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

BTGModeratorVI wrote: ↑

Sun Feb 23, 2020 7:00 am

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

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

BTGModeratorVI wrote: ↑

Sun Feb 23, 2020 7:00 am

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

We are given that over the course of 31 days a shop can rent out an unlimited number of tuxedos, with a different number of tuxedos rented each day. We are also given that the rental of 55 tuxedos, which was a store record, occurred on May 23rd. We must determine the maximum number of tuxedos rented in May.

Since 55 was the maximum number of tuxedos rented on a single day and we need to determine the maximum number of tuxedos in the whole month, we want 54 to be the second highest number sold, 53 to be the third highest number sold, so on and so forth. We need to follow this pattern for a total of 31 days. To most easily determine the lowest number of tuxedos rented in a single day, we can use the quantity formula of consecutive numbers:

quantity = last number - first number + 1

Since we know that the total number of days is 31, we can use the following formula to determine the lowest number of tuxedos rented:

31 = 55 - (lowest number of tuxedos rented) + 1

lowest number of tuxedos rented = 25

Next we want to determine the average number of tuxedos rented. Since we have an evenly spaced set of integers we can use the formula:

(1st number + last number)/2 = avg

(25 + 55)/2 = 80/2 = 40

Lastly, we can calculate the maximum number of tuxedos rented using the formula:

sum = avg x quantity

sum = 40 x 31 = 1,240

Answer: A