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# If M is the set of all consecutive multiples of 9 between 100 and 500,

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Math Expert
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If M is the set of all consecutive multiples of 9 between 100 and 500,  [#permalink]

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17 Sep 2015, 01:27
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If M is the set of all consecutive multiples of 9 between 100 and 500, what is the median of M?

A. 300
B. 301
C. 301.5
D. 302.5
E. 306

Kudos for a correct solution.

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Re: If M is the set of all consecutive multiples of 9 between 100 and 500,  [#permalink]

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17 Sep 2015, 04:25
8
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Bunuel wrote:
If M is the set of all consecutive multiples of 9 between 100 and 500, what is the median of M?

A. 300
B. 301
C. 301.5
D. 302.5
E. 306

Kudos for a correct solution.

Without getting into too much of complication, we can see one simple issue in this q..
1) all numbers are in arith prog, so the median will be =average of (1st number + last number)..
first number is 108 and last number is 495...
answer = $$\frac{(108 + 495)}{2}=\frac{603}{2}$$
ans C 301.5
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Re: If M is the set of all consecutive multiples of 9 between 100 and 500,  [#permalink]

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17 Sep 2015, 02:28
1
Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.

If M is the set of all consecutive multiples of 9 between 100 and 500, what is the median of M?

A. 300
B. 301
C. 301.5
D. 302.5
E. 306

Since 100/9 =11.11, 500/9=55.55, M can be written as {9*12, 9*13, ...., 9*54, 9*55}. The number of elements of M is, therefore, 44(=55-12+1). Since the number of elements is even number the median is the average of the 22nd element(=9*33) and 23rd element(=9*34). So the median is (9*33 + 9*34)/2 = 301.5. The answer is C.
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Re: If M is the set of all consecutive multiples of 9 between 100 and 500,  [#permalink]

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17 Sep 2015, 03:55
2
Bunuel wrote:
If M is the set of all consecutive multiples of 9 between 100 and 500, what is the median of M?

A. 300
B. 301
C. 301.5
D. 302.5
E. 306

Kudos for a correct solution.

M is a set of number multiples of 9 (100-500)
First, by this statement you will get the first number of this set 108 and the last number 495
Second, you will find that 108 is a 108/9 = 12th and 495/9 = 55th
Third, the question is asked about the median so (12th+55th)/2 = 33.5th
Hence, (33th+34th)/2 is the answer = [(9*33)+(9*34)]/2 = 301.5
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Re: If M is the set of all consecutive multiples of 9 between 100 and 500,  [#permalink]

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17 Sep 2015, 16:10
1
Bunuel wrote:
If M is the set of all consecutive multiples of 9 between 100 and 500, what is the median of M?

A. 300
B. 301
C. 301.5
D. 302.5
E. 306

Kudos for a correct solution.

First multiple of 9 after 100 is 108 and multiple of 9 just before 500 is 495.
Now,
For evenly spaced sets, Mean = Median = (first term + last term)/ 2 = (108+495)/2 = 301.5

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Re: If M is the set of all consecutive multiples of 9 between 100 and 500,  [#permalink]

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18 Sep 2015, 04:20
Bunuel wrote:
If M is the set of all consecutive multiples of 9 between 100 and 500, what is the median of M?

A. 300
B. 301
C. 301.5
D. 302.5
E. 306

Kudos for a correct solution.

First Multiple of 9 in the given range = 108
Last Multiple of 9 in the given range = 495

All consecutive multiples of 9 make and Arithmetic progressions

CONCEPT: In any Arithmetic Progression in which terms are equally spaced Mean = Median = Average of First and Last term of the series

i.e. Median = (108+495)/2 = 301.5

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Re: If M is the set of all consecutive multiples of 9 between 100 and 500,  [#permalink]

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19 Sep 2015, 19:25
1

First multiple of 9 after 100 is 108 = 9*12
Last multiple of 9 before 500 is 55 (since 500/9 = 55.x)

So, the sequence now has 9(12), 9(13), ... 9(55)

No. of terms = 55 - 12 + 1 = 44

=> Median is the mean of 22nd and 23rd terms
22nd term = 9*(22+11) = 9*33 = 297
23rd term = 297 + 9 = 306

(306 + 297)/2 = 301.5 (C)
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Re: If M is the set of all consecutive multiples of 9 between 100 and 500,  [#permalink]

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19 Sep 2015, 22:05
4
Bunuel wrote:
If M is the set of all consecutive multiples of 9 between 100 and 500, what is the median of M?

A. 300
B. 301
C. 301.5
D. 302.5
E. 306

Kudos for a correct solution.

