jain2016 wrote:During the 31-day month of May, a tuxedo shop rents a different number of tuxedos each day, including a store-record 55 tuxedos on May 23rd. Assuming that the shop had an unlimited inventory of tuxedos to rent, what is the maximum number of tuxedos the shop could have rented during May?

A) 1240

B) 1295

C) 1650

D) 1705

E) 1760

To maximize the sum for all 31 days, we must maximize the number of tuxedos rented per day.
A different number of tuxedos is rented each day, with the maximum number -- 55 -- rented on May 23.
Thus, the number of tuxedos rented per day must be a descending set of CONSECUTIVE INTEGERS, starting with 55:
55, 54, 53, 52...

If exactly 50 tuxedos are rented each today, the sum for all 31 days = 31*50 = 1550.
Since FEWER than 50 tuxedos are actually rented on most of the days, the sum for all 31 days must be LESS than 1550.
Eliminate C, D and E.

For any set of consecutive integers, median = (sum)/(number of integers).
Here, there are 31 consecutive integers, implying the following:
median = sum/31.
The median of an ODD number of consecutive integers will always be an INTEGER value.
Thus, the correct answer choice must be a multiple of 31.
Test A and B:
A: 1240/31 = 40.
B: 1295/31 = non-integer.

The correct answer is A.

Another trick here:

We have 31 days in the month, so our sum should be

(55 - 0) + (55 - 1) + (55 - 2) + ... + (55 - 30)

or

55*31 - (1 + 2 + ... + 30)

or

55*31 - (sum from 1 to 30)

Since the sum of the integers from 1 to n is n*(n+1)/2, we now have

55*31 - (30*31/2)

or

31 * (55 - (30/2))

or

31 * (55 - 15)

or

31 * 40

and we're set!