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One night a certain hotel rented 3/4 of its rooms. including [#permalink]

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28 Sep 2010, 00:01

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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?

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;

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 Not-rented | | AC 24 12 | 36 | | Non AC | ............................................................... 45 15 | 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

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 air-conditioned rooms is 3/5, thus we can say that 1/5 of the total rooms in the hotel were air-conditioned 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 air-conditioned rooms to the non-rented rooms: (1/5 x T) / (1/4 x T) = 4/5, and thus non air-conditioned rooms make 80% of the total non-rented rooms.

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 ???

Re: One night a certain hotel rented 3/4 of its rooms. including [#permalink]

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11 Aug 2014, 19:42

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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?

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 [#permalink]

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22 Aug 2015, 23:25

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: One night a certain hotel rented 3/4 of its rooms. including [#permalink]

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08 Jul 2016, 21:56

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

Got the answer but the wording could use some improvement..
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One night a certain hotel rented 3/4 of its rooms. including [#permalink]

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22 Sep 2016, 12:55

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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 step-by-step 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 air-conditioned So, 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 rooms 3/4 of 60 = 45, so 45 rooms are rented and the remaining 15 rooms are not rented.

...a certain motel rented ....2/3 of its air-conditioned rooms So 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 air-conditioned? There were 15 unrented rooms and 12 of them were air conditioned. 12/15 = 4/5 = 80%

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