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

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During the 31-day month of May, a tuxedo shop rents a different number  [#permalink]

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New post 21 Sep 2015, 03:33
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A
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E

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Question Stats:

60% (01:31) correct 40% (01:01) wrong based on 362 sessions

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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
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Re: During the 31-day month of May, a tuxedo shop rents a different number  [#permalink]

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New post 06 Jan 2018, 18:08
One quick way to solve this problem is to leverage the concept of average for evenly-spaced sets. Watch this "GMAT Jujitsu"...

The question gives us three handholds we can use to take it down quickly:
(1) The tuxedo shop rented "a different number of tuxedos each day".
(2) The highest number was 55.
(3) Our target is the "maximum number" for the entire month.

The maximum number starts with a 55-tuxedo day, but since we must have "a different number of tuxedos" each day, we would need to rent one less tuxedo per day for 30 days after the 55-max day. The "smallest" number (while still maximizing the total) would be \(55-30 = 25\).

Remember that the Sum of a set of numbers is equal to the number of terms in the set, multiplied by the set's average. For a 31-day month, the "# of terms" is easily 31. Thus,

\(Sum = Avg * (31)\)

Now, for evenly-spaced sets, the average of a set is equal to the average of the largest and smallest numbers in the set. For the month of May, the maximum average would be equal to:

\(\frac{55+25}{2} = \frac{80}{2} = 40\)

Plugging this into our equation gives us: \(Sum = 40 * 31\)

Now, before you multiply this out longhand, break the "31" into smaller components for one, final time-saver. (I call this "Easy Math + Chunk" in my classes.)

\(Sum = 40*(30 + 1) = 1200 + 40 = 1240\)

The answer is A.
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Re: During the 31-day month of May, a tuxedo shop rents a different number  [#permalink]

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New post 21 Sep 2015, 04:26
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amatya 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

I was unable to solve this question can anyone please guide how to solve it ?


Press +1 if you like the post


There are 31 days in may. On each day shop rents different number of tuxedo.
Store record is of 55 rented tuxedo on a day i.e Maximum number of tuxedos rented on any day in may was 55.
To maximize the sum of tuxedos rented, we must maximize the number of tuxedos rented per day.
This can be done by taking consecutive numbers starting with 55 and decreasing in number by 1 for each day. (55,54,53,52...)
Now,
the lowest number of tuxedos sold on any day will be (55-31)+1 = 25
So the sum of tuxedos will be
\(Average*Total Number Of Terms\) or \(Total Number Of Terms*(first+last)/2\)
= 31*(25+55)/2
= 31*40
=1240

Answer:- A
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Re: During the 31-day month of May, a tuxedo shop rents a different number  [#permalink]

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New post 16 Feb 2016, 01:04
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amatya 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

I was unable to solve this question can anyone please guide how to solve it ?


Press +1 if you like the post



I failed it too the first time, but i later realized how i should have analysed the question better.


For a 31 day period there are 31 different numbers to add up: the sum is the solution to the problem.
this is because it asks you the MOST(maximum) amount (not just any amount) and no amount is repeated.
you biggest number is already given- 55.
so you count down from 55 and get the other 30 integers. long process.
for short is 55 - 30 equals 25.
Sum of consecutive integers = the middle number(median) \(*\) N(number of digits).
middle number from 25 thru 55 is 40 (say, 25, 30, 35, 40, 45, 50, 55)
Sum = 40 * 31 (31 is number of days)
1240 - answer

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Re: During the 31-day month of May, a tuxedo shop rents a different number  [#permalink]

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New post 17 Jun 2016, 03:01
another approach !!

since the number tuxedos has to be different on each day.

max tuxedos = maximum (55) + (multiply 54 with the remaining no of days) - ( subtract the difference (Arithmetic progression) from the 30 days of 54 tuxedos)

max tuxedos = 55 + (54*30) - 29(30)/2 = 1675 - 435 = 1240
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Re: During the 31-day month of May, a tuxedo shop rents a different number  [#permalink]

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New post 23 Jun 2016, 16:01
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Since the maximum is 55 and everyday is distinct , just list(not literally) for the 31 days, 55 + 54 + ..... 25 (for 31 days) . Use the formula for Sum of series (AP) and it will be 1240.
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Re: During the 31-day month of May, a tuxedo shop rents a different number  [#permalink]

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New post 22 Aug 2017, 12:01
can anyone tell me which word states that "55" is the maximum?
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Re: During the 31-day month of May, a tuxedo shop rents a different number  [#permalink]

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New post 06 Sep 2017, 01:06
pclawong wrote:
can anyone tell me which word states that "55" is the maximum?


"a store-record 55 tuxedos on May 23rd."
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Re: During the 31-day month of May, a tuxedo shop rents a different number  [#permalink]

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New post 06 Sep 2017, 03:01
amatya 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

I was unable to solve this question can anyone please guide how to solve it ?


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Sure I would be happy to help ..
Different number of tuxedos was rented each day.
Max no. rented is 55
Also we have to take max value for each other days to maximize the overall value.

So, other days can have 54, 53, 52,.......,25 tuxedos rented.
So, we need calculate a total of 55+54+53+52+.... +26+25 which is sum A.P , 1st term = 25 c.d. = 1 and last term is 55
Sum = n/2*(first term + last term ) = 31/2 * (25+55) = 31*40 = 1240

Answer A
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Re: During the 31-day month of May, a tuxedo shop rents a different number  [#permalink]

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New post 05 Dec 2017, 19:40
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amatya 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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Re: During the 31-day month of May, a tuxedo shop rents a different number  [#permalink]

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New post 07 Jan 2018, 07:14
31 days in the month of May
23rd day had record sales of 55 => indicating other sales would be less than 55
To calculate maximum, 2nd highest would be 54.

Hence the sum of series, 55, 54 .... 25 => 25 * 31 + [0+...30]= 1240
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Re: During the 31-day month of May, a tuxedo shop rents a different number  [#permalink]

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New post 23 Jan 2018, 14:50
Hi All,

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

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Re: During the 31-day month of May, a tuxedo shop rents a different number  [#permalink]

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New post 06 Jun 2018, 01:04
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I used POE after a close investigation of the answer choices. The answer cannot be 1705 (ans choice D) because that is what we will get if the store sells 55 tuxedo's everyday. Eliminate E as it is greater than 1705. Between A,B and C, only A provides an integer when divided by 31, and the store obviously cannot sell a fraction of a tux. Therefore I picked A. Let me know if this approach is valid. Thanks!
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Re: During the 31-day month of May, a tuxedo shop rents a different number  [#permalink]

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New post 06 Jun 2018, 18:53
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Hi yashkanoongo,

Your approach does work here, but only because we know that we're dealing with 31 CONSECUTIVE INTEGERS - so we know that the average will be an integer (the median number - in this case, the 16th number in the sequence) and the total MUST be a multiple of 31. If the numbers were not necessarily consecutive and/or we had any 'duplicate' numbers, then that approach might not have led to the correct answer.

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Re: During the 31-day month of May, a tuxedo shop rents a different number &nbs [#permalink] 06 Jun 2018, 18:53
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