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Re: The average (arithmetic mean) of 10 distinct odd integers in a set is [#permalink]
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14 Jul 2017, 01:58
Bunuel wrote:
The average (arithmetic mean) of 10 distinct odd integers in a set is 55. What is the median of these 10 integers?
(1) The largest integer is 65. (2) The smallest integer is 41.
The total sum of 10 distinct odd integers is 550
(1) the largest integer is 65...not sufficient
( Longer Way-Making Sets is to write 55 ten times and then play with numbers) 65,63,61,59,57,55,51,49,47,43 Mean is 55 and Median is 56 65,63,61,59,57,53,51,49,47,45 Mean is 55 and Median is 55
(2) the smallest integer is 41.....not sufficient
41,43,45,47,53,57,63,65,67,69 Mean is 55 and Median is 55 41,43,45,47,55,57,61,65,67,69 Mean is 55 and Median is 56
ON combining 41,45,51,53,55,57,59,61,63,65 Mean is 55 and Median is 56 41,47,49,53,55,57,59,61,63,65 Mean is 55 and Median is 56
Re: The average (arithmetic mean) of 10 distinct odd integers in a set is [#permalink]
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14 Jul 2017, 07:48
1
Bunuel wrote:
The average (arithmetic mean) of 10 distinct odd integers in a set is 55. What is the median of these 10 integers?
(1) The largest integer is 65. (2) The smallest integer is 41.
1) Largest Int = 65 Insufficient. Multiple values possible.
2) Smallest int = 41 Insufficient. Multiple values possible.
1+2) 41 < 8 odd integers < 65 mean = 55 The mean is closer to 65, and we've lost leverage in terms of going below 41, or above 65 => the only place where I can switch numbers around is near 41. So the median remains the same at 56.
Below are the cases. Case 1 - 41,45,51,53,55,57,59,61,63,65 Case 2 - 41,47,49,53,55,57,59,61,63,65 Sufficient. Answer is C IMO.
My question I took around 3.05 minutes to solve this, and I feel it's way too long. Is there a shorter way to do this question? Is there something I missed i.e. a shortcut?
_________________
Re: The average (arithmetic mean) of 10 distinct odd integers in a set is [#permalink]
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14 Jul 2017, 11:25
akshayk wrote:
Bunuel wrote:
The average (arithmetic mean) of 10 distinct odd integers in a set is 55. What is the median of these 10 integers?
(1) The largest integer is 65. (2) The smallest integer is 41.
1) Largest Int = 65 Insufficient. Multiple values possible.
2) Smallest int = 41 Insufficient. Multiple values possible.
1+2) 41 < 8 odd integers < 65 mean = 55 The mean is closer to 65, and we've lost leverage in terms of going below 41, or above 65 => the only place where I can switch numbers around is near 41. So the median remains the same at 56.
Below are the cases. Case 1 - 41,45,51,53,55,57,59,61,63,65 Case 2 - 41,47,49,53,55,57,59,61,63,65 Sufficient. Answer is C IMO.
My question I took around 3.05 minutes to solve this, and I feel it's way too long. Is there a shorter way to do this question? Is there something I missed i.e. a shortcut?
akshayk how do u say that mean remains at 56,,on what basis???
Re: The average (arithmetic mean) of 10 distinct odd integers in a set is [#permalink]
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14 Jul 2017, 11:47
mohshu wrote:
akshayk wrote:
Bunuel wrote:
The average (arithmetic mean) of 10 distinct odd integers in a set is 55. What is the median of these 10 integers?
(1) The largest integer is 65. (2) The smallest integer is 41.
1) Largest Int = 65 Insufficient. Multiple values possible.
2) Smallest int = 41 Insufficient. Multiple values possible.
1+2) 41 < 8 odd integers < 65 mean = 55 The mean is closer to 65, and we've lost leverage in terms of going below 41, or above 65 => the only place where I can switch numbers around is near 41. So the median remains the same at 56.
Below are the cases. Case 1 - 41,45,51,53,55,57,59,61,63,65 Case 2 - 41,47,49,53,55,57,59,61,63,65 Sufficient. Answer is C IMO.
My question I took around 3.05 minutes to solve this, and I feel it's way too long. Is there a shorter way to do this question? Is there something I missed i.e. a shortcut?
akshayk how do u say that mean remains at 56,,on what basis???
Case 1 - 41,45,51,53,55,57,59,61,63,65 Case 2 - 41,47,49,53,55,57,59,61,63,65
The middle 2 numbers in both cases are 55 and 57 (55 + 57)/2 = 112/2 = 56 So the median is 56. I couldn't come up with another possibility after combining the 2 statements.
_________________
I took the long road and evaluated the values. Is there an algebraic approach/ a more logical approach to this? It took me 4 minutes to do this! :(
_________________
Re: The average (arithmetic mean) of 10 distinct odd integers in a set is [#permalink]
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18 Aug 2017, 22:16
Sum = AVG*number of terms (n) Sum of the sequence =10(first term +last term)/2 using the first term/ last term we can find the median using arithmetic progression
Yes, this is a data sufficiency question. Options for DS questions are always the same.
The data sufficiency problem consists of a question and two statements, labeled (1) and (2), in which certain data are given. You have to decide whether the data given in the statements are sufficient for answering the question. Using the data given in the statements, plus your knowledge of mathematics and everyday facts (such as the number of days in July or the meaning of the word counterclockwise), you must indicate whether—
A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked. B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked. C. BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked. D. EACH statement ALONE is sufficient to answer the question asked. E. Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.
close
41% (01:42) correct 
