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One night a certain hotel rented 3/4 of its rooms. including
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28 Sep 2010, 00:01
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54% (02:20) correct 46% (02:29) wrong based on 1233 sessions
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One night a certain hotel rented 3/4 of its rooms. including 2/3 of their air conditioned rooms. if 3/5 of its rooms were air conditioned, what percent of the rooms that were not rented were air conditioned? A. 20 B. 33 1/3 C. 35 D. 40 E. 80
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Re: OG 10 Q 248  % rented
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28 Sep 2010, 00:09
vanidhar wrote: One night a certain hotel rented 3/4 of its rooms. including 2/3 of their air conditioned rooms. if 3/5 of its rooms were air conditioned, what percent of the rooms that were not rented were air conditioned?
20 33 1/3 35 40 80
please explain the logic Consider total # of rooms to be 100; As 3/5 of the rooms are air conditioned then # of rooms that are air conditioned is 3/5*100=60; 3/4 rooms were rented > 1/4*100=25 were NOT rented; 2/3 of air conditioned rooms were rented > 1/3*60=20 air conditioned room were NOT rented; 20/25=4/5=80%. Answer: E.
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Re: OG 10 Q 248  % rented
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28 Sep 2010, 20:56
The problem will be very easy to solve if you take a 2x2 matrix and work with the given data. Given fractions  2/3, 3/4, 3/5  L.C.M  60 [consider this to be the total number of rooms.]
Rooms   Rented  Available  Total  AirCon  24 (Given 2/3 of total aircon)  12  36 (Given  3/5 of total)  NonAircon  21  3  24  Total  45 (Given  3/4 of total)  15  60 (Total considered) 
Hence % of nonrented (available) that were aircon  12/15  4/5  80% (E).
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Re: OG 10 Q 248  % rented
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28 Sep 2010, 00:13
This can be easily solved using the matrix method as the % is asked i can choose a good # for total rooms..let me choose 60 as all the 3 fractions given have 3,4&5 in the denominator and 60 is divisible by all thease 3 #s.
Given: Total Rented=(60)3/4 = 45 Total AC rooms = (60)3/5 = 36 Total AC rooms rented = (36)2/3 = 24
Plug in: Rented Notrented   AC 24 12  36   Non AC  ............................................................... 45 15  60
qtn is what % is 12 in 15 = 1200/15 = 80%



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Re: OG 10 Q 248  % rented
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28 Sep 2010, 20:48
conditined un condition total
rented 24 21 45
not rent 12 3 15 total 36 24 60
total rooms 60 total rented 60*3/4 =45 total conditioned 60*3/5 =36 total conditioned 36*2/3 24 1200/15=80% (rooms that were not rented were air conditioned



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Re: rooms in hotel
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05 Mar 2011, 12:54
Rooms Rented: 3/4 Rooms Not Rented: 1/4 Total AC: 3/5 AC Rented: (2/3)*(3/5) = 2/5 AC Not rented: 3/5  2/5 = 1/5. % of AC that is yet to be rented per total of not rented room; ((1/5)/(1/4))*100 = (4/5)*100 = 80% of remaining(nonrented) rooms are AC. Ans: "E"
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Re: rooms in hotel
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05 Mar 2011, 12:59
Let's consider T as the total number of rooms. If 2/3 of the air conditioned rooms were rented, than 1/3 of them were not rented. We know that the total number of airconditioned rooms is 3/5, thus we can say that 1/5 of the total rooms in the hotel were airconditioned and not rented (1/3 x 3/5 = 1/5). (a)
We know that 3/4 of the total rooms were rented, thus 1/4 were not rented. (b)
Using (a) and (b), we can calculate the ratio of airconditioned rooms to the nonrented rooms: (1/5 x T) / (1/4 x T) = 4/5, and thus non airconditioned rooms make 80% of the total nonrented rooms.



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Re: OG 10 Q 248  % rented
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03 Nov 2011, 03:17
vanidhar wrote: One night a certain hotel rented 3/4 of its rooms. including 2/3 of their air conditioned rooms. if 3/5 of its rooms were air conditioned, what percent of the rooms that were not rented were air conditioned?
20 33 1/3 35 40 80
please explain the logic Can someone please explain where am i going wrong here? Let total rooms be 60 Rented rooms = 3/4 * 60 = 45 Rented AC rooms = 2/3 * 45 = 30 Total AC rooms = 3/5* 60 = 36 % Rooms that are AC and not rented = [(36  30) /36] * 100 = 16 * (2/3) I am not sure where is my mistake... Can someone please correct me ???



