Question

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

• A. $31$
• B. $33$
• C. $29$
• D. $27$

Answer 1

The correct option is A $31$

Dividing the numbers into sets of $9.$
Median is the middle number when $9$ numbers are arranged in ascending or descending order.
Median is the $5^{\text{th}}$ number and we will have $4$ numbers that are greater than the $5^{\text{th}}$ number and remaining $4$ numbers should be as small as possible because they are not contributing to increase median.
Largest possible median is $41$
So, possible set representation would be $\{1,2,3,4,41,42,43,44,45\},\{5,6,7,8,36,37,38,39,40\},\{9,10,11,12,31,32,33,34,35\},\{13,14,15,16,26,27,28,29,30\},\{17,18,19,20,21,22,23,24,25\}$
Corresponding median of above sets are $41,37,31,26,21$
Now, mean of these medians is
$\frac{21+26+31+36+41}{5}=\frac{155}{5}=31$