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