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Re: One night a certain hotel rented 3/4 of its rooms. including
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11 Aug 2014, 21:14
siddhans wrote: vanidhar wrote: One night a certain hotel rented 3/4 of its rooms. including 2/3 of their air conditioned rooms. if 3/5 of its rooms were air conditioned, what percent of the rooms that were not rented were air conditioned?
20 33 1/3 35 40 80
please explain the logic Can someone please explain where am i going wrong here? Let total rooms be 60 Rented rooms = 3/4 * 60 = 45 Rented AC rooms = 2/3 * 45 = 30 Total AC rooms = 3/5* 60 = 36 % Rooms that are AC and not rented = [(36  30) /36] * 100 = 16 * (2/3) I am not sure where is my mistake... Can someone please correct me ??? "including 2/3rd of their air conditioned rooms" implies that 2/3rd of their air conditioned rooms were a part of the rented rooms. So if total air conditioned rooms = 36 (as found by you above), 2/3rd of these i.e. 24 of these were rented. So 12 AC rooms were rented. Total number of rooms that were not rented = 60  45 = 15 So AC rooms form what percent of rooms not rented? 12/15 * 100 = 80%
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Re: One night a certain hotel rented 3/4 of its rooms. including
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12 Aug 2014, 00:59
Refer diagram below (Solved using matrix method) Going as per serial no (Marked in red) in the matrix 1. Let the total no of rooms = 100 2. 75% rented = 75 3. Not rented = 10075 = 254. \(\frac{3}{5}\) th of 100 are air conditioned \(= \frac{3}{5} * 100 = 60\) 5. \(\frac{2}{3}\) rd of air conditioned rooms are AC\(= \frac{2}{3} * 60 = 40\) 6. Not Rented AC rooms = 6040 = 20Percentage of Not rented AC rooms \(= \frac{20}{25}* 100 = 80%\) Answer = E
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One night a certain hotel rented 3/4 of its rooms. including
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Updated on: 28 Apr 2018, 06:56
vanidhar wrote: One night a certain hotel rented 3/4 of its rooms. including 2/3 of their air conditioned rooms. if 3/5 of its rooms were air conditioned, what percent of the rooms that were not rented were air conditioned?
A. 20 B. 33 1/3 C. 35 D. 40 E. 80 Here's a stepbystep approach using the Double Matrix method. Here, we have a population of motel rooms, and the two characteristics are:  air conditioning or no air conditioning  rented or not rented So, we can set up our matrix like this: Now since we're asked to find a certain PERCENTAGE, let's choose a nice value for the total number of motel rooms. Notice that the question includes the fractions 3/4, 2/3 and 3/5. So, let's choose a number that works well with all of these fractions. Since 60 is the least common denominator of 3/4, 2/3 and 3/5, let's say that there are 60 motel rooms altogether. Now that we're set up, we can use the given information to determine the number of rooms to place in each of the four boxes. If 3/5 of its rooms were airconditionedSo, 3/5 of the 60 rooms are air conditioned. 3/5 of 60 = 36, which means 36 rooms have AC and the remaining 24 rooms do not have AC... ...motel rented 3/4 of its rooms3/4 of 60 = 45, so 45 rooms are rented and the remaining 15 rooms are not rented. ...a certain motel rented ....2/3 of its airconditioned roomsSo of the 36 rooms with AC, 2/3 were rented. 2/3 of 36 = 24. So, 24 rooms had AC AND were rented. Now that we know the number of rooms in 1 box, and we know the sums of the rows and columns, we can fill in the remaining boxes. Question: what percent of the rooms that were not rented were airconditioned? There were 15 unrented rooms and 12 of them were air conditioned. 12/15 = 4/5 = 80% Answer: E Cheers, Brent
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Re: One night a certain hotel rented 3/4 of its rooms. including
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01 Dec 2016, 18:01
Say we have 60 rooms total: Rented = 45 Not Rented = 15
Further, we are told that 3/5 of its rooms have A/C A/C = 36 > 2/3 of this are included in the rented rooms (above) = 24. So we have 24 Rented A/C Rooms & 12 Rented NonA/C rooms No A/C = 24
Overall > There will be RENTED > 24 A/C & 12 no A/C > NOT RENTED > 12 A/C & 3 no A/C
12/15 = 80%
E.



