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# Problem Solving Pack 3, Question 3 A set of 11 positive integers...

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EMPOWERgmat Instructor
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Problem Solving Pack 3, Question 3 A set of 11 positive integers...  [#permalink]

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04 Nov 2015, 18:11
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55% (hard)

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62% (02:07) correct 38% (02:17) wrong based on 195 sessions

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QUANT 4-PACK SERIES Problem Solving Pack 3 Question 3 A set of 11 positive integers...

A set of 11 positive integers has an average of 25. Which of the following is the greatest possible value for the median of this set?

A) 25
B) 30
C) 36
D) 45
E) 46

48 Hour Window Answer & Explanation Window
OA, and explanation will be posted after the 48 hour window closes.

This question is part of the Quant 4-Pack series

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Re: Problem Solving Pack 3, Question 3 A set of 11 positive integers...  [#permalink]

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04 Nov 2015, 19:54
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EMPOWERgmatRichC wrote:
QUANT 4-PACK SERIES Problem Solving Pack 3 Question 3 A set of 11 positive integers...

A set of 11 positive integers has an average of 25. Which of the following is the greatest possible value for the median of this set?

A) 25
B) 30
C) 36
D) 45
E) 46

48 Hour Window Answer & Explanation Window
OA, and explanation will be posted after the 48 hour window closes.

This question is part of the Quant 4-Pack series

Scroll Down For Official Explanation

hello,
for the median to be max , we should take all the values below meadian to be the lowest and the remaining values to be the same..
here 1 is the lowest value(positive int) so the lowest 5 int will equal 5..
remaining 6= 25*11-5=270..
therefore each valueof these 6 int =270/6=45
ans D
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Re: Problem Solving Pack 3, Question 3 A set of 11 positive integers...  [#permalink]

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05 Nov 2015, 00:07
Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.

A set of 11 positive integers has an average of 25. Which of the following is the greatest possible value for the median of this set?

A) 25
B) 30
C) 36
D) 45
E) 46

In order to make the median maximum the 6th large number(=median number) should be as great as possible. That means the greater numbers than 6th number should be as small as possible. So all the numbers which are greater than 6th number should be equal to 6th number.

Moreover the numbers which are less than 6th number should be as small as possible and they should be all equal with similar reason. That is first 5 numbers should be 1.

So the sum of remaining 6 numbers is 11*25-5 = 270. That is remaining numbers should be (270/6=)45.

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Re: Problem Solving Pack 3, Question 3 A set of 11 positive integers...  [#permalink]

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07 Nov 2015, 14:57
EMPOWERgmatRichC wrote:
QUANT 4-PACK SERIES Problem Solving Pack 3 Question 3 A set of 11 positive integers...

A set of 11 positive integers has an average of 25. Which of the following is the greatest possible value for the median of this set?

A) 25
B) 30
C) 36
D) 45
E) 46

48 Hour Window Answer & Explanation Window
OA, and explanation will be posted after the 48 hour window closes.

This question is part of the Quant 4-Pack series

Scroll Down For Official Explanation

Hi All,

When dealing with Quant questions involving Statistical terms (mean, median, mode, range, standard deviation), it's important to understand the specific definitions of the terms involved.

We're told that a set of 11 POSITIVE integers has an average of 25. We're asked for the GREATEST possible MEDIAN of that set.

To find the median, we have to first put the numbers in order from least to greatest. One of the 'keys' to this question is that the prompt does NOT state that the 11 terms have to be distinct (re: different), so duplicates can be used. To maximize the median, we need to minimize ALL of the other numbers though...

For the first 5 numbers, the smallest we can make each of them is 1. Once we do that, we can make the median (and each of the 5 other numbers) all X...

1 1 1 1 1 X X X X X X

Since the average of the 11 numbers is 25, the SUM of those numbers is 11(25) = 275.

Subtracting the five 1's from the sum gives us 275 - 5(1) = 270. Thus, the remaining 6 values are all the SAME and sum to 270. The largest possible median would be....

270/6 = 45

GMAT assassins aren't born, they're made,
Rich
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Re: Problem Solving Pack 3, Question 3 A set of 11 positive integers...  [#permalink]

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19 Mar 2020, 04:56
MathRevolution wrote:
Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.

A set of 11 positive integers has an average of 25. Which of the following is the greatest possible value for the median of this set?

A) 25
B) 30
C) 36
D) 45
E) 46

In order to make the median maximum the 6th large number(=median number) should be as great as possible. That means the greater numbers than 6th number should be as small as possible. So all the numbers which are greater than 6th number should be equal to 6th number.

Moreover the numbers which are less than 6th number should be as small as possible and they should be all equal with similar reason. That is first 5 numbers should be 1.

So the sum of remaining 6 numbers is 11*25-5 = 270. That is remaining numbers should be (270/6=)45.

Don't you think that the question has an error ?
You have mentioned that 11 no.s are in a set. A no. never repeats in a set. Given that, we cannot take 2 nos to be same.
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Re: Problem Solving Pack 3, Question 3 A set of 11 positive integers...  [#permalink]

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19 Mar 2020, 15:27
1
Hi Sejal16,

Unless the question specifically states that there are no 'duplicate' numbers (for example, by describing the numbers as "distinct" or "different"), then a number can appear more than once in a set of numbers. Here, the 11 numbers have an average of 25, meaning that all 11 numbers COULD be 25s - meaning that the median would also be 25. The question asks for the GREATEST possible median though; to find that value, we'll need two 'groups' of duplicate values (one for the smallest possible number - in this case, "1" - and one for the largest).

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Rich
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Re: Problem Solving Pack 3, Question 3 A set of 11 positive integers...  [#permalink]

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02 Apr 2020, 02:43
We can solve this by another approach. ( Standard deviation approach)
If Median = 46. to get a avg 25, Right hand side numbers to the median shall be equal to 46
that is 21 more than the avg. And 6 numbers have 21 more than the avg so 21*6 = 126 more than avg
So below the median we need 5 numbers that will balance out this 126 i.e -126 additive.
126/6 > 25. So any numbers you take the average will be more than 25.

Same trial with 45 and you will give you the possibility.
IMO D
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Re: Problem Solving Pack 3, Question 3 A set of 11 positive integers...   [#permalink] 02 Apr 2020, 02:43