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by Brent@GMATPrepNow » Mon Oct 31, 2016 6:15 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
To MAXIMIZE the total sum of the 31 days, we must MAXIMIZE the number of tuxedos rented each day.
We know that, on May 23rd, 55 tuxedos were rented.

Since a DIFFERENT number of tuxedos were rented each day, the next highest value is 54 tuxedos rented, then 53 tuxedos rented, etc. This will MAXIMIZE the total sum of the 31 days,

Now we must find the sum of 55 + 54 + 53 + ...... + 25

Aside: How did I know that the sum goes down to 25?
Well, a nice rule says: the number of integers from x to y inclusive equals y - x + 1
We know that there are 31 integers altogether, and we know that the y-value in this formula is 55
We, we get: 31 = 55 - x + 1
Solve to get x = 25

There are many ways to find the sum of 55 + 54 + 53 + ...... + 25
A quick way is to add the sum TWICE in a special way.

55 + 54 + 53 + ...... + 26 + 25
25 + 26 + 27 + .......+ 54 + 55

Now add the values in pairs, taking one value in the TOP sum and adding it to the value BENEATH it.
Notice that each pair adds to 80.
So, we now must add the following: 80 + 80 + 80 + 80 .... + 80 + 80

How many 80's are there in this sum? 31, since there are 31 days in the month.
So, (31)(80) = 2480
Since we've added each value TWICE, we must divide 2480 by 2 to get 1240

Answer: A

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by GMATGuruNY » Mon Oct 31, 2016 7:23 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
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.
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by [email protected] » Mon Oct 31, 2016 7:39 am
Hi baalok88,

This prompt is essentially about addition, but it presents a number of "rules" that we have to follow, so that we can figure out what numbers to "add up."

We're told that we'll have 31 numbers AND that they're all DIFFERENT. The highest number is 55 (since it was a "store-record"). We're asked for the maximum total possible, which means that we'll have to choose the 31 "biggest" numbers, starting at 55 and "working down."

55, 54, 53, 52.....

The first task is to figure out what would be the "smallest" number of this group. Since there are 55 positive integers from 1 to 55 (inclusive) and we want the 31 biggest numbers, we need to remove the 24 smallest. So, remove the numbers 1 to 24 (inclusive) and you're left with 25 to 55, inclusive.

Now, we have to add up the numbers from 25 to 55, inclusive. You can approach this process in a couple of different ways.

One way is to take the average of the largest and smallest, then multiply this value by the total number of terms:

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

40 x 31 terms = 1240

Final Answer: A

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by Scott@TargetTestPrep » Tue Nov 01, 2016 5:15 am
baalok88 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
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

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