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Math Expert V
Joined: 02 Sep 2009
Posts: 59722
Positive integers from 1 to 45, inclusive are placed in 5 groups of 9  [#permalink]

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19 00:00

Difficulty:   95% (hard)

Question Stats: 42% (02:48) correct 58% (02:28) wrong based on 221 sessions

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Positive integers from 1 to 45, inclusive are placed in 5 groups of 9 each. What is the highest possible average of the medians of these 5 groups?

A. 15
B. 23
C. 25
D. 26
E. 31

_________________
Director  D
Joined: 05 Mar 2015
Posts: 978
Re: Positive integers from 1 to 45, inclusive are placed in 5 groups of 9  [#permalink]

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5
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Bunuel wrote:
Positive integers from 1 to 45, inclusive are placed in 5 groups of 9 each. What is the highest possible average of the medians of these 5 groups?

A. 15
B. 23
C. 25
D. 26
E. 31

for getting highest average of medians we have to maximize medians
also the median is 5th term of every set(since set is 0f 9 integers)
let the last 5 terms of first set is (1st, 2nd,3rd,4th , 41,42,43,44,45)

similarly 2nd set (1st, 2nd,3rd,4th ,36,37,38,39,40)

3rd set (1st, 2nd,3rd,4th ,31,32,33,34,35)

4th set (1st, 2nd,3rd,4th ,26,27,28,29,30)

5th set (1st, 2nd,3rd,4th ,21,22,23,24,25)

sum of all medians = 41+36+31+26+21= 155
avg = 155/5 =31
Ans E
##### General Discussion
Manager  Joined: 25 Dec 2012
Posts: 116
Positive integers from 1 to 45, inclusive are placed in 5 groups of 9  [#permalink]

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1
Bunuel wrote:
Positive integers from 1 to 45, inclusive are placed in 5 groups of 9 each. What is the highest possible average of the medians of these 5 groups?

A. 15
B. 23
C. 25
D. 26
E. 31

To have the highest possible median in all 5 groups,

1,2,3,4,_,42,43,44,45.

From the above one we can deduce that,

45 - 4 = 41
40 - 4 = 36 like this it will go on in a sequence.

Since the above sequence forms a AP, we can say that 45-24 = 21

Range of the median 21 to 41, 21 being the least and 41 being the max.

(21+41)/2 = 62/2 = 31

IMO.E
Director  P
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Posts: 591
Location: India
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Re: Positive integers from 1 to 45, inclusive are placed in 5 groups of 9  [#permalink]

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1
Bunuel wrote:
Positive integers from 1 to 45, inclusive are placed in 5 groups of 9 each. What is the highest possible average of the medians of these 5 groups?

A. 15
B. 23
C. 25
D. 26
E. 31

Form sets --take first 4 number (1,2,3,4) and last 5 (41,42,43,44,45)..Repeat the process ...
values of median we will get --41,36 ....so on ..
the sum of median will come 155.
Avg =31...
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Re: Positive integers from 1 to 45, inclusive are placed in 5 groups of 9  [#permalink]

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Highest possible will depend upon how we distribute the integers in a group to maximize the median value in each group.
total integers in a group are 9
so,
take least 4 and max 4 and take the 5th value be the highest available.

1,2,3,4, 41 ,42,43,44,45 Median =41
5,6,7,8, 36 ,37,38,39,40 M= 36
9,10,11,12, 31 , 32,33,34,35 M=31
Note that the difference is 5,so next M = 26 And the fifth M=21
41+36+31+26+21= 155
average = 155/5 =31
E
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Re: Positive integers from 1 to 45, inclusive are placed in 5 groups of 9  [#permalink]

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1
can someone please assist me with this question
Since the question is stating that there are 5 groups of 9 terms each the 5th value of the group is the median.
therefore the actual median for 5 groups are G1- 5
G2 -14
G3=23
G4=32
G5=41
these are the actual median and the actual average of these median is 23
Now from here what is the next possible step ??Can someone Assist
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Re: Positive integers from 1 to 45, inclusive are placed in 5 groups of 9  [#permalink]

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longhaul123 wrote:
can someone please assist me with this question
Since the question is stating that there are 5 groups of 9 terms each the 5th value of the group is the median.
therefore the actual median for 5 groups are G1- 5
G2 -14
G3=23
G4=32
G5=41
these are the actual median and the actual average of these median is 23
Now from here what is the next possible step ??Can someone Assist

hello pal,
The question asks you the maximum possible value. Surely the answer you arrived at is a possible avg of the medians of the gr,oup. But it is not the maximum avg.
Whenever you are asked to maximize something you have to minimize other elements as much as possible.
Median of a set is derived by arranging the set in ascending order and then if the number of elements in the set is odd then the middle term is the median. Here as each group is divided in 9 elements therefore the median is the 5th element. So far so good.

Now we have to maximize. >> we have all distinct numbers. Note that itis not mandatory that we choose the element sequentially. We just have to arrange the elements of the set in ascending order.

So to maximize start from the greatest number. As we have to arrange in ascending order , the largest will come at the last >> (...,41,42,43,44,45) now AS WE HAVE TO MAXIMIZE DO NOT WASTE THE LARGER NUMBERS. USE THE SMALLER NUMBERS BEFORE THE MEDIAN>> ( 1,2,3,4,41,42,43,44,45). This is 1st set with median 41.

Now remember th,at elements can be used only once>> we have used up the above set so start again from the unused largest ,number>> (...,36,37,38,39,40) and now choose the lowest of the remaining >> (5,6,7,8,36,37,38,39,40)>> median 36.

similarly you get M3=31 M4=26 M5=21.

avg= (41+36+31+26+21)/5 = 31 E
Math Expert V
Joined: 02 Aug 2009
Posts: 8309
Re: Positive integers from 1 to 45, inclusive are placed in 5 groups of 9  [#permalink]

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Bunuel wrote:
Positive integers from 1 to 45, inclusive are placed in 5 groups of 9 each. What is the highest possible average of the medians of these 5 groups?

A. 15
B. 23
C. 25
D. 26
E. 31

There are 5 groups and a median of each of these 5 groups.
Now let them be $$M_1<M_2<M_3<M_4<M_5$$, so we are looking for the MAX possible average of all 5..

So, we have to maximize all the Ms. How do we ensure?

a) The group containing $$M_5$$ will have FOUR numbers more and FOUR numbers less than $$M_2$$, so 4 numbers are greater than $$M_5=45-4=41$$.

b) The group containing $$M_4$$ will have FOUR numbers greater than $$M_4$$, and largest of this group will be <$$M_5$$so at least 4+(4+1) or 9 numbers are greater than $$M_4=45-9=36$$.
Similarly $$M_3=45-(4+5+5)=31...M_2=31-5=26...M_1=26-5=21$$

Medians are 21, 26, 31, 36 and 41, so an AP with a difference of 5 each. Thus the middle number will be the average = 31

E
_________________ Re: Positive integers from 1 to 45, inclusive are placed in 5 groups of 9   [#permalink] 29 Nov 2019, 06:44
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