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Re: One night a certain hotel rented 3/4 of its rooms. including [#permalink]
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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: One night a certain hotel rented 3/4 of its rooms. including [#permalink]
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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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Re: One night a certain hotel rented 3/4 of its rooms. including [#permalink]
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 [#permalink]
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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 [#permalink]
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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 \(= \frac{20}{25}* 100 = 80%\)

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


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%

Answer: E

This question type is VERY COMMON on the GMAT, so be sure to master the technique.

To learn more about the Double Matrix Method, watch this video:



EXTRA PRACTICE QUESTION



More questions to practice with:
EASY: https://gmatclub.com/forum/of-the-120-p ... 15386.html
MEDIUM: https://gmatclub.com/forum/in-a-certain ... 21716.html
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KILLER: https://gmatclub.com/forum/a-certain-hi ... 32899.html

Originally posted by BrentGMATPrepNow on 22 Sep 2016, 12:55.
Last edited by BrentGMATPrepNow on 23 Feb 2022, 07:14, edited 6 times in total.
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Re: One night a certain hotel rented 3/4 of its rooms. including [#permalink]
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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


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 [#permalink]
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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


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 \(\frac{2}{3}*60= 40\)

\(60-40 = 20\)

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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Re: One night a certain hotel rented 3/4 of its rooms. including [#permalink]
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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 \(\frac{2}{3}*60= 40\)

\(60-40 = 20\)

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. :lol:

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!
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Re: One night a certain hotel rented 3/4 of its rooms. including [#permalink]
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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 \(\frac{2}{3}*60= 40\)

\(60-40 = 20\)

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 dave13

generis, 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 \(= 60-45=15\)

Rooms with AC \(=\frac{3}{5}*60=36\)

AC rooms rented \(= \frac{2}{3}*36=24\)

Hence AC rooms not rented \(= 36-24=12\)

Therefore % of rooms that were not rented and were air conditioned \(= \frac{12}{15}*100 = 80\)%
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Re: One night a certain hotel rented 3/4 of its rooms. including [#permalink]
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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?

let's assume that the total no of rooms in the hotel = 60 (LCM of the denominators of the fractions in the question i.e 4,3,5 = 60)

No of rooms rented = 3/4 * 60 = 45
No of air conditioned rooms = 3/5 * 60 = 36
No of air conditioned rooms rented = 2/3 * 36 = 24
No of rooms not rented= 60 - 45 = 15
No of air conditioned rooms not rented = 36-24 = 12.

Percent of the rooms that were not rented were air conditioned = 12/15 *100 = 4/5 * 100 = 80 %

Option E is the correct answer.

Thanks,
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Re: One night a certain hotel rented 3/4 of its rooms. including [#permalink]
Given: One night a certain hotel rented 3/4 of its rooms. including 2/3 of their air conditioned rooms.

Asked: if 3/5 of its rooms were air conditioned, what percent of the rooms that were not rented were air conditioned?


Rented Not RentedTotal
Airconditioned2y/3=24xy/3=12xy=(3/5)*60x =36x
Non Air conditioned
Total (3/4)*60x = 45x60x - 45x = 15x60x


The percent of the rooms that were not rented were air conditioned = 12x/15x * 100% = 80%

IMO E
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One night a certain hotel rented 3/4 of its rooms. including [#permalink]
Bunuel 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's consider the total number of rooms to be 100.

Since "3/5 of the rooms were air conditioned", the number of air conditioned rooms = 3/5*100 = 60.

Since "3/4 of the rooms were rented", the number of rooms that were NOT rented = 1/4*100 = 25.

Since "2/3 of the air conditioned rooms were rented", the number of air conditioned rooms that were NOT rented = 1/3*60 = 20.

The question asks to find what percent of the rooms that were not rented were air conditioned:

(air conditioned rooms that were NOT rented)/(rooms that were NOT rented)*100 = 20/25*100 = 4/5*100 = 80%.

Answer: E.


The 20% value is of airconed rooms not rented.
The 25% value is of total rooms not rented.
How can we just ratio these two completely different percentages without either of them having any relationship between them?
KarishmaB, IanStewart, chetan2u, MartyTargetTestPrep, Thanks :)
I got the answer right as well but I merely guessed that the value has to be more than 50%. I reached the 20% airconed rooms not rented value but never thought to ratio the two percentages. :(
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Re: One night a certain hotel rented 3/4 of its rooms. including [#permalink]
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 no of rooms in the hotel be 60 (LCM of 4,3 & 5)
AC rooms in hotel is 36 (ie 60*3/5)
Rooms rented on a night is 45 (ie 60*3/4) ; rooms not rented is 15
AC rooms rented on that nigh is = 24 (ie36*2/3) ;AC rooms not rented is 12

So, percent of the rooms that were not rented were air conditioned is \(\frac{12}{15}*100 = 80\)%, Answer is clearly (E) 80
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Re: One night a certain hotel rented 3/4 of its rooms. including [#permalink]
Let's assume the total number of rooms in the hotel is represented by "x".

Given:
3/4 of the rooms were rented, which means (3/4) * x rooms were rented.
2/3 of the air-conditioned rooms were rented, which means (2/3) * (3/5) * x = (2/5) * x air-conditioned rooms were rented.

We need to find the percentage of the rooms that were not rented and were air-conditioned. Let's represent this by "p".

The number of air-conditioned rooms that were not rented is given by:
Number of air-conditioned rooms that were not rented = (3/5) * x - (2/5) * x = (1/5) * x

We can now calculate the percentage "p" using the formula:
p = [(Number of air-conditioned rooms that were not rented) / (Total number of rooms that were not rented)] * 100

Since the total number of rooms that were not rented is (1 - 3/4) * x = (1/4) * x, we have:
p = [(1/5) * x / (1/4) * x] * 100 = (1/5) * (4/1) * 100 = 80%

Therefore, (E) 80% of the rooms that were not rented were air-conditioned.
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Re: One night a certain hotel rented 3/4 of its rooms. including [#permalink]
the number of rooms is x.

\(\frac{3x}{4}\)-> rented
\(\frac{x}{4}\)-> not rented

of all the x rooms:
\(\frac{3x}{5}\) -> with AC
of which rented:
\(\frac{3x}{5}*\frac{2}{3}=\frac{2x}{5}\)
implying that:
\(\frac{3x}{5}-\frac{2x}{5}=\frac{x}{5}\) -> proportion of total AC rooms, which are not rented

Since we want (not rented ac rooms/not rented rooms):

\(\frac{x}{5}/\frac{x}{4}=\frac{x}{5}*\frac{4}{x}=\frac{4}{5}=\frac{80}{100}\)
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One night a certain hotel rented 3/4 of its rooms. including [#permalink]
Please find the diagram attached herewith. Let , X be the total room. R= rented rooms . 

Required percentage = (x/ 5) / (x/4 ) * 100
= x/5 * 4/x * 100 = 4*20 =80 %­
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