Roland2rule wrote:A set of numbers has an average of 50. If the largest element is 5 greater than 3 times the smallest element and if the median equals the mean . what is largest possible value in the set
A. 85
B. 86
C. 88
D. 91
E. 92
OA is E
The breakdown here is kind of hard, pls i need the help of an expert. Thanks in anticipation
This is a very time-consuming question. You may leave this one. The GMAT will not ask this kind of question.
Say the smallest number is x, thus, the largest number = 3x + 5.
Case 1: Say the set has odd numbers of elements ( = n)
Thus, the set is {x, ...
50, ... (3x + 5)}; we are given that median = 50, thus, the middle-most number = 50.
Since we need to maximize the value of the largest number, other numbers in the set must be as small as possible.
Thus, the set is {x, x, x, ... x,
50, 50, ... 50, 50, (3x + 5)}
This means that must be (n - 1)/2 numbers of x and (n - 1)/2 numbers of 50s.
The average of the set = [x*(n - 1)/2 + 50*(n - 1)/2 + (3x + 5)] / n
=> [x*(n - 1)/2 + 50*(n - 1)/2 + (3x + 5)] / n = 50; we are given that Mean = Median = 50
Upon solving this, we get xn - 51n - 40 + 5x = 0
We have two unknowns x and n. We only know that n is a positive integer. By glancing at the options, we get that x is also a positive integer. But we can't solve for x.
Let's plug-in the option values from the options. Let's start with option E as 92 is the highest value.
Option E: We have 3x + 5 = 92 => x = 29
We plug-in x = 29 in xn - 51n - 40 + 5x = 0.
xn - 51n - 40 + 5x = 0 => 29n - 51n - 40 + 5*29 = 0 => n = 105/22 = not an integer value
This does not mean that 92 is 100% incorrect answer. It is possible that the set has even numbers of elements.
Case 2: Say the set has even numbers of elements (= n)
Thus, the set is {x, ...
50, 50, ... (3x + 5)}; we are given that median = 50; the median = (50 + 50)/2 = 50.
Since we need to maximize the value of the largest number, other numbers in the set must be as small as possible.
Thus, the set is {x, x, x, ... x,
50, 50, 50, ... 50, 50, (3x + 5)}
This means that there are (n/2 - 1) numbers of x and n/2 numbers of 50s.
The average of the set = [x*(n/2 - 1) + 50*(n /2) + (3x + 5)] / n
=> [x*(n/2 - 1) + 50*(n /2) + (3x + 5)] / n = 50; we are given that Mean = Median = 50
Upon solving this, we get xn - 50n + 4x + 10 = 0
Again, we have two unknowns x and n.
Let's plug-in the option values from the options. Again, let's start with option E as 92 is the highest value.
Option E: We have 3x + 5 = 92 => x = 29
We plug-in x = 29 in xn - 50n + 4x + 10 = 0.
xn - 50n + 4x + 10 = 0 => 29n - 50n + 4*29 + 10 = 0 => n = 126/21 = 6 (a positive integer). Thus, option E is the correct answer.
The set would be {29, 29, 50, 50, 50, 92}.
The correct answer:
E
Hope this helps!
-Jay
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