Number Set

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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