One night a certain motel rented 3/4 th of its rooms, including 2/3rd of its air-conditioned rooms. If 3/5th 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%
Q/A-E
motel
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This question is a great candidate for the Double Matrix method, whichcan be used for most questions featuring a population in which each member has two characteristics associated with it.Ankitaverma wrote:One night a certain motel rented 3/4 th of its rooms, including 2/3rd of its air-conditioned rooms. If 3/5th 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, 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 = [spoiler]80% = E[/spoiler]
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Brent
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Let T be the total number of rooms in the motel.Ankitaverma wrote:One night a certain motel rented 3/4 th of its rooms, including 2/3rd of its air-conditioned rooms. If 3/5th 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%
Q/A-E
Total number of air-conditioned rooms = (3/5)T
The motel rented out (3/4)T rooms
The motel did not rent out T/4 rooms.
2/3 of its air-conditioned rooms were rented out = 2/3*3/5*T = (2/5)T
1/3 of its air-conditioned rooms were not rented out = 1/3*3/5*T = (1/5)T
Percentage of rooms not rented out and air-conditioned = ((1/5)T) / (T/4) * 100
= 80%
Choose E
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Question is pretty simple but including 2/3rd of its air-conditioned rooms statement is little complex.
2/3 of rooms rented or 2/3 of AC rooms.
2/3 of rooms rented or 2/3 of AC rooms.
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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.)Ankitaverma wrote:One night a certain motel rented 3/4 th of its rooms, including 2/3rd of its air-conditioned rooms. If 3/5th 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 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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