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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 ## integers----> nice question ##### This topic has 4 expert replies and 5 member replies ## integers----> nice question a, b, and c are integers and a < b < c. S is the set of all integers from a to b, inclusive. Q is the set of all integers from b to c, inclusive. The median of set S is (3/4) b. The median of set Q is (7/8) c. If R is the set of all integers from a to c, inclusive, what fraction of c is the median of set R? (A) 3/8 (B) Â½ (C) 11/16 (D) 5/7 (E) Â¾ ### GMAT/MBA Expert GMAT Instructor Joined 25 May 2010 Posted: 15385 messages Followed by: 1872 members Upvotes: 13060 GMAT Score: 790 atulmangal wrote: a, b, and c are integers and a < b < c. S is the set of all integers from a to b, inclusive. Q is the set of all integers from b to c, inclusive. The median of set S is (3/4) b. The median of set Q is (7/8) c. If R is the set of all integers from a to c, inclusive, what fraction of c is the median of set R? (A) 3/8 (B) Â½ (C) 11/16 (D) 5/7 (E) Â¾ We can plug in a value for c. Let c=16. Median of Q = (7/8)*c = (7/8)*16 = 14. Since 14 is halfway between b and 16, b=12. Median of S = (3/4)*b = (3/4)*12 = 9. Since 9 is halfway between a and 12, a=6. Thus, one combination that satisfies all the conditions given is a=6, b=12, c=16. Since R is the set of all the integers from a to c, inclusive, R = {6,7,8,9,10,11,12,13,14,15,16}. Median of R = 11. (Median of R)/c = 11/16. The correct answer is C. _________________ 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 Last edited by GMATGuruNY on Mon Dec 05, 2011 6:58 pm; edited 4 times in total 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. Newbie | Next Rank: 10 Posts Joined 09 Jan 2011 Posted: 5 messages hi Mitch, As usual a great solution from your side. However I dont understand 1 thing: In set R , the median should be the middle value i.e 12..rite..?? Master | Next Rank: 500 Posts Joined 08 Mar 2011 Posted: 150 messages Upvotes: 5 GMATGuruNY wrote: atulmangal wrote: a, b, and c are integers and a < b < c. S is the set of all integers from a to b, inclusive. Q is the set of all integers from b to c, inclusive. The median of set S is (3/4) b. The median of set Q is (7/8) c. If R is the set of all integers from a to c, inclusive, what fraction of c is the median of set R? (A) 3/8 (B) Â½ (C) 11/16 (D) 5/7 (E) Â¾ We can plug in a value for c. Let c=16. Median of Q = (7/8)*c = (7/8)*16 = 14. Since 14 is halfway between b and 16, b=12. Median of S = (3/4)*b = (3/4)*12 = 9. Since 9 is halfway between a and 12, a=6. Thus, one combination that works for set R is a=6, b=12, c=16. Given any set of consecutive integers: Median = average = (biggest+smallest)/2. Thus, median of R = (16+6)/2 = 11. (Median of R)/c = 11/16. The correct answer is C. I did not get the point. Is the rule " Given any set of consecutive integers: Median = average = (biggest+smallest)/2." valid for any consecutive numbers. What about the set of numbers:4,5,7,10??? ### GMAT/MBA Expert GMAT Instructor Joined 08 Jan 2008 Posted: 3225 messages Followed by: 613 members Upvotes: 1710 GMAT Score: 800 finance wrote: Quote: a, b, and c are integers and a < b < c. S is the set of all integers from a to b, inclusive. Q is the set of all integers from b to c, inclusive. The median of set S is (3/4) b. The median of set Q is (7/8) c. If R is the set of all integers from a to c, inclusive, what fraction of c is the median of set R? (A) 3/8 (B) Â½ (C) 11/16 (D) 5/7 (E) Â¾ I did not get the point. Is the rule " Given any set of consecutive integers: Median = average = (biggest+smallest)/2." valid for any consecutive numbers. What about the set of numbers:4,5,7,10??? Hi! 4, 5, 7 and 10 are not consecutive numbers, so the rule doesn't apply. "Consecutive" means that the numbers are a fixed distance apart. Some examples of sets of consecutive numbers to which the rule applies: {1, 2, 3, 4} {3, 6, 9, 12} {-10, -5, 0, 5, 10} {11, 14, 17, 20} In your set, there's a space of 1 between 4&5, a space of 2 between 5&7 and a space of 3 between 7&10; since the spacing varies (even though it follows a rule), the numbers aren't considered consecutive. Free GMAT Practice Test under Proctored Conditions! - Find a practice test near you or live and online in Kaplan's Classroom Anywhere environment. Register today! Master | Next Rank: 500 Posts Joined 08 Mar 2011 Posted: 150 messages Upvotes: 5 Stuart Kovinsky wrote: finance wrote: Quote: a, b, and c are integers and a < b < c. S is the set of all integers from a to b, inclusive. Q is the set of all integers from b to c, inclusive. The median of set S is (3/4) b. The median of set Q is (7/8) c. If R is the set of all integers from a to c, inclusive, what fraction of c is the