenly spaced integers, we know the median is equal to the mean. Thus, the difference between the median and the mean is zero.

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Re: Set K contains every multiple of 6 from 18 to 306 inclusive. If w is [#permalink]

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31 Mar 2018, 07:06
we have an evenly spaced set so mean and median are always equal so C is correct

with algebra:
Set K contains every multiple of 6 from 18 to 306 inclusive so we have an evenly spaced set with increment=6 so $$n=\frac{(last-first)}{increment}+1$$ both 18 and 306 are multiples of 6 (both divided by 2 and 3) so $$n=\frac{(306-18)}{(6)}+1=49$$. To find the mean since its odd we can do this : $$\frac{n+1}{2}=25$$ so the 25th number must be $$(25-1+3)6=27*6=162$$ since its an evenly spaced set $$\frac{first+last}{2}=Avg=\frac{18+306}{2}=162$$ so $$mean-median=0$$.
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Re: Set K contains every multiple of 6 from 18 to 306 inclusive. If w is   [#permalink] 31 Mar 2018, 07:06
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