November 22, 2018 November 22, 2018 10:00 PM PST 11:00 PM PST Mark your calendars  All GMAT Club Tests are free and open November 22nd to celebrate Thanksgiving Day! Access will be available from 0:01 AM to 11:59 PM, Pacific Time (USA) November 24, 2018 November 24, 2018 07:00 AM PST 09:00 AM PST Attend this webinar to learn how to leverage Meaning and Logic to solve the most challenging Sentence Correction Questions.
Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 01 Oct 2014
Posts: 41

During the 31day month of May, a tuxedo shop rents a different number
[#permalink]
Show Tags
21 Sep 2015, 02:33
Question Stats:
60% (01:28) correct 40% (01:01) wrong based on 420 sessions
HideShow timer Statistics
During the 31day month of May, a tuxedo shop rents a different number of tuxedos each day, including a storerecord 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
Official Answer and Stats are available only to registered users. Register/ Login.




Veritas Prep GMAT Instructor
Joined: 01 Jul 2017
Posts: 60
Location: United States
Concentration: Leadership, Organizational Behavior

During the 31day month of May, a tuxedo shop rents a different number
[#permalink]
Show Tags
Updated on: 10 Sep 2018, 13:09
One quick way to solve this problem is to leverage the concept of average for evenlyspaced 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 55tuxedo 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 55max day. The "smallest" number (while still maximizing the total) would be \(5530 = 25\). 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 31day month, the "# of terms" is easily 31. Thus, \(Sum = Avg * (31)\) Now, for evenlyspaced 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 timesaver. (I call this " Easy Math + Chunk" in my classes.) \(Sum = 40*(30 + 1) = 1200 + 40 = 1240\) The answer is A.
_________________
Aaron J. Pond Veritas Prep EliteLevel Instructor
Hit "+1 Kudos" if my post helped you understand the GMAT better. Look me up at https://www.veritasprep.com/gmat/aaronpond/ if you want to learn more GMAT Jujitsu.
Originally posted by AaronPond on 06 Jan 2018, 17:08.
Last edited by AaronPond on 10 Sep 2018, 13:09, edited 1 time in total.




Manager
Joined: 29 Jul 2015
Posts: 158

Re: During the 31day month of May, a tuxedo shop rents a different number
[#permalink]
Show Tags
21 Sep 2015, 03:26
amatya wrote: During the 31day month of May, a tuxedo shop rents a different number of tuxedos each day, including a storerecord 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 (5531)+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




Senior Manager
Joined: 15 Oct 2015
Posts: 322
Concentration: Finance, Strategy
GPA: 3.93
WE: Account Management (Education)

Re: During the 31day month of May, a tuxedo shop rents a different number
[#permalink]
Show Tags
16 Feb 2016, 00:04
amatya wrote: During the 31day month of May, a tuxedo shop rents a different number of tuxedos each day, including a storerecord 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 give some kudos



Director
Joined: 20 Feb 2015
Posts: 796
Concentration: Strategy, General Management

Re: During the 31day month of May, a tuxedo shop rents a different number
[#permalink]
Show Tags
17 Jun 2016, 02:01
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



Senior Manager
Joined: 23 Apr 2015
Posts: 309
Location: United States
Concentration: General Management, International Business
WE: Engineering (Consulting)

Re: During the 31day month of May, a tuxedo shop rents a different number
[#permalink]
Show Tags
23 Jun 2016, 15:01
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.



Manager
Joined: 07 Jun 2017
Posts: 103

Re: During the 31day month of May, a tuxedo shop rents a different number
[#permalink]
Show Tags
22 Aug 2017, 11:01
can anyone tell me which word states that "55" is the maximum?



Intern
Joined: 20 Sep 2015
Posts: 20

Re: During the 31day month of May, a tuxedo shop rents a different number
[#permalink]
Show Tags
06 Sep 2017, 00:06
pclawong wrote: can anyone tell me which word states that "55" is the maximum? "a storerecord 55 tuxedos on May 23rd."



Director
Joined: 13 Mar 2017
Posts: 632
Location: India
Concentration: General Management, Entrepreneurship
GPA: 3.8
WE: Engineering (Energy and Utilities)

Re: During the 31day month of May, a tuxedo shop rents a different number
[#permalink]
Show Tags
06 Sep 2017, 02:01
amatya wrote: During the 31day month of May, a tuxedo shop rents a different number of tuxedos each day, including a storerecord 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 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
_________________
CAT 2017 99th percentiler : VA 97.27  DILR 96.84  QA 98.04  OA 98.95 UPSC Aspirants : Get my app UPSC Important News Reader from Play store.
MBA Social Network : WebMaggu
Appreciate by Clicking +1 Kudos ( Lets be more generous friends.) What I believe is : "Nothing is Impossible, Even Impossible says I'm Possible" : "Stay Hungry, Stay Foolish".



Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 4170
Location: United States (CA)

Re: During the 31day month of May, a tuxedo shop rents a different number
[#permalink]
Show Tags
05 Dec 2017, 18:40
amatya wrote: During the 31day month of May, a tuxedo shop rents a different number of tuxedos each day, including a storerecord 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
_________________
Scott WoodburyStewart
Founder and CEO
GMAT Quant SelfStudy Course
500+ lessons 3000+ practice problems 800+ HD solutions



Manager
Joined: 23 Oct 2017
Posts: 64

Re: During the 31day month of May, a tuxedo shop rents a different number
[#permalink]
Show Tags
07 Jan 2018, 06:14
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



EMPOWERgmat Instructor
Status: GMAT Assassin/CoFounder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 12895
Location: United States (CA)

Re: During the 31day month of May, a tuxedo shop rents a different number
[#permalink]
Show Tags
23 Jan 2018, 13:50
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 "storerecord"). 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: GMAT assassins aren't born, they're made, Rich
_________________
760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com
Rich Cohen
CoFounder & GMAT Assassin
Special Offer: Save $75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/
*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****



Intern
Joined: 30 Oct 2017
Posts: 1

Re: During the 31day month of May, a tuxedo shop rents a different number
[#permalink]
Show Tags
06 Jun 2018, 00:04
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!



EMPOWERgmat Instructor
Status: GMAT Assassin/CoFounder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 12895
Location: United States (CA)

Re: During the 31day month of May, a tuxedo shop rents a different number
[#permalink]
Show Tags
06 Jun 2018, 17:53
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. GMAT assassins aren't born, they're made, Rich
_________________
760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com
Rich Cohen
CoFounder & GMAT Assassin
Special Offer: Save $75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/
*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****



Intern
Joined: 05 Jan 2017
Posts: 40

Re: During the 31day month of May, a tuxedo shop rents a different number
[#permalink]
Show Tags
18 Oct 2018, 01:40
Hi BunuelCan you please help me with few more questions of similar type. Thanks.
_________________
 Kudos if my post helps!




Re: During the 31day month of May, a tuxedo shop rents a different number &nbs
[#permalink]
18 Oct 2018, 01:40






