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

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21 Sep 2015, 03:33

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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

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

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

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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

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

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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

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

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16 Jun 2016, 10:51

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

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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)

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

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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.

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

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

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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 ?

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

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