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Re: One night a certain hotel rented 3/4 of its rooms. including
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19 Mar 2018, 16:09
vanidhar wrote: One night a certain hotel rented 3/4 of its rooms. including 2/3 of their air conditioned rooms. if 3/5 of its rooms were air conditioned, what percent of the rooms that were not rented were air conditioned?
A. 20 B. 33 1/3 C. 35 D. 40 E. 80 We can let the number of rooms = 60. (We choose this convenient number 60 because it is the least common multiple (LCM) of the denominators 3, 4, and 5.) We see that 3/4 x 60 = 45 rooms were rented and 60  45 = 15 rooms were not rented. We also see that 3/5 x 60 = 36 rooms are air conditioned, and 60  36 = 24 rooms are not air conditioned. Thus, 2/3 x 36 = 24 air conditioned rooms were rented, so 36  24 = 12 air conditioned rooms were not rented. Thus, the percent of rooms that were not rented were air conditioned is 12/15 = 4/5 = 80%. Answer: E
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Re: One night a certain hotel rented 3/4 of its rooms. including
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23 Apr 2018, 02:27
vanidhar wrote: One night a certain hotel rented 3/4 of its rooms. including 2/3 of their air conditioned rooms. if 3/5 of its rooms were air conditioned, what percent of the rooms that were not rented were air conditioned?
A. 20 B. 33 1/3 C. 35 D. 40 E. 80 Let the no. of rooms in the hotel be x. Total rooms rented = 3/4 x Total no. of airconditioned rooms = 3/5 x Total no. of airconditioned room that is rented = (2/3)(3/5 x) = 2/5 x Total no. of airconditioned room that were not rented = 3/5 x  2/5 x = 1/5 x Total no. of room that were not rented = 1/4 x Percentage of rooms that were not rented were air conditioned = (1/5 x)/(1/4 x) * 100 = 4/5 * 100 = 80% Answer E
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One night a certain hotel rented 3/4 of its rooms. including
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26 Apr 2018, 14:26
vanidhar wrote: One night a certain hotel rented 3/4 of its rooms. including 2/3 of their air conditioned rooms. if 3/5 of its rooms were air conditioned, what percent of the rooms that were not rented were air conditioned?
A. 20 B. 33 1/3 C. 35 D. 40 E. 80 let total number of rooms be \(x\) total number of rented rooms \(\frac{3}{4}\)x let total number of AC rooms be \(y\) AC rented rooms \(\frac{2}{3}\)y if total number of AC rooms is \(\frac{3}{5} *100 = 60\) then rented AC rooms is \(\frac{2}{3}*60= 40\) \(6040 = 20\) Hence 20 AC rooms were not rented (perhaps air conditioning was noisy ) now since \(\frac{1}{4} *100 = 25\) rooms were not rented then To find what percent of the rooms that were not rented were air conditioned we just divide number of AC not rented rooms by total number of not rented rooms \(\frac{20}{25}=0.8\) i made it !!!!!!!!!!!!!' pushpitkc, niks18 , generis is my approach correct ?



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One night a certain hotel rented 3/4 of its rooms. including
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26 Apr 2018, 15:08
dave13 wrote: vanidhar wrote: One night a certain hotel rented 3/4 of its rooms. including 2/3 of their air conditioned rooms. if 3/5 of its rooms were air conditioned, what percent of the rooms that were not rented were air conditioned?
A. 20 B. 33 1/3 C. 35 D. 40 E. 80 let total number of rooms be \(x\) total number of rented rooms \(\frac{3}{4}\)x let total number of AC rooms be \(y\) AC rented rooms \(\frac{2}{3}\)y if total number of AC rooms is \(\frac{3}{5} *100 = 60\) then rented AC rooms is \(\frac{2}{3}*60= 40\) \(6040 = 20\) Hence 20 AC rooms were not rented (perhaps air conditioning was noisy ) now since \(\frac{1}{4} *100 = 25\) rooms were not rented then To find what percent of the rooms that were not rented were air conditioned we just divide number of AC not rented rooms by total number of not rented rooms \(\frac{20}{25}=0.8\) i made it !!!!!!!!!!!!!' pushpitkc, niks18 , generis is my approach correct ? dave13 , well done! Spot on. Excellent logical reasoning. +1 Quote: (perhaps air conditioning was noisy ) Hilarious aside. P.S. I owe you answers . . . I have not forgotten . . . P.P.S. Here is how I checked your math . . . Categories, binary choices, fractions and percents? Fast check? Maxtrix! Attachment:
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Re: One night a certain hotel rented 3/4 of its rooms. including
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27 Apr 2018, 10:24
dave13 wrote: vanidhar wrote: One night a certain hotel rented 3/4 of its rooms. including 2/3 of their air conditioned rooms. if 3/5 of its rooms were air conditioned, what percent of the rooms that were not rented were air conditioned?
A. 20 B. 33 1/3 C. 35 D. 40 E. 80 let total number of rooms be \(x\) total number of rented rooms \(\frac{3}{4}\)x let total number of AC rooms be \(y\) AC rented rooms \(\frac{2}{3}\)y if total number of AC rooms is \(\frac{3}{5} *100 = 60\) then rented AC rooms is \(\frac{2}{3}*60= 40\) \(6040 = 20\) Hence 20 AC rooms were not rented (perhaps air conditioning was noisy ) now since \(\frac{1}{4} *100 = 25\) rooms were not rented then To find what percent of the rooms that were not rented were air conditioned we just divide number of AC not rented rooms by total number of not rented rooms \(\frac{20}{25}=0.8\) i made it !!!!!!!!!!!!!' pushpitkc, niks18 , generis is my approach correct ? Hi dave13generis, has already explained you the matrix method. Here's a small tip from my side  Whenever you see fractions, it's convinient to use SMART Numbers. Here Smart numbers are usually the LCM of denominators. So we have three fractions here 3/4, 2/3, & 3/5, hence the denominators are 4, 3, 5 whose LCM is 60. So let the total rooms be 60 So rooms rented \(= \frac{3}{4}*60=45\) Hence rooms not rented \(= 6045=15\) Rooms with AC \(=\frac{3}{5}*60=36\) AC rooms rented \(= \frac{2}{3}*36=24\) Hence AC rooms not rented \(= 3624=12\) Therefore % of rooms that were not rented and were air conditioned \(= \frac{12}{15}*100 = 80\)%



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