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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]

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30 Oct 2009, 15:24
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65% (hard)

Question Stats:

57% (01:57) correct 43% (01:54) wrong based on 271 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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Senior Manager
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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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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! Kudos [?]: 15 [0], given: 11 Senior Manager Joined: 18 Aug 2009 Posts: 318 Kudos [?]: 351 [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, $$x=13%$$ Kudos [?]: 351 [1], given: 13 Intern Joined: 26 Oct 2009 Posts: 10 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 Kudos [?]: 1 [1], given: 13 Intern Joined: 10 Nov 2009 Posts: 7 Kudos [?]: [0], given: 7 Re: parking garage [#permalink] ### Show Tags 17 Nov 2009, 04:19 Let parking rate = 100 =(15/100+15)*100 =13.04% Kudos [?]: [0], given: 7 Manager Status: Last few days....Have pressed the throttle Joined: 20 Jun 2010 Posts: 67 Kudos [?]: 62 [9], given: 27 WE 1: 6 years - Consulting Re: parking garage [#permalink] ### Show Tags 13 Aug 2010, 06:45 9 This post received KUDOS 3 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 Kudos [?]: 62 [9], given: 27 Director Joined: 29 Nov 2012 Posts: 866 Kudos [?]: 1454 [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 Kudos [?]: 1454 [0], given: 543 Current Student Joined: 06 Sep 2013 Posts: 1970 Kudos [?]: 743 [0], given: 355 Concentration: Finance Re: parking garage [#permalink] ### Show Tags 10 Oct 2013, 13:21 2 This post was BOOKMARKED 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 Kudos [?]: 743 [0], given: 355 Manager Joined: 13 Aug 2012 Posts: 114 Kudos [?]: 93 [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

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Re: A particular parking garage is increasing its rates by 15 pe [#permalink]

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30 Aug 2017, 21:57
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: A particular parking garage is increasing its rates by 15 pe [#permalink]

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02 Sep 2017, 07:23
tejal777 wrote:
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%

We can let the per-month rate = p and the number of days in a month Bob uses the garage = n. We can also let x = the percentage decrease. Thus:

(1.15p)(n)((100-x/100) = np

(115/100)(100-x/100) = 1

(100-x/100) = 20/23

23(100-x) = 2000

2300 - 23x = 2000

300 = 23x

x ≈ 13

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Re: A particular parking garage is increasing its rates by 15 pe [#permalink]

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11 Sep 2017, 03:51
Formula
In case of increase use formula- $$\frac{r}{(100+r)}$$X100
In case of decrease use formula-$$\frac{r}{(100-r)}$$X100
r is the percentage change.

Now to the question

r is the % increase

$$\frac{15}{(100+15)}$$X100=$$\frac{1500}{115}$$=13%

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Re: A particular parking garage is increasing its rates by 15 pe   [#permalink] 11 Sep 2017, 03:51
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