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Re: D0121 [#permalink]
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11 Jan 2018, 07:25
dave13 wrote: Bunuel wrote: Official Solution:
A set of 11 different integers has a median of 25 and a range of 50. What is the greatest possible integer that could be in this set?
A. 65 B. 70 C. 75 D. 80 E. 85
Consider 11 numbers in ascending order to be \(x_1\), \(x_2\), \(x_3\), ..., \(x_{11}\). The median of a set with odd number of elements is the middle number (when arranged in ascending or descending order), so the median of given set is \(x_{6}=25\); The range of a set is the difference between the largest and the smallest numbers of a set, so the range of given set is \(50=x_{11}x_{1}\) \(\rightarrow\) \(x_{11}=50+x_{1}\); We want to maximize \(x_{11}\), hence we need to maximize \(x_{1}\). Since all integers must be distinct then the maximum value of \(x_{1}\) will be \(median5=255=20\) and thus the maximum value of \(x_{11}\) is \(x_{11}=50+20=70\). The set could be {20, 21, 22, 23, 24, 25, 26, 27, 26, 29, 70}
Answer: B Hello Bunuel, here is my solution: I simply plugged in values from answer choices 6525 =40 7025 = 45 7525 = 50 voila, so I chose C, because 25 can be both the smallest and median 25, 25, 25, 25, 25, 25 25, 25, 25, 25, 75 isnt it correct ? why this solution is not correct, pls explain P.S. its only now i noticed that integers must be different!!!! ok forget my qurestion



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Re: D0121 [#permalink]
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09 Mar 2018, 01:45
I don't agree with the explanation. consider this set (25,25,25,25,25,25,26,27,28,29,75)



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Re: D0121 [#permalink]
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09 Mar 2018, 03:05
suraj_patil wrote: I don't agree with the explanation. consider this set (25,25,25,25,25,25,26,27,28,29,75) We are given that the set consists of 11 different integers .
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Re: D0121 [#permalink]
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10 Mar 2018, 00:12
I did not agree with this question it was written that 11 distinct integer not 11 distinct positive integers so smallest integer can be negative to any extent. please comment on this question Bunuel



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Re: D0121 [#permalink]
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10 Mar 2018, 00:14
rishabhmishra wrote: I did not agree with this question it was written that 11 distinct integer not 11 distinct positive integers so smallest integer can be negative to any extent. please comment on this question BunuelPlease read the whole thread before posting. We want to find the greatest possible integer in the set. It turns out that the greatest possible integer is 70, and in this case the set is {20, 21, 22, 23, 24, 25, 26, 27, 26, 29, 70}, so no negative integers could be in the set if the greatest integer is 70, median is 25 and the range is 50.
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Re: D0121 [#permalink]
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10 Mar 2018, 00:26
Bunuel wrote: rishabhmishra wrote: I did not agree with this question it was written that 11 distinct integer not 11 distinct positive integers so smallest integer can be negative to any extent. please comment on this question BunuelPlease read the whole thread before posting. We want to find the greatest possible integer in the set. It turns out that the greatest possible integer is 70, and in this case the set is {20, 21, 22, 23, 24, 25, 26, 27, 26, 29, 70}, so no negative integers could be in the set if the greatest integer is 70, median is 25 and the range is 50. sorry i was totally confused just because i was trying to solve this question with plugging no. not by algebra i am really sorry to disturb you but thanks.



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Re: D0121 [#permalink]
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14 Apr 2018, 16:24
dave13 wrote: Bunuel wrote: Official Solution:
A set of 11 different integers has a median of 25 and a range of 50. What is the greatest possible integer that could be in this set?
A. 65 B. 70 C. 75 D. 80 E. 85
Consider 11 numbers in ascending order to be \(x_1\), \(x_2\), \(x_3\), ..., \(x_{11}\). The median of a set with odd number of elements is the middle number (when arranged in ascending or descending order), so the median of given set is \(x_{6}=25\); The range of a set is the difference between the largest and the smallest numbers of a set, so the range of given set is \(50=x_{11}x_{1}\) \(\rightarrow\) \(x_{11}=50+x_{1}\); We want to maximize \(x_{11}\), hence we need to maximize \(x_{1}\). Since all integers must be distinct then the maximum value of \(x_{1}\) will be \(median5=255=20\) and thus the maximum value of \(x_{11}\) is \(x_{11}=50+20=70\). The set could be {20, 21, 22, 23, 24, 25, 26, 27, 26, 29, 70}
Answer: B Hello Bunuel, here is my solution: I simply plugged in values from answer choices 6525 =40 7025 = 45 7525 = 50 voila, so I chose C, because 25 can be both the smallest and median 25, 25, 25, 25, 25, 25 25, 25, 25, 25, 75 isnt it correct ? why this solution is not correct, pls explain Pls note that the question says that ALL 11 NUMBERS ARE DIFFERENT so numbers less than median .i.e. 25 has to be <25 and different. Hope its clear.



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Re: D0121 [#permalink]
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16 Apr 2018, 08:17
typo in the official solution: "The set could be {20, 21, 22, 23, 24, 25, 26, 27, 26, 29, 70}" should be... "The set could be {20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 70}"



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Re: D0121 [#permalink]
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16 Apr 2018, 08:20
benejo wrote: typo in the official solution: "The set could be {20, 21, 22, 23, 24, 25, 26, 27, 26, 29, 70}" should be... "The set could be {20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 70}" Edited the typo. Thank you.
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Re: D0121 [#permalink]
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16 May 2018, 20:16
Hi Bunuel,
I have a doubt. why are we assuming the numbers are in AP. Why can't the first number be 1 or 2?



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Re: D0121 [#permalink]
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16 May 2018, 21:44
nishitraj wrote: Hi Bunuel,
I have a doubt. why are we assuming the numbers are in AP. Why can't the first number be 1 or 2? This is exaplined couple of times in this thread. 1. We are not assuming that the numbers are consecutive. 2. We want to find the greatest possible integer in the set. It turns out that the greatest possible integer is 70, and in this case the set is {20, 21, 22, 23, 24, 25, 26, 27, 26, 29, 70}, so no negative integers could be in the set if the greatest integer is 70, median is 25 and the range is 50.
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Re: D0121 [#permalink]
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05 Jun 2018, 08:58
This is a poor answer. All the numbers dont have to be distinct before the median. It can as well be 25 all through before the median.



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Re: D0121 [#permalink]
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05 Jun 2018, 09:02







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