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

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

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27 Sep 2010, 23:01
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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?

A. 20
B. 33 1/3
C. 35
D. 40
E. 80
[Reveal] Spoiler: OA
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Re: OG 10 Q 248 - % rented [#permalink]

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27 Sep 2010, 23:09
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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?

20
33 1/3
35
40
80

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%.

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Re: OG 10 Q 248 - % rented [#permalink]

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27 Sep 2010, 23:13
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 Not-rented |
|
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 [#permalink]

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28 Sep 2010, 19:48
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: OG 10 Q 248 - % rented [#permalink]

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28 Sep 2010, 19:56
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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 non-rented (available) that were aircon -- 12/15 -- 4/5 -- 80% (E).
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05 Mar 2011, 11:54
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(non-rented) rooms are AC.

Ans: "E"
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05 Mar 2011, 11:59
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 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.
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Re: OG 10 Q 248 - % rented [#permalink]

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03 Nov 2011, 02:17
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

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 [#permalink]

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11 Aug 2014, 18:42
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Re: One night a certain hotel rented 3/4 of its rooms. including [#permalink]

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11 Aug 2014, 20:14
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

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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11 Aug 2014, 23:59
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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 = 100-75 = 25

4. $$\frac{3}{5}$$ 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 = 60-40 = 20

Percentage of Not rented AC rooms $$= \frac{20}{25}* 100 = 80%$$

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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, 22:25
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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, 20:56
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, 11:55
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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%

[Reveal] Spoiler:
E

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

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01 Dec 2016, 17:01
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 Non-A/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.
Re: One night a certain hotel rented 3/4 of its rooms. including   [#permalink] 01 Dec 2016, 17:01
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