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# A particular parking garage is increasing its rates by 15 pe

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Director
Joined: 25 Oct 2008
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A particular parking garage is increasing its rates by 15 pe [#permalink]

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30 Oct 2009, 15:24
2
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Difficulty:

75% (hard)

Question Stats:

58% (02:54) correct 42% (02:02) wrong based on 219 sessions

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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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30 Oct 2009, 15:52
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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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Intern
Affiliations: University of Florida Alumni
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30 Oct 2009, 16:48
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 _________________ Kudos are greatly appreciated and I'll always return the favor on one of your posts. Thanks! Senior Manager Joined: 18 Aug 2009 Posts: 322 Followers: 9 Kudos [?]: 317 [1] , given: 13 Re: parking garage [#permalink] ### Show Tags 30 Oct 2009, 17:01 1 This post received KUDOS 1 This post was BOOKMARKED $$R*D=E$$ $$1.15R*(1-x)D=E$$ $$R*D=1.15R*(1-x)D$$ $$1-\frac{1}{1.15}=x$$ Solving, $$x=13%$$ Intern Joined: 26 Oct 2009 Posts: 11 Followers: 0 Kudos [?]: 1 [1] , given: 13 Re: parking garage [#permalink] ### Show Tags 30 Oct 2009, 21:13 1 This post received KUDOS I got 13 too, thanks sri and others for explanations Intern Joined: 10 Nov 2009 Posts: 7 Followers: 0 Kudos [?]: 0 [0], given: 7 Re: parking garage [#permalink] ### Show Tags 17 Nov 2009, 04:19 Let parking rate = 100 =(15/100+15)*100 =13.04% Manager Status: Last few days....Have pressed the throttle Joined: 20 Jun 2010 Posts: 69 WE 1: 6 years - Consulting Followers: 3 Kudos [?]: 49 [8] , given: 27 Re: parking garage [#permalink] ### Show Tags 13 Aug 2010, 06:45 8 This post received KUDOS 2 This post was BOOKMARKED 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! _________________ Consider giving Kudos if my post helps in some way Director Joined: 29 Nov 2012 Posts: 885 Followers: 15 Kudos [?]: 1194 [0], given: 543 Re: A particular parking garage is increasing its rates by 15 pe [#permalink] ### Show Tags 13 Jun 2013, 02:02 Plugging numbers is a better approach here or algebraic? _________________ Click +1 Kudos if my post helped... Amazing Free video explanation for all Quant questions from OG 13 and much more http://www.gmatquantum.com/og13th/ GMAT Prep software What if scenarios http://gmatclub.com/forum/gmat-prep-software-analysis-and-what-if-scenarios-146146.html Current Student Joined: 06 Sep 2013 Posts: 2004 Concentration: Finance Followers: 68 Kudos [?]: 643 [0], given: 355 Re: parking garage [#permalink] ### Show Tags 10 Oct 2013, 13:21 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 Manager Joined: 13 Aug 2012 Posts: 114 Followers: 1 Kudos [?]: 70 [0], given: 118 Re: A particular parking garage is increasing its rates by 15 pe [#permalink] ### Show Tags 11 Oct 2013, 08:07 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]

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08 Jul 2015, 02:58
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Re: A particular parking garage is increasing its rates by 15 pe [#permalink]

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27 Jul 2015, 10:30
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, 10:30
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