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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## OG 18, question-211 tagged by: Brent@GMATPrepNow ##### This topic has 4 expert replies and 0 member replies ## OG 18, question-211 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? A. 82 B. 118 C. 120 D. 134 E. 152 ### GMAT/MBA Expert GMAT Instructor Joined 25 May 2010 Posted: 15362 messages Followed by: 1866 members Upvotes: 13060 GMAT Score: 790 Quote: 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? A) 82 B) 118 C) 120 D) 134 E) 152 When given an average, calculate the SUM. Sum = (number)(average). In the problem above: Sum of the lengths = (number of lengths)(average length) = 7*68 = 476. Let the smallest piece = x. Then the length of the longest piece = 4x+14. Median piece = 84. Let the remaining pieces be a, b, c, d. Here are the 7 pieces, in ascending order: x, a, b, 84, c, d, 4x+14. To MAXIMIZE the value of 4x+14, we must MINIMIZE the values of a, b, c, and d. The least possible value for a and b is x. The least possible value for c and d is 84. Here are the 7 pieces: x, x, x, 84, 84, 84, 4x+14. Since the sum of the lengths is 476, we get: x + x + x + 84 + 84 + 84 + 4x+14 = 476 7x + 266 = 476 7x = 210 x = 30. Thus: Greatest possible value for the longest piece = 4x+14 = 4*30 + 14 = 134. The correct answer is D. _________________ Mitch Hunt Private Tutor for the GMAT and GRE GMATGuruNY@gmail.com If you find one of my posts helpful, please take a moment to click on the "UPVOTE" icon. Available for tutoring in NYC and long-distance. For more information, please email me at GMATGuruNY@gmail.com. Student Review #1 Student Review #2 Student Review #3 Free GMAT Practice Test How can you improve your test score if you don't know your baseline score? Take a free online practice exam. Get started on achieving your dream score today! Sign up now. ### GMAT/MBA Expert GMAT Instructor Joined 08 Dec 2008 Posted: 13001 messages Followed by: 1250 members Upvotes: 5254 GMAT Score: 770 Quote: 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? (A) 82 (B) 118 (C) 120 (D) 134 (E) 152 We have 7 rope lengths. If the median is 84, then the lengths (arranged in ascending order) look like this: {_, _, _, 84, _, _, _} The length of the longest piece of rope is 14 cm more than 4 times the length of the shortest piece of rope. Let x = length of SHORTEST piece. This means that 4x+14 = length of LONGEST piece. So, we now have: {x, _, _, 84, _, _, 4x+14} Our task is the MAXIMIZE the length of the longest piece. To do this, we need to MINIMIZE the other lengths. So, we'll make the 2nd and 3rd lengths have length x as well (since x is the shortest possible length) We get: {x, x, x, 84, _, _, 4x+14} Since 84 is the middle-most length, the 2 remaining lengths must be greater than or equal to 84. So, the shortest lengths there are 84. So, we get: {x, x, x, 84, 84, 84, 4x+14} Now what? At this point, we can use the fact that the average length is 68 cm. There's a nice rule (that applies to MANY statistics questions) that says: the sum of n numbers = (n)(mean of the numbers) So, if the mean of the 7 numbers is 68, then the sum of the 7 numbers = (7)(68) = 476 So, we now now that x+x+x+84+84+84+(4x+14) = 476 Simplify to get: 7x + 266 = 476 7x = 210 x=30 If x=30, then 4x+14 = 134 So, the longest piece will be 134 cm long. Answer = D Cheers, Brent _________________ Brent Hanneson â€“ Creator of GMATPrepNow.com Use my video course along with Sign up for free Question of the Day emails And check out all of these free resources GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMATâ€™s FREE 60-Day Study Guide and reach your target score in 2 months! ### GMAT/MBA Expert Legendary Member Joined 14 Jan 2015 Posted: 2667 messages Followed by: 122 members Upvotes: 1153 GMAT Score: 770 A popular Q-Type. See here for another good example: http://www.beatthegmat.com/statistics-problem-2-median-t62108.html _________________ Veritas Prep | GMAT Instructor Veritas Prep Reviews Save$100 off any live Veritas Prep GMAT Course

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vaibhav101 wrote:
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?

A. 82
B. 118
C. 120
D. 134
E. 152
We need to first recognize that we are working with a maximum problem. This means that of the 7 pieces of rope, we must make the lengths of 6 of those pieces as small as we possibly can, within the confines of the given information, and doing so will maximize the length of the seventh piece.

We are first given that 7 pieces of rope have an average (arithmetic mean) length of 68 centimeters. From this we can determine the sum.

sum = average x quantity

sum = 68 x 7 = 476

Next we are given that the median length of a piece of rope is 84 centimeters. Thus, when we arrange the pieces of rope from least to greatest length, the middle length (the fourth piece) will have a length of 84 centimeters. We also must keep in mind that we can have pieces of rope of the same length. Let's first label our 7 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.

piece 1: x

piece 2: x

piece 3: x

piece 4: 84

piece 5: 84

piece 6: 84

piece 7: m

Notice that the median (the fourth rope) is 84 cm 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 lengths 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 were also given that the length of the longest piece of rope is 14 centimeters 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 in 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 = 134 centimeters.

_________________

Jeffrey Miller
jeff@targettestprep.com

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