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A set of number has an average of 50. If the largest element

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A set of number has an average of 50. If the largest element

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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

OA E

Source: Veritas Prep

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BTGmoderatorDC 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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BTGmoderatorDC 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

Source: Veritas Prep
\[?\,\,\,:\,\,\,{\text{not}}\,\,{\text{a}}\,\,{\text{member}}\]
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:
\[x < 50 < 3x + 4\]
(*) If x = 3x+4 and we have all numbers equal to -2, their average would not be 50.

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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BTGmoderatorDC 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

OA E

Source: Veritas Prep
The largest element is 4 greater than 3 times the smallest element
Let x = the smallest number in the set.
So, 3x + 4 = the largest number in the set.
So, the set looks something like this {x, ?, ?, . . . . ?, 3x+4}

The key to this question is that the AVERAGE = 50
Since all of the answer choices are greater than 50, there's a good likelihood that the correct answer will be the biggest number.
So, let's start by checking answer choice E

(E) 155
There are two possibilities here:
i) 155 is the biggest number in the set
ii) 155 is NOT the biggest number in the set

If 155 IS the biggest number in the set, we can use the given information to determine the smallest number in the set
That is, 3x +4 = 155
So, 3x = 152
So, x = 152/3 = 50 2/3
Hmmmm, the "smallest value" is GREATER than 50 (the average value on the set)
This is impossible; if all of the values in the set are greater than 50, the average cannot equal 50.
So, 155 CANNOT be the LARGEST number in the set (case i).

This also means that 155 CANNOT be in the set at all (case ii), because if 155 is NOT the biggest number, then the biggest number is EVEN BIGGER, which would also make it impossible for the average to be 50.

Answer: E

Cheers,
Brent

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Let L the largest and S 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 the 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. E is the correct option.

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BTGmoderatorDC 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
Let S = the smallest number of L = the largest number.
Since the largest number is 4 more than 3 times the smallest number, we get:
L = 3S + 4
L-4 = 3S
(L-4)/3 = S.

Since the average is 50 -- and the numbers are not all the same -- S must be LESS THAN 50, while L must be greater than 50.
Substituting S = (L-4)/3 into S < 50, we get;
(L-4)/3 < 50
L-4 < 150
L < 154.

Since the largest number must be less than 154, option E -- 155 -- cannot be in the set.

The correct answer is E.

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