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A set of numbers has an average of 50. If the largest element is 4 gre
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19 Aug 2015, 01:15
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Re: A set of numbers has an average of 50. If the largest element is 4 gre
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23 Aug 2015, 12:02
Bunuel wrote: A set of numbers has an average of 50. If the largest element is 4 greater than 3 times the smallest element, which of the following values cannot be in the set?
(A) 85 (B) 90 (C) 123 (D) 150 (E) 155
Kudos for a correct solution. VERITAS PREP OFFICIAL SOLUTION:This question might look a little ominous but it isn’t very tough, really! The set has an average of 50 so that already tells us that we can represent each element of the set by 50. If there is an element which is a little less than 50, there will be another element which is a little more than 50. The largest element is 4 greater than 3 times the smallest element so L = 4 + 3S. The smallest element must be less than 50 and the largest must be greater than 50. Say, if the smallest element is 20, the largest will be 4 + 3*20 = 64. Is there any limit imposed on the largest value of the largest element? Yes, because there is a limit on the largest value of the smallest element. The smallest element must be less than 50. The smallest member of the set can be 49.9999… The limiting value of the smallest number is 50. As long as the smallest number is a tiny bit less than 50, you can have the greatest number a tiny bit less than 4 + 3*50 = 154. The number 154 and all numbers greater than 154 cannot be a part of the set. Say if the smallest element is 49, the largest element will be 4 + 3*49 = 151. So the set could look something like this: S = {49, 49, 49, 49, … (101 times to balance out the extra 101 in 151), 50, 50, 151} Only option (E) cannot be a part of the set.
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Re: A set of numbers has an average of 50. If the largest element is 4 gre
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19 Aug 2015, 01:47
Bunuel wrote: A set of numbers has an average of 50. If the largest element is 4 greater than 3 times the smallest element, which of the following values cannot be in the set?
(A) 85 (B) 90 (C) 123 (D) 150 (E) 155
Kudos for a correct solution. Ans: E Solution: we are given the relation between smallest and the largest term. so let the smallest a and largest be 3a+4 so the avg = 50 which tells us that any value of a must be less than 50 so a<50 means, largest value 3a+4 <(3*50)+4 =largest value must be <154 so 155 can not be the value in the set.
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Re: A set of numbers has an average of 50. If the largest element is 4 gre
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19 Aug 2015, 02:10
Bunuel wrote: A set of numbers has an average of 50. If the largest element is 4 greater than 3 times the smallest element, which of the following values cannot be in the set?
(A) 85 (B) 90 (C) 123 (D) 150 (E) 155
Kudos for a correct solution. Let L be the largest and S be the smallest of the numbers L=3s+4, given If 155 is a number between L and S, then L>155 when s=51, L=157 when s=50, L=154 If the smallest number is equal to average of all numbers, then all the numbers will be equal to 50. That means L cannot be greater than or equal to 154. L has to be less than 154 Therefore, 155 cannot be in the set. The correct option is E



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Re: A set of numbers has an average of 50. If the largest element is 4 gre
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19 Aug 2015, 03:52
O A is E Largest no should be less than 154 L = 3*s=4 S can be maximized as 50 so L needs to be < 154



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Re: A set of numbers has an average of 50. If the largest element is 4 gre
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19 Aug 2015, 04:01
Although I reached to OA by method explained above
But had strange results when I tried this approach:
AM= (Largest term + Smallest Term)/2 L= 3*S + 4
AM= (3*S +4 +S)/2 = 50 S=24 L= 76
This does not lead us anywhere.
Experts kindly let me know what am I missing here



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Re: A set of numbers has an average of 50. If the largest element is 4 gre
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19 Aug 2015, 04:21
kanigmat011 wrote: Although I reached to OA by method explained above
But had strange results when I tried this approach:
AM= (Largest term + Smallest Term)/2 L= 3*S + 4
AM= (3*S +4 +S)/2 = 50 S=24 L= 76
This does not lead us anywhere.
Experts kindly let me know what am I missing here This method is applicable ONLY to arithmetic progression series (ie series in which \(a_n\) = \(a_{n1}\) \(\pm\) constant). The formula you have mentioned can not be applied to any generic set.



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Re: A set of numbers has an average of 50. If the largest element is 4 gre
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20 Aug 2015, 13:22
I think answer is E, question of maximising the smallest number by making it equal to 50 constraining largest to 3(50) + 4 = 154 see asetofnumbershasanaverageof50ifthelargest128996.htmlThanks for everything you do Bunuel!



