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tricky PS when using numbers

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tricky PS when using numbers

by ildude02 » Thu Jul 10, 2008 1:26 pm
One night a certain motel renetd 3/4 of its rooms, including 2/3 of its AC rooms. If 3/5 of its rooms were AC, what % of the rooms that were not renetd were AC?

1. 20
2. 33 1/3
3 35
4 40
5 80

If I tried picking numbers and solving it, I went wrong with it. But I was able to solve it drawing a grid. I would like to see someone solve this with the help of picking numbers. I wanted to see where I went wrong with my numbers approach.
Last edited by ildude02 on Thu Jul 10, 2008 2:23 pm, edited 2 times in total.
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by lion147 » Thu Jul 10, 2008 1:57 pm
Might there be something wrong with this question?

I ask because I picked a number of rooms - 100 - and got this:

100 rooms total
3/4 * 100 = 75 rented rooms
3/5 * 100 = 60 rooms with a/c
2/5 * 60 = 24 rented rooms with a/c

So, 25 total rooms were not rented, but 36 a/c rooms were not rented either.

Maybe I went wrong somewhere?
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by ildude02 » Thu Jul 10, 2008 2:04 pm
Sorry, I updated the question . It should be 2/3 of it's air conditioned rooms.
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by lion147 » Thu Jul 10, 2008 2:13 pm
ildude02 wrote:Sorry, I updated the question . It should be 2/3 of it's air conditioned rooms.
ah, OK, so now the rented rooms with a/c is (2/3)*60 = 40, and 60-40=20 rooms with a/c were not rented, therefore the answer to the question is:

( a/c rooms not rented / total rooms not rented ) * 100 =
(20/25)*100 = 80
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by ildude02 » Thu Jul 10, 2008 2:22 pm
The mistake I made was, I calculated "2/3 of rented rooms" for AC rented rooms. Somehow, "2/3 of it's airconditioned rooms" got me thinking it's "2/3 of the rented rooms that are AC". That's where I went wrong and came up with 40% instead of 80%.

Thanks for your response.
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