During the 31-day month of May, a tuxedo shop rents

BTGModeratorVI wrote: ↑

Fri Mar 06, 2020 5:54 pm

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

Answer: A
Source: Veritas Prep

Since 55 tuxedos is the maximum, in order to find out the maximum number of tuxedos the shop could have rented during May, we can assume that it rented 54, 53, 52, and onwards for the remaining 30 days.

Minimum no. of tuxedos = 55 – 31 + 1 = 25

Thus, the total no. of tuxedos = [(55 + 25)/2]*31 = 1,240.

The correct answer: A

Hope this helps!

-Jay
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BTGModeratorVI wrote: ↑

Fri Mar 06, 2020 5:54 pm

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

Answer: A
Source: Veritas Prep

We're told that the tuxedo shop rents a different number of tuxedos each day, including a store-record 55 tuxedos

To MAXIMIZE the number of tuxedos rented, we want the rentals for the other 30 days to be a big as possible (while still being DIFFERENT from the other rental numbers).
So, on one day, we can say 54 tuxedos were rented
On one day, 53 tuxedos were rented
On one day, 52 tuxedos were rented
etc

So, the maximum number of rentals = 55 + 54 + 53 + 52 + . . . . + 25

-------------------------------------------------
ASIDE: How do I know that the last number is 25?
We want a total of 31 consecutive numbers from 55 to x

A nice rule says: the number of integers from x to y inclusive equals y - x + 1

So, for this question, we want 55 - x + 1 = 31 (for the 31 days of May)

Solve to get: x = 25
-------------------------------------------------

One way to find the sum 55 + 54 + 53 + 52 + . . . . + 25 is to recognize that the AVERAGE value = (first + last)/2
= (55 + 25)/2
= 80/2
= 40

Since there are 31 numbers in total, the sum = (40)(31) = 1240

Answer: A

Cheers,
Brent