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There are 100 rooms in a hotel. The room charges for all the

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There are 100 rooms in a hotel. The room charges for all the

by Mo2men » Thu Mar 30, 2017 7:47 pm
There are 100 rooms in a hotel. The room charges for all the rooms are always the same and the hotel's sales only include room charge. In November, an average of 80 rooms were occupied every day. In a month of November, how many days were the rooms all occupied?

1) An average of 40 rooms were occupied on days that the rooms weren't all occupied.
2) The sales of days that the rooms were all occupied is 5 times greater than the sales of days that the rooms were not all occupied.

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OA:D
Last edited by Mo2men on Fri Mar 31, 2017 6:35 pm, edited 1 time in total.
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by [email protected] » Fri Mar 31, 2017 9:50 am
Hi Mo2men,

This question revolves around 'system' Algebra, but you'll likely have to play around with the question a bit to realize it.

We're told that there are 100 rooms in a hotel (so there's a LIMIT to how many rooms can be rented out at once). We're also told in November, an average of 80 rooms were rented each day. Since November has 30 days, that means a total of (80)(30) = 2400 room-days of rentals occurred in November. We're asked for the number of days in November that had ALL 100 rooms rented.

1) An average of 40 rooms were occupied on days that the rooms weren't all occupied.

With this Fact, we can create two 'groups' of numbers: X non-full days (at 40 rooms each) and Y full days (at 100 rooms each). With those variables, we can create the following equations:

X + Y = 30
40X + 100Y = 2400

This is a 'system' of equations, which we can solve and get the value of Y (so we CAN answer the question and there will be just one solution: X=10 and Y=20).
Fact 1 is SUFFICIENT

2) The sales of days that the rooms were all occupied is 5 times greater than the sales of days that the rooms were not all occupied.

This Fact is a little 'quirkier' to deal with, but you can use your work from Fact 1 and a bit of simple arithmetic to prove the solution without too much trouble.

From Fact 1, we know that 20 full days and 10 non-full days 'fits' the given prompt. Let's see how that information fits Fact 2:

20 full days = 20(100) = 2000 room-days of rentals

2000 is five times 400, so there would be 400 additional room-days of rentals. This matches the total (2400 total room-days in November), so this answer clearly fits what we're told. Is there any other way to add a number to 5 TIMES that number and get 2400?

X + 5X = 2400
6X = 2400
X = 400
Thus, 400 and 2000 is the ONLY way to get to 2400 here, so this is the only possible answer. There MUST be 20 full-days.
Fact 2 is SUFFICIENT

Final Answer:D

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by GMATGuruNY » Fri Mar 31, 2017 2:55 pm
Mo2men wrote:There are 100 rooms in a hotel. The room charges for all the rooms are always the same and the hotel's sales only include room charge. In November, an average of 80 rooms were occupied every day. In a month of November, how many days were the rooms all occupied?

1) An average of 40 rooms were occupied on days that the rooms weren't all occupied.
2) The sales of days that the rooms were all occupied is 5 times greater than the sales of days that the rooms were not all occupied.
Let N = the non-full days and F = the full days.

Statement 1:
Occupation rate for the non-full days = 40%.
Occupation rate for the full days = 100%.
Occupation rate for the MIXTURE of all the days = 80%.

This is a MIXTURE problem.
To determine the ratio of N to F, we can use ALLIGATION.

Step 1: Plot the 3 percentages on a number line, with the percentages for N and F on the ends and the percentage for the mixture in the middle.
N 40----------80-----------100 F

Step 2: Calculate the distances between the percentages.
N 40----40----80----20----100 F

Step 3: Determine the ratio in the mixture.
The ratio of N to F is equal to the RECIPROCAL of the distances in red.
N:F = 20:40 = 10:20.
Thus, the 30 days of November are composed 10 non-full days and 20 full days.
SUFFICIENT.

Statement 2:
Let the room rate = $1 per day, implying that each full day yields $100.
Since an average of 80 rooms are occupied each day -- and there are a total of 30 days in November -- the total revenue for November = 80*30 = $2400.

Let $F = the revenue yielded by the full days and $N = the revenue yielded by the non-full days.
Since $F : $N = 5:1 = 500:100 = 2000:400, the full days yield $2000, while the non-full days yield $400, for a total of $2400.
Since each full day yields $100 of income, the total number of full days = (total income for the full days)/(income for each full-day) = 2000/100 = 20.
SUFFICIENT.

The correct answer is D.
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