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

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Senior Manager
Joined: 25 Oct 2008
Posts: 460
Location: Kolkata,India
A particular parking garage is increasing its rates by 15 pe  [#permalink]

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30 Oct 2009, 15:24
6
00:00

Difficulty:

75% (hard)

Question Stats:

60% (02:23) correct 40% (02:40) wrong based on 218 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%

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Manager
Status: Last few days....Have pressed the throttle
Joined: 20 Jun 2010
Posts: 50
WE 1: 6 years - Consulting

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13 Aug 2010, 06:45
10
3
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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##### General Discussion
Manager
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Joined: 13 Oct 2009
Posts: 210

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30 Oct 2009, 15:52
3
1

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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Affiliations: University of Florida Alumni
Joined: 25 Oct 2009
Posts: 34
Schools: Wharton, Booth, Stanford, HBS

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30 Oct 2009, 16:48
1
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! Manager Joined: 18 Aug 2009 Posts: 249 Re: parking garage [#permalink] ### Show Tags 30 Oct 2009, 17:01 1 1 $$R*D=E$$ $$1.15R*(1-x)D=E$$ $$R*D=1.15R*(1-x)D$$ $$1-\frac{1}{1.15}=x$$ Solving, $$x=13%$$ Intern Joined: 26 Oct 2009 Posts: 7 Re: parking garage [#permalink] ### Show Tags 30 Oct 2009, 21:13 1 I got 13 too, thanks sri and others for explanations Intern Joined: 10 Nov 2009 Posts: 7 Re: parking garage [#permalink] ### Show Tags 17 Nov 2009, 04:19 Let parking rate = 100 =(15/100+15)*100 =13.04% Director Joined: 29 Nov 2012 Posts: 705 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? SVP Joined: 06 Sep 2013 Posts: 1571 Concentration: Finance Re: parking garage [#permalink] ### Show Tags 10 Oct 2013, 13:21 2 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: 90 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?
Manager
Joined: 06 Mar 2014
Posts: 83
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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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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Manager
Joined: 04 May 2014
Posts: 152
Location: India
WE: Sales (Mutual Funds and Brokerage)
Re: A particular parking garage is increasing its rates by 15 pe  [#permalink]

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11 Sep 2017, 03:51
1
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]

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03 Mar 2019, 09:29
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Re: A particular parking garage is increasing its rates by 15 pe   [#permalink] 03 Mar 2019, 09:29
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