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

[BTGModeratorVI]

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

[ezmodest]

We want to max(# of tuxes sold in may), which translates to max(# of tuxes sold per day). Knowing that the most amount of tuxes sold was 55 and having the constraint that every day be a different amount leads to the following logic:

Least # of tuxes sold on any given day = max # of tuxes sold in a day - days in month + 1
25 = 55 - 30 +1

This gives us a set of consecutive integers from [25, 55] . Since the set is an odd number of integers we know that we can use the median to get the sum of the total set, therefore:

median = $$\frac{\left(55-25\right)}{2}$$ + 25 = 40
Sum of the set = 40 * 31 = 1240

[Brent@GMATPrepNow]

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

[quichopipe]

Remember that for consecutive integers, we can use the formula
\(\frac{sum}{#\ of\ terms}=\ average\)

From the problem, we know that the store sold a DIFFERENT number of tuxes each day and the max number of tuxes it sold was 55. So on the other days, it could have sold 54, 53, 52, 51, 50, etc tuxes.

There are 31 days in May so the 31 numbers that are less than 55 are: 25 through 55

Using the bookend method, we know that the average of consecutive integers is
\(\frac{\left(first\ number\ +\ last\ number\right)}{2}\)

So the average of 25 through 55 is
\(\frac{\left(55+25\right)}{2}\ =\ 40\)

So the sum of tuxes becomes rearranging our formula from the beginning:
sum = (# of terms)*(average)
sum = 31*40 = 1240

Answer choice A

[Scott@TargetTestPrep]

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

Solution:

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

[i256]

Number of days left in the month = 30 ( 1 day already chosen for highest = 55 tuxs)

Since each days sold different numbers, for maximizing this, we need numbers closest to 55

55 (max) - 30 (days left) = 25 = least sold in the month

Use sum of sequence formula to calculate --> \frac{n}{2}x\ \left(2a+\left(n\ -1\right)d\right)
n= total terms (31), a = initial term (25), d = difference b/w terms (1)

So, sum = 31 x 80 /2 = 1240