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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 ## CARS ##### This topic has 2 expert replies and 3 member replies ## CARS The number of defects in the first five cars to come through a new production line are 9, 7, 10, 4, and 6, respectively. If the sixth car through the production line has either 3, 7, or 12 defects, for which of theses values does the mean number of defects per car for the first six cars equal the median? I. 3 II. 7 III. 12 A. I only B. II only C. III only D. I and III only E. I, II, and III CORRECT ANSWER D Legendary Member Joined 15 Apr 2011 Posted: 1085 messages Followed by: 21 members Upvotes: 158 by arranging the set {4,6,7,9,10} we know that mean=median when the values are consecutive; so must be {4,five,6,7,eight,9,10} or {three,4,6,7,9,10,twelve,thirteen} where 3,4 and 6,7 and 9,10 and 12,13 are considered as clued, One unit expressions -> 7, 13, 19, 25 (the difference is 6, so consecutive properties). Thirteen is out BUT we must keep it in mind as possible number in this set. Option d 3 and 12 phelps wrote: The number of defects in the first five cars to come through a new production line are 9, 7, 10, 4, and 6, respectively. If the sixth car through the production line has either 3, 7, or 12 defects, for which of theses values does the mean number of defects per car for the first six cars equal the median? I. 3 II. 7 III. 12 A. I only B. II only C. III only D. I and III only E. I, II, and III CORRECT ANSWER D _________________ Success doesn't come overnight! Legendary Member Joined 07 Nov 2008 Posted: 1255 messages Followed by: 90 members Upvotes: 312 The sum of the first five cars is 36. If the sixth car has 3, the sum of the first 6 is 39, which makes the mean 6.5 arranging them in order 3,4,6,7,9,10 - the median is the average of the two middle numbers = 6.5 "I" works. Rule out B and C Since III appears twice we test it next. the new sum = 36+12 = 48 average = 8. again arranging 4,6,7,9,10, 13 = the median is the average of 7 and 9 = 8 "III" works - rule out A. We still have to check II. New sum = 45. Average = 7.5 arranging 4,6,7,7,9,10 - the median is 7. II doesn't work, the answer is D _________________ Tani Wolff ### Top Member Senior | Next Rank: 100 Posts Joined 12 Jan 2019 Posted: 38 messages The sum of defects in the first five cars is 9 + 7 + 10 + 4 + 6 = 36. With six cars, the median will be the average of third and fourth ranked defects, when arranged in ascending order. Since number of defects is always an integer, hence the median must either be an integer or integer plus 0.5. Now if there are X defects in the sixth car, then the mean is obtained as M = (36 + X)/6 Since 36 is already divisible by 6, hence to satisfy the median condition, X must either be a multiple of 6 or a multiple of 3. From the given options, 7 does not satisfy the condition, hence it is out. We now need to check for both 3 and 12. With X = 3, the mean is M = 39/6 = 6.5; and the values arranged in ascending order are {3, 4, 6, 7, 9, 10} for which the median is (6 + 7)/2 = 6.5 = M. So it matches for I. With X = 12, the mean is M = 48/6 = 8; and the values arranged in ascending order are {4, 6, 7, 9, 10, 12} for which the median is (7 + 9)/2 = 8 = M. So it matches for III. Both I and III match, hence D. ### GMAT/MBA Expert Elite Legendary Member Joined 23 Jun 2013 Posted: 10020 messages Followed by: 494 members Upvotes: 2867 GMAT Score: 800 Hi All, We're told that the number of defects in the first five cars to come through a new production line are 9, 7, 10, 4, and 6, respectively and that the sixth car through the production line has either 3, 7, or 12 defects. We're asked which of the following three values does the MEAN number of defects per car for the first six cars equal the MEDIAN. This question comes down to organizing the data (to find the Median) and using the Average Formula. I. 3 defects IF... the 6th car has 3 defects, then the six numbers would be: 3, 4, 6, 7, 9 and 10 The Median would be (6+7)/2 = 6.5 The Average would be (3+4+6+7+9+10)/6 = 39/6 = 6.5 The Average and the Median ARE equal. Eliminate Answers B and C. II. 7 defects IF... the 6th car has 7 defects, then the six numbers would be: 4, 6, 7, 7, 9 and 10 The Median would be (7+7)/2 = 7 The Average would be (4+6+7+7+9+10)/6 = 43/6 = 7 1/6 The Average and the Median are NOT equal. Eliminate Answer E III. 12 defects IF... the 6th car has 12 defects, then the six numbers would be: 4, 6, 7, 9, 10 and 12 The Median would be (7+9)/2 = 8 The Average would be (4+6+7+9+10+12)/6 = 48/6 = 8 The Average and the Median ARE equal. Eliminate Answer A. Final Answer: D GMAT assassins aren't born, they're made, Rich _________________ Contact Rich at Rich.C@empowergmat.com ### GMAT/MBA Expert GMAT Instructor Joined 25 Apr 2015 Posted: 1808 messages Followed by: 14 members Upvotes: 43 phelps wrote: The number of defects in the first five cars to come through a new production line are 9, 7, 10, 4, and 6, respectively. If the sixth car through the production line has either 3, 7, or 12 defects, for which of theses values does the mean number of defects per car for the first six cars equal the median? I. 3 II. 7 III. 12 A. I only B. II only C. III only D. I and III only E. I, II, and III Letâ€™s check each number in the given Roman numerals. I. 3 The average is (3 + 4 + 6 + 7 + 9 + 10)/6 = 6.5. The median is (6 + 7)/2 = 13/2 = 6.5. We see that I is correct. II. 7 The average is (4 + 6 + 7 + 7 + 9 + 10)/6 â‰ˆ 7.17. The median is (7 + 7)/2 = 14/2 = 7. We see that II is not correct. III. 12 The average is (4 + 6 + 7 + 9 + 10 + 12)/6 = 8. The median is (7 + 9)/2 = 16/2 = 8. We see that III is correct. Answer: D _________________ Scott Woodbury-Stewart Founder and CEO • Free Veritas GMAT Class Experience Lesson 1 Live Free Available with Beat the GMAT members only code • FREE GMAT Exam Know how you'd score today for$0

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