Seven pieces of rope have an average (arithmetic mean) length of 68 centimeters and a median length of 84 centimeters. If the length of the longest pieces of rope is 14 centimeters more than 4 times the length of the shortest piece of rope, what is the maximum possible length, in centimeters, of the longest piece of rope?

(A) 82

(B) 118

(C) 120

(D) 134

(E) 152

This is a classic example of a min-max problem. Let’s start by drawing seven blanks to represent the seven pieces of rope in order of length and putting 84 (the median) in the middle blank.

, , , 84 , , ,

Let’s call the shortest piece x. Consequently, the longest piece is 4x + 14.

x , , , 84 , , , 4x + 14

In order to maximize the length of the longest piece and maintain the overall average of 68, we need to minimize the length of all the other pieces. Because there’s no restriction on repeated lengths (this is an important wrinkle to look out for in min-max problems) we can fill out the other blanks like so:

x , x , x , 84 , 84 , 84 , 4x + 14

We’ll use the expression for the arithmetic mean to solve for x, and then calculate the length of the longest piece of rope:

\(\frac{x + x + x + 84 + 84 + 84 + (4x + 14)}{7} = 68\)

Simplify a little bit, but not too much – leave things in factored form as much as possible… you never know, something might cancel!

\(\frac{7x + 3(84) + 14}{7} = 68\)

It’s a miracle! Seven, 84, and 14 are all divisible by 7, so we can simplify by dividing everything on the left-hand-side by 7, and avoid calculating \({3}\times{84}\)

\(x + 3(12) + 2 = 68\)

So, \(x = 30\) and \(4x + 14 = 4(30) + 14 = 134\).

**The correct answer is D.**