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Bunuel
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

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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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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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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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kanigmat011
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 \(a_n\) = \(a_{n-1}\) \(\pm\) constant). The formula you have mentioned can not be applied to any generic set.
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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 a-set-of-numbers-has-an-average-of-50-if-the-largest-128996.html


Thanks for everything you do Bunuel!
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IMO E,
Because only 155-3 = 152 is not devisible by 3
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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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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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Bunuel
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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my simple approach was
50(n-2)+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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Bunuel
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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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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Bunuel
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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