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Re: A set of numbers has an average of 50. If the largest element is 4 gre
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20 Aug 2015, 18:43
IMO E, Because only 1553 = 152 is not devisible by 3
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Re: A set of numbers has an average of 50. If the largest element is 4 gre
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10 Sep 2016, 09:41
The average of the series is 50. The elements of the set are spread across the two segments of less than or greater than 50. If there are not less than 50, they have to be at least equal to 50. While the smallest number that could be part of this could be  infinity, but the greatest smallest number that definitely has to be a part in case all elements are same is 50. There cannot be a series that has the smallest number as 51 and an average as 50. Hence, considering 50  the alrgest number would be three times the smallest number (150) + four = 154. Hence, 155 cannot be a part of the series. Answer: E
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Re: A set of numbers has an average of 50. If the largest element is 4 gre
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09 Dec 2016, 20:56
Here is my solution to this one. Given data => Mean =50 Let the smallest number be x so the largest number will be 3x+4 AS they are different => x<50 So the maximum value of 3a+4=> 154 Hence All the values must be less than 154
Also 3a+4>50 Hence a must be greater than 15.3
Hence the range of values of data set will be => (15.3,154)
Clearly E is out of Bound here. Hence E
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Re: A set of numbers has an average of 50. If the largest element is 4 gre
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01 May 2018, 07:44
Bunuel wrote: A set of numbers has an average of 50. If the largest element is 4 greater than 3 times the smallest element, which of the following values cannot be in the set?
(A) 85 (B) 90 (C) 123 (D) 150 (E) 155
Kudos for a correct solution. This question cannot have two right answers, so the highest has to the right one. lol. In case of this question, an answer option cannot be greater than 155. so the highest has to be the right one.



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Re: A set of numbers has an average of 50. If the largest element is 4 gre
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01 May 2018, 08:53
my simple approach was 50(n2)+x+3x+4=50n where n is the number of elements and x be the smallest. above equation solves to x= 24 thus smallest equals to 24 and greatest is 76. where did i go wrong



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Re: A set of numbers has an average of 50. If the largest element is 4 gre
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03 May 2018, 15:22
Bunuel wrote: A set of numbers has an average of 50. If the largest element is 4 greater than 3 times the smallest element, which of the following values cannot be in the set?
(A) 85 (B) 90 (C) 123 (D) 150 (E) 155 We can let the smallest element = n and the largest = 3n + 4. Since the average is 50, then the smallest element, n, must be less than 50. Since n < 50, the largest element, 3n + 4, must be less than 3(50) + 4 = 154. We see that all the numbers in the choices are less than 154 except 155. Thus, 155 can’t be in the set. Answer: E
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Re: A set of numbers has an average of 50. If the largest element is 4 gre
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17 May 2018, 01:38
Smallest element in a given set cannot exceeds the given average.So, the Max value of smallest one could be 50.
Largest=3*Smallest+4 L=3*S+4 L=3*50+4=154
So the list should not contain any number greater than 154.
Option E doesn't meet our desired criterion so so Ans E



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A set of numbers has an average of 50. If the largest element is 4 gre
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26 Sep 2018, 07:55
Bunuel wrote: A set of numbers has an average of 50. If the largest element is 4 greater than 3 times the smallest element, which of the following values cannot be in the set?
(A) 85 (B) 90 (C) 123 (D) 150 (E) 155
\(?\,\,\,:\,\,\,{\text{not}}\,\,{\text{a}}\,\,{\text{member}}\) Let´s call the smallest number x, so that the largest number is 3x+4 and we know each element of the set must belong to the [x, 3x+4] interval. From the fact that 50 is the average of all the elements of this set, and not all of them are equal (*), we have: (*) If x = 3x+4 and we have all numbers equal to 2, their average would not be 50. \(x < 50 < 3x + 4\) Hence: \(x < 50\,\,\,\, \Rightarrow \,\,\,\,{\text{largest}} = 3x + 4 < 3 \cdot 50 + 4 = 154\,\,\,\,\mathop \Rightarrow \limits^{{\text{alternatives}}\,!} \,\,\,? = 155\) This solution follows the notations and rationale taught in the GMATH method. Regards, Fabio.
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