For consecutive intergers -
mean = median

So, first term is - next multiple of 9 > 100 = 108
last term is - multiple of 9 just prior to 500 = 495

So, mean = median = (First term + Last term)/2
=> (108 + 495)/2 = 301.5
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Re: If M is the set of all consecutive multiples of 9 between 100 and 500,  [#permalink]

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19 Sep 2015, 23:48
1
Bunuel wrote:
If M is the set of all consecutive multiples of 9 between 100 and 500, what is the median of M?

A. 300
B. 301
C. 301.5
D. 302.5
E. 306

Kudos for a correct solution.

We know,In any evenly spaced set the arithmetic mean is equal to the median and can be calculated by the formula,Mean=Median=$$\frac{First Term+Last Term}{2}$$

Now from the question we can find that,First Term of Consecutive multiples of 9 between 100 and 500=108 (9*12) and the Last Term of Consecutive multiples of 9 between 100 and 500 =495 (9*55)

Since it is an evenly spaced set,so the mean=mediam=$$\frac{108+495}{2}$$ =301.5

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Re: If M is the set of all consecutive multiples of 9 between 100 and 500,  [#permalink]

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20 Sep 2015, 01:22
GMATinsight wrote:
Bunuel wrote:
If M is the set of all consecutive multiples of 9 between 100 and 500, what is the median of M?

A. 300
B. 301
C. 301.5
D. 302.5
E. 306

Kudos for a correct solution.

First Multiple of 9 in the given range = 108
Last Multiple of 9 in the given range = 495

All consecutive multiples of 9 make and Arithmetic progressions

CONCEPT: In any Arithmetic Progression in which terms are equally spaced Mean = Median = Average of First and Last term of the series

i.e. Median = (108+495)/2 = 301.5

GMATinsight can you please explain what do you mean by equally spaced. A small example would be very helpfull.
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Posts: 56304
Re: If M is the set of all consecutive multiples of 9 between 100 and 500,  [#permalink]

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20 Sep 2015, 21:41
Bunuel wrote:
If M is the set of all consecutive multiples of 9 between 100 and 500, what is the median of M?

A. 300
B. 301
C. 301.5
D. 302.5
E. 306

Kudos for a correct solution.

ECONOMIST GMAT TUTOR OFFICIAL SOLUTION:

The median of an arithmetic sequence is also its average. The average of an arithmetic sequence is the average of the smallest and largest terms. In set M, the smallest multiple of 9 is 108, and the largest is 495. Thus the average (and the median) is (108+495)/2 = 603/2 = 301.5.

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Re: If M is the set of all consecutive multiples of 9 between 100 and 500,  [#permalink]

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26 Oct 2015, 15:57
Did a little bit different...maybe it is more time consuming...
108 is 12*9
495 is 55*9

55-12+1 = 44, so we have to find mean of the 12+22 and 55-22 multiples of 9, or 34 and 33
(34+33)(9)/(2)= (69*9)/2, rewrite as 67*10-67 / 2 = 301.5
C.
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Re: If M is the set of all consecutive multiples of 9 between 100 and 500,  [#permalink]

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16 Mar 2016, 02:22
In any AP series
Median = Mean = 1/2 [First term +last term]
thus median =301.5
excellent question to learn concepts..
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Re: If M is the set of all consecutive multiples of 9 between 100 and 500,  [#permalink]

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16 Mar 2016, 02:36
Bunuel wrote:
If M is the set of all consecutive multiples of 9 between 100 and 500, what is the median of M?

A. 300
B. 301
C. 301.5
D. 302.5
E. 306

Kudos for a correct solution.

There are 44 terms between 108 and 495
The median is 22.5.
Therefore, 99+22.5*9 = 301.5
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Re: If M is the set of all consecutive multiples of 9 between 100 and 500,  [#permalink]

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10 Jun 2016, 13:50
multiples of 9 between 100 and 500 begin from 108 and ends at 495 that is from 9*12 to 9*55

so number of terms from 12 to 55 = 55-12+1 = 44 terms. With 21 to left and 21 to right the the average of 22nd and 23rd terms will be the Median.

22nd term = (12+22) - 1 [subtratcing 1 as this includes 12] = 33 * 9 = 297
23rd term will then be 34*9 = 306
.
(306+297)/2 = 301.5
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Re: If M is the set of all consecutive multiples of 9 between 100 and 500,  [#permalink]

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11 Jun 2016, 05:46
Bunuel wrote:
If M is the set of all consecutive multiples of 9 between 100 and 500, what is the median of M?

A. 300
B. 301
C. 301.5
D. 302.5
E. 306

Kudos for a correct solution.

M = { 108, 117..........486, 495}

M has 44 terms.........

Sum of the Series is 13266

Average ( Mean ) = 13266/44 =>301.5

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Re: If M is the set of all consecutive multiples of 9 between 100 and 500,  [#permalink]

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16 Jul 2018, 22:00
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Re: If M is the set of all consecutive multiples of 9 between 100 and 500,   [#permalink] 16 Jul 2018, 22:00
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