Question

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 piece 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?

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Solution

The correct option is **C** 134

We are first given that seven pieces of rope have an average (arithmetic mean) length of 68 centimeters. From this we can determine the sum.

⇒ Average=sumquantity

⇒ sum=68×7=476

Next we are given that the median length of a piece of rope is 84cm. Thus when we arrange the pieces of rope from least length to greatest, the middle length (the 4th piece) will have a length of 84cm. We also must keep in mind that we can have pieces of rope of the same length. Let’s first label our seven pieces of rope with variables or numbers, starting with the shortest piece and moving to the longest piece. We can let x equal the shortest piece of rope, and m equal the longest piece of rope.

⇒ Seven pieces are x,x,x,84,84,84,m

Notice that the median (the 4th rope) is 84cm long. Thus, pieces 5 and 6 are either equal to the median, or they are greater than the median. In keeping with our goal of minimizing the length of the first 6 pieces, we will assign 84 to pieces 5 and 6 to make them as short as possible. Similarly, we have assigned a length of x to pieces 1,2 and 3.

We can plug these variables into our sum equation:

∴ x+x+x+84+84+84+m=476

∴ 3x+252+m=476

∴ 3x+m=224

We also given that the length of the longest piece of rope is 14cm more than 4 times the length of the shortest piece of rope. So we can say:

⇒ m=14+4x

We can now plug 14+4x in for m into the equation 3x+m=224. So we have:

⇒ 3x+14+4x=224

⇒ 7x=210

⇒ x=30

Thus, the longest piece of rope is 4(30)+14=134cm

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