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If M is the set of all consecutive multiples of 9 between 100 and 500,
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17 Sep 2015, 01:27
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Re: If M is the set of all consecutive multiples of 9 between 100 and 500,
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17 Sep 2015, 04:25
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,
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17 Sep 2015, 02:28
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(=5512+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,
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17 Sep 2015, 03:55
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 (100500) 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,
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17 Sep 2015, 16:10
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 Answer: C



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Re: If M is the set of all consecutive multiples of 9 between 100 and 500,
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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 seriesi.e. Median = (108+495)/2 = 301.5 Answer: option C
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Re: If M is the set of all consecutive multiples of 9 between 100 and 500,
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19 Sep 2015, 19:25
Start with identifying first and last term.
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,
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19 Sep 2015, 22:05
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,
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19 Sep 2015, 23:48
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 So,the Correct Answer is C
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Re: If M is the set of all consecutive multiples of 9 between 100 and 500,
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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 seriesi.e. Median = (108+495)/2 = 301.5 Answer: option C GMATinsight can you please explain what do you mean by equally spaced. A small example would be very helpfull.



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Re: If M is the set of all consecutive multiples of 9 between 100 and 500,
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20 Sep 2015, 21:41



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Re: If M is the set of all consecutive multiples of 9 between 100 and 500,
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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
5512+1 = 44, so we have to find mean of the 12+22 and 5522 multiples of 9, or 34 and 33 (34+33)(9)/(2)= (69*9)/2, rewrite as 67*1067 / 2 = 301.5 C.



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Re: If M is the set of all consecutive multiples of 9 between 100 and 500,
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16 Mar 2016, 02:22



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Re: If M is the set of all consecutive multiples of 9 between 100 and 500,
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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,
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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 = 5512+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,
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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 Hence answer will be (C)
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Re: If M is the set of all consecutive multiples of 9 between 100 and 500,
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