List S consists of the positive integers that are multiples of 9 and

lets see what are the extreme multiples of 9..
smallest = 9 itself
and largest = 9*11=99
so total are 11 numbers..

now two ways for median
1) average of numbers
since the numbers are consecutive multiples of 9, and hence Arithmetic progression average = \(\frac{(smallest+largest)}{2}=\frac{(9+99)}{2}=\frac{108}{2}=54\)

2) finding the middle number
since there are 11 numbers, the middle number = \(\frac{(11+1)}{2}=6\)
6th number = 9*6=54

D
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The smallest multiple of 9 is 9, and the largest is 99. We recall that both the mean and the median of a set of evenly-spaced integers is given by the formula: (smallest + largest)/2, so the median of List S is (99 + 9)/2 = 54.

Answer: D
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\(S = { 9, 18 , 27 , 36 ,45 , 54 , 63 , 72 , 81, 90 , 99}\)

Median here will be 6th term which is 54, Answer must be (D) 54
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The way the list is defined, I would think that the elements are all distinct.

List S is made up of positive multiples of 9 less than 100. That means it is made up of 9, 18, 27 ... 99

Had the question said that every element of list S is a positive integer divisible by 9 and less than 100, I would worry about repetition.

Also, note how the terms "set" and "list" are defined.

In a list, order of elements is important and repetitions are allowed.
In a set, by definition, order is irrelevant and all elements are distinct.

Positive multiples of 9 less than 100 are in the form 9x, where x ranges from 1 to 11.
So, we have 11 number of elements in the set S.
Median of a set of 11 elements is the (\(\frac{11+1}{2}\))th or 6th term.
So, Median=9*6=54

Ans. (D)
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The multiples of 9 within 100:
[(Last multiple -first multiple)/9]+1
[(99-9)/9]+1 = 11
So, the sixth term will be median
6th term = 9+(6-5)9 = 54
Ans. D

Another way is;
It is a series of consecutive numbers. So the median and average will be equal and the average is the average of the first number and the last number of the series.
so. (9+99)/2 = 54.

Multiples of 9 = 9 ,18 , 27 , 36 , 45 , 54 , 63 , 72 , 81 , 90 , 99.

Median for Odd no. of terms = N + 1 / 2 = 11 + 1 / 2 = 12/2 = 6

Median = 6th term = 54

chetan2u

Why are we taking all different multiples of 9, I mean s could consist of five 9s and six 99s. I mean we can have any multiple of 9 as the median of the set.

Why are we assuming that each element of the set is unique/distinct?

What am I missing ?

Regards,

Posted from my mobile device

Multiples of 9 that are less than 100: 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99
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Hey!

I know the multiples of 9 that are less than 100. My question is something else. Where in the question it is mentioned that all the multiples elements in the set s must be distinct/unique?

The precise language is “List S consists of the positive integers that are multiples of 9 and are less than 100” ... where is it written that all elements must be unique/ distinct?

From what I have understood, max possible element is 99. So we have 11 elements at max. Which means that (11+1)/2 th term will be the median ie. the sixth term. Now, the sixth term can be any multiple, even repetition can be allowed.

Set s can be (9, 9, 18, 27, 36, 36, 36, 54, 81, 81, 99) — in this case the median ie the 6th term is 36.

Set s can also be (9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99) — in this case the median ie the 6th term is 54.

Where in the question is it mentioned that we must have distinct elements/multiples of 9 in S or 1st 11 multiples of 9?

VeritasKarishma chetan2u Bunuel

ScottTargetTestPrep

I may have skipped a few names of the experts, nothing against you lovely people.. please feel free to comment.

Regards,

Yes, the question could have been better written by adding ‘distinct’ with ‘positive integers’.
But here in PS, there is no way that we can have two different answers, and the only way everyone will get the same answer is by taking this question to mean that we are looking at distinct multiples.
Maybe the data given would be insufficient if it were a DS question.
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Multiples of 9 that are less than 100 are 11 numbers.

The Median of 11 numbers will be \(6^{th}\) number which is 54.

Answer D

The exact words in the question are "List S consists of the positive integers that are multiples of 9 and are less than 100.". If the question instead said "List S consists of some of the multiples of 9" or "The elements of list S are multiples of 9", then it would be possible to omit some multiples of 9 or list some multiples of 9 more than once. However, "the positive integers that are multiples of 9 and less than 100" are 9, 18, 27, ... , 99. If we consider the list (9, 9, 18, 27, 36, 36, 36, 54, 81, 81, 99), this is not a list that consists of the positive integers that are multiples of 9 because 45 is a multiple of 9, but it is not contained in this list. Same goes for 63 and 72. If the list did include 45, 63 and 72 in addition to the above numbers, it would still not be "the list of multiples of 9 less than 100". It is the same reason as why (2, 2, 2, 3, 3, 5, 5, 5, 5, 5, 7, 7) is not "the list of prime numbers less than 10".
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