Integers from 1 to 25 (both inclusive) are split into 5 groups of 5

Question

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 smallest values m can take?

A. 7
B. 8
C. 9
D. 10
E. 11

A. 7
B. 8
C. 9
D. 10
E. 11

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GMATinsight · Accepted Answer

25 terms to be split among 5 sets

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

Strategy: 
- Median to be lowest which is third term of each set
- So let's minimize the third term of each set by taking first three terms lowest

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

Average of Medians = median of Medians = 9

Answer: Option C

ScottTargetTestPrep · Answer

Since each of the 5 groups contains numbers different from one another, the medians of these 5 groups are different as well. Therefore, we can let A < B < C < D < E. Since we want the smallest value of m, we want A, B, C, D, and E to be as small as possible, too. Let’s start with A. The smallest value of A is 3 (when 3 is in the same group along with 1 and 2). Then the smallest value of B is 6 (when 6 is in the same group along with 4 and 5). Then the smallest value of C is 9 (when 9 is in the same group along with 7 and 8). Then the smallest value of D is 12 (when 12 is in the same group along with 10 and 11). Finally, the smallest value of E is 15 (when 15 is in the same group along with 13 and 14). Therefore, we have m = (3 + 6 + 9 + 12 + 15)/5 = 9.

Answer: C

Kinshook · Answer

Given: 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.

Asked: If the average of these medians is m, what is the smallest values m can take?

5 Groups formed with least medians: -
{1,2,3,24,25}
{4,5,6,22,23}
{7,8,9,20,21}
{10,11,12,18,19}
{13,14,15,16,17}

Medians = {3,6,9,12,15}

Average of medians = 9

IMO C

Nipunh1991 · Answer

To keep the median of the 5 groups as low as possible we need the first 3 digits of each group to be the lowest since the third will be the median in each case

Hence

1,2, 3 ,24,25

4,5, 6 ,23,22

7,8, 9 ,20,21

10,11, 12 ,18,19

13,14, 15 ,16,17

Average of the medians = (3+6+9+12+15) / 5 = 9

Hope this helps a little

davidbeckham · Answer

Bunuel , the question does not say that the numbers in each set have to be different - why can't we take the same numbers for all 5 sets or why can't we take numbers in the following form:

A - 1, 2, 3, 4, 5
B- 2, 3, 4, 5, 6
C - ..
D - ..

MBAB123 · Answer

Bunuel , the question does not say that the numbers in each set have to be different - why can't we take the same numbers for all 5 sets or why can't we take numbers in the following form:

A - 1, 2, 3, 4, 5
B- 2, 3, 4, 5, 6
C - .. 
D - ..

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 smallest values m can take?

A. 7
B. 8
C. 9
D. 10
E. 11

A. 7
B. 8
C. 9
D. 10
E. 11

Hey, you can't repeat the integers since the question is splitting is the first 25 integers into 5 groups. If "2" is going into group 1, it cannot go into group 2 again.

HamP95 · Answer

Key here is understanding that to get the median average as low as possible, we need to use the lowest possible numbers in the for the first 3 numbers of each set to reach the 5 medians and make them as low as we can, like so:

Set 1: 1 2 3
Set 2: 4 5 6
Set 3: 7 8 9
Set 4 : 10 11 12
Set 5: 13 14 15

Medians in these sets are 3, 6, 9, 12, 15, average of 9 - B.

Set 1: 1 2 3
Set 2: 4 5 6
Set 3: 7 8 9
Set 4 : 10 11 12
Set 5: 13 14 15

Medians in these sets are 3, 6, 9, 12, 15, average of 9 - B.

steffe440 · Answer

We’re dividing the integers 1–25 into 5 groups of 5 numbers each.
Each group’s median is its 3rd smallest number, and we want the smallest possible average of those 5 medians.

To make a group’s median as small as possible, we should use the three smallest available numbers for that group, since the median depends only on the first three values. The remaining two numbers in the group can be large—they don’t affect the median.

So the optimal grouping is:

Group 1: 1, 2, 3, 24, 25 → median = 3
Group 2: 4, 5, 6, 22, 23 → median = 6
Group 3: 7, 8, 9, 20, 21 → median = 9
Group 4: 10, 11, 12, 18, 19 → median = 12
Group 5: 13, 14, 15, 16, 17 → median = 15

Average of medians = (3 + 6 + 9 + 12 + 15) / 5 = 45 / 5 = 9

Therefore, the smallest possible value of m is 9.

Integers from 1 to 25 (both inclusive) are split into 5 groups of 5

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Integers from 1 to 25 (both inclusive) are split into 5 groups of 5

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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