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Integers from 1 to 25 (both inclusive) are split into 5 groups of 5 nu 15 Apr 2020, 04:24
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Integers from 1 to 25 (both inclusive) are split into 5 groups of 5 numbers each. The medians of these 5 groups are A, B, C, D and E. If the average of these medians is m, what is the largest value m can take?

A. 16
B. 17
C. 18
D. 19
E. 20


Are You Up For the Challenge: 700 Level Questions­
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B
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Schools: IIM (A) ISB '24
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Integers from 1 to 25 (both inclusive) are split into 5 groups of 5 nu 15 Apr 2020, 04:35
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Bunuel
Integers from 1 to 25 (both inclusive) are split into 5 groups of 5 numbers each. The medians of these 5 groups are A, B, C, D and E. If the average of these medians is m, what is the largest value m can take?

A. 16
B. 17
C. 18
D. 19
E. 20


Are You Up For the Challenge: 700 Level Questions
25 terms to be split among 5 sets

i.e. terms in each set = 25/5 = 5

Strategy:
- Median to be Largest which is third term of each set
- i.e. Last three terms of each set must be largest possible (let's set aside 3*5 = 15 terms from 11 to 25)



First set {1, 2, 11, 12, 13} Median = 11
Second set {3, 4, 14, 15, 16} Median = 14
Third set {5, 6, 17, 18, 19} Median = 17
Forth set {7, 8, 20, 21, 22} Median = 20
Fifth set {9, 10, 23, 24, 25} Median = 23

Average of Medians = median of Medians = 17

Answer: Option B­
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Integers from 1 to 25 (both inclusive) are split into 5 groups of 5 nu 15 Apr 2020, 04:46
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Bunuel
Integers from 1 to 25 (both inclusive) are split into 5 groups of 5 numbers each. The medians of these 5 groups are A, B, C, D and E. If the average of these medians is m, what is the largest value m can take?

A. 16
B. 17
C. 18
D. 19
E. 20


So, to make m the maximum, we have to make A, B, C, D and E the maximum....
Let the value of median be a<B<C<D<E
Let us therefore, take rest of the numbers other than the medians and two numbers greater than the medians, smaller than A.

We can. therefore distribute numbers as..
25, 24, 23(E), 22, 21, 20(D), 19, 18, 17(C), 16, 15, 14(B), 13, 12, 11(A), 10....
so groups could be
1) 25, 24, 23, 2, 1
2) 22, 21, 20(D), 4, 3
3) 19, 18, 17(C), 6, 5
4) 16, 15, 14(B), 8, 7
5) 13, 12, 11(A), 10, 9

Average of medians - 23, 20, 17, 14, 11 is 17

B
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Integers from 1 to 25 (both inclusive) are split into 5 groups of 5 nu 16 Apr 2020, 00:21
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Each group contains five numbers, so the median is the 3rd position
Let smallest 10 of the numbers (1-10) be the 2 smallest of each group, you can arrange them whatever you want.

Next put 3 largest consecutive numbers in a group, so we have
#,#,11,12,13
#,#,14,15,16
#,#,17,18,19
#,#,20,21,22
#,#,23,24,25

Those bold are the medians. And the average of medians is 17
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Integers from 1 to 25 (both inclusive) are split into 5 groups of 5 nu 22 Nov 2020, 22:23
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Bunuel
Integers from 1 to 25 (both inclusive) are split into 5 groups of 5 numbers each. The medians of these 5 groups are A, B, C, D and E. If the average of these medians is m, what is the largest value m can take?

A. 16
B. 17
C. 18
D. 19
E. 20

Are You Up For the Challenge: 700 Level Questions
25 terms to be split among 5 sets

i.e. terms in each set = 25/5 = 5

Strategy:
- Median to be Largest which is third term of each set
- i.e. Last three terms of each set must be largest possible (let's set aside 3*5 = 15 terms from 11 to 25)



First set {1, 2, 11, 12, 13} Median = 11
Second set {3, 4, 14, 15, 16} Median = 14
Third set {5, 6, 17, 18, 19} Median = 17
Forth set {7, 8, 20, 21, 22} Median = 20
Fifth set {9, 10, 23, 24, 25} Median = 23

Average of Medians = median of Medians = 17

Answer: Option B
can this be done under 2 mins? any shortcut?­
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Integers from 1 to 25 (both inclusive) are split into 5 groups of 5 nu 23 Nov 2020, 00:43
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MEDIAN is the middle value of a group of numbers.
MEAN/AVERAGE is the sum of a group of numbers divided by the number of numbers.
The intent is to MAXIMIZE the median of each set by arranging the numbers, so that they have the Maximum average.

For example, the highest possible median is 23 with the numbers 21,22,(23),a1,a2 [FIRST SET]
Now, there needs to be 3 numbers in between the highest median & the next possible one.
(2 numbers above the median and the median itself making 3 nos.)

Hence we have 23-3=20 as our next median for 18,19,(20),a3,a4 [SECOND SET]
Similarly 20-3=17 as the next for 15,16,(17),a5,a6 [THIRD]
14 for 12,13,(14),a7,a8 [FOURTH] and finally
9,10,(11),a9,a10 [FIFTH]
Subscripts on "a" are interchangeable.

Hence, average of these medians is 23+20+17+14+11/5 = 17 (B)

Hope this helps. Happy studying all :)
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Integers from 1 to 25 (both inclusive) are split into 5 groups of 5 nu 27 Feb 2025, 21:07
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