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A particular parking garage is increasing its rates by 15 pe

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Director
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A particular parking garage is increasing its rates by 15 pe [#permalink] New post 30 Oct 2009, 14:24
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Difficulty:

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Question Stats:

52% (02:49) correct 48% (01:55) wrong based on 132 sessions
A particular parking garage is increasing its rates by 15 percent per month. Bob decides to reduce the number of days he uses the garage per month so that the amount he spends at the garage per month remains unchanged. Which of the following is closest to Bob’s percentage reduction in the number of days he uses the garage each month?

A. 10%
B. 11%
C. 12%
D. 13%
E. 14%
[Reveal] Spoiler: OA

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Re: parking garage [#permalink] New post 30 Oct 2009, 14:52
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is 13% the answer?

let m be the rate per month , then 1.15m is the rate per month after increase

let d be the number of days he used garage, then total cost assuming 30 days in month = md/30

let n be days to be reduced to remain at same cost, so we get

we need to find n*100/d

md/30 = 1.15m (d-n)/30

==>

d = 1.15(d-n)

divide both sides by d ==>
1 = 1.15 - 1.15n/d
==>1.15n/d = .15
==>100n/d = 15/1.15 = 13
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Re: parking garage [#permalink] New post 30 Oct 2009, 15:48
Assuming Current Cost of $1, next month it will cost $1.15 for same services

X * 1.15 = $1 with X being the percentage of full previous services

X = 86.956% so you must use roughly 13% less at the higher rate to keep your costs the same.

Note if it asked you in Month 3 or X amount of decrease, you would just compound this rate.

Ex. .86956 ^2 = Usage in Month 3 or just 1 / 1.15^2
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Re: parking garage [#permalink] New post 30 Oct 2009, 16:01
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\(R*D=E\)
\(1.15R*(1-x)D=E\)
\(R*D=1.15R*(1-x)D\)
\(1-\frac{1}{1.15}=x\)

Solving, \(x=13%\)
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Re: parking garage [#permalink] New post 30 Oct 2009, 20:13
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I got 13 too, thanks sri and others for explanations
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Re: parking garage [#permalink] New post 17 Nov 2009, 03:19
Let parking rate = 100

=(15/100+15)*100
=13.04%
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Re: parking garage [#permalink] New post 13 Aug 2010, 05:45
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Best way to solve such Questions:

Remember - If Value of an item goes UP by x%, then the REDUCTION to be made to bring it to original value is calculated by formula:

x*100/(100+x)%

If Value of an item goes DOWN by x%, then the INCREMENT to be made to bring it to original value is calculated by formula:

x*100/(100-x)%

Hope this helps!
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Re: A particular parking garage is increasing its rates by 15 pe [#permalink] New post 13 Jun 2013, 01:02
Plugging numbers is a better approach here or algebraic?
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Re: parking garage [#permalink] New post 10 Oct 2013, 12:21
hgp2k wrote:
\(R*D=E\)
\(1.15R*(1-x)D=E\)
\(R*D=1.15R*(1-x)D\)
\(1-\frac{1}{1.15}=x\)

Solving, \(x=13%\)


This is the best approach. Always try to use 1 if possible. Makes lifes easier
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Re: A particular parking garage is increasing its rates by 15 pe [#permalink] New post 11 Oct 2013, 07:07
Can someone help me here? I did it by the following method but the ans is coming out to be wrong

So let the original rate be 100$ and assuming that the no of days for that month is 30, The Per day rate comes out to be \(\frac{100}{30}=\frac{10}{3}$\)
Now the rate increases 15 percent per month, so the the next month's rate would be 115$. The per day rate comes out to be \(\frac{115}{30}$\).

So what i did was that let the no of days in the new month that would equate to 100$ be x. The equation comes out to be
\(\frac{115x}{30}=100\)
\(x=\frac{3000}{115} days\)

The percentage change would be \([30-(3000/115)]/30\)\(\approx{10%}\)
What am i doing wrong?
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Re: A particular parking garage is increasing its rates by 15 pe [#permalink] New post 08 Jul 2015, 01:58
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Re: A particular parking garage is increasing its rates by 15 pe [#permalink] New post 27 Jul 2015, 09:30
jlgdr
hgp2k

What is x in ur solution.
And how can it be(1-x), how can u subtract 1- number of days???

Thanks
Re: A particular parking garage is increasing its rates by 15 pe   [#permalink] 27 Jul 2015, 09:30
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