median of set R? (A) 3/8 (B) Â½ (C) 11/16 (D) 5/7 (E) Â¾ I did not get the point. Is the rule " Given any set of consecutive integers: Median = average = (biggest+smallest)/2." valid for any consecutive numbers. What about the set of numbers:4,5,7,10??? Hi! 4, 5, 7 and 10 are not consecutive numbers, so the rule doesn't apply. "Consecutive" means that the numbers are a fixed distance apart. Some examples of sets of consecutive numbers to which the rule applies: {1, 2, 3, 4} {3, 6, 9, 12} {-10, -5, 0, 5, 10} {11, 14, 17, 20} In your set, there's a space of 1 between 4&5, a space of 2 between 5&7 and a space of 3 between 7&10; since the spacing varies (even though it follows a rule), the numbers aren't considered consecutive. Thank you for your explanation! I know "consecutive integers" just as you described above, but I got a little bit confused from the provided solution because the given numbers a=6, b=12 and c=16 are said to be consecutive, but they have not equal distances among each other?? ### GMAT/MBA Expert GMAT Instructor Joined 25 May 2010 Posted: 15385 messages Followed by: 1872 members Upvotes: 13060 GMAT Score: 790 finance wrote: Stuart Kovinsky wrote: finance wrote: Quote: a, b, and c are integers and a < b < c. S is the set of all integers from a to b, inclusive. Q is the set of all integers from b to c, inclusive. The median of set S is (3/4) b. The median of set Q is (7/8) c. If R is the set of all integers from a to c, inclusive, what fraction of c is the median of set R? (A) 3/8 (B) Â½ (C) 11/16 (D) 5/7 (E) Â¾ I did not get the point. Is the rule " Given any set of consecutive integers: Median = average = (biggest+smallest)/2." valid for any consecutive numbers. What about the set of numbers:4,5,7,10??? Hi! 4, 5, 7 and 10 are not consecutive numbers, so the rule doesn't apply. "Consecutive" means that the numbers are a fixed distance apart. Some examples of sets of consecutive numbers to which the rule applies: {1, 2, 3, 4} {3, 6, 9, 12} {-10, -5, 0, 5, 10} {11, 14, 17, 20} In your set, there's a space of 1 between 4&5, a space of 2 between 5&7 and a space of 3 between 7&10; since the spacing varies (even though it follows a rule), the numbers aren't considered consecutive. Thank you for your explanation! I know "consecutive integers" just as you described above, but I got a little bit confused from the provided solution because the given numbers a=6, b=12 and c=16 are said to be consecutive, but they have not equal distances among each other?? The problem states that R is the set of all the integers from a to c, inclusive. Thus, if a=6 and c=16, R = {6,7,8,9,10,11,12,13,14,15,16}. _________________ 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 Last edited by GMATGuruNY on Tue May 10, 2011 12:03 pm; edited 1 time in total 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. Master | Next Rank: 500 Posts Joined 08 Mar 2011 Posted: 150 messages Upvotes: 5 Thank you GMATGuruNY! I enjoyed your solution, but contrary to what you stated in your last reply, R is given as the set of all integers from a to c, inclusive, without mentioning it to be "consecutive". I thought that you found out that those terms are consecutive, because it is not mentioned in the question. ### GMAT/MBA Expert GMAT Instructor Joined 08 Jan 2008 Posted: 3225 messages Followed by: 613 members Upvotes: 1710 GMAT Score: 800 finance wrote: Thank you GMATGuruNY! I enjoyed your solution, but contrary to what you stated in your last reply, R is given as the set of all integers from a to c, inclusive, without mentioning it to be "consecutive". I thought that you found out that those terms are consecutive, because it is not mentioned in the question. Hi, if R is a set of all integers between two specific integers, inclusive, then by definition R will be a set of consecutive integers. For example, if a=2 and c=10, then R = {2, 3, 4, 5, 6, 7, 8, 9, 10}. Free GMAT Practice Test under Proctored Conditions! - Find a practice test near you or live and online in Kaplan's Classroom Anywhere environment. Register today! Master | Next Rank: 500 Posts Joined 08 Mar 2011 Posted: 150 messages Upvotes: 5 Stuart Kovinsky wrote: finance wrote: Thank you GMATGuruNY! I enjoyed your solution, but contrary to what you stated in your last reply, R is given as the set of all integers from a to c, inclusive, without mentioning it to be "consecutive". I thought that you found out that those terms are consecutive, because it is not mentioned in the question. Hi, if R is a set of all integers between two specific integers, inclusive, then by definition R will be a set of consecutive integers. For example, if a=2 and c=10, then R = {2, 3, 4, 5, 6, 7, 8, 9, 10}. Ohh I see! I have to read the questions more carefully. 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