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tejal777
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A particular parking garage is increasing its rates by 15 pe 30 Oct 2009, 15:24
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65% (hard)

Question Stats:

60% (02:26) correct 40% (02:35) wrong based on 525 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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D
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oldstudent
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A particular parking garage is increasing its rates by 15 pe 13 Aug 2010, 06:45
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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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A particular parking garage is increasing its rates by 15 pe 30 Oct 2009, 15:52
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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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A particular parking garage is increasing its rates by 15 pe 30 Oct 2009, 16:48
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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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A particular parking garage is increasing its rates by 15 pe 30 Oct 2009, 17:01
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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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A particular parking garage is increasing its rates by 15 pe 17 Nov 2009, 04:19
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Let parking rate = 100

=(15/100+15)*100
=13.04%
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A particular parking garage is increasing its rates by 15 pe 10 Oct 2013, 13:21
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hgp2k
\(R*D=E\)
\(1.15R*(1-x)D=E\)
\(R*D=1.15R*(1-x)D\)
\(1-\frac{1}{1.15}=x\)

Solving, \(x=13%\)
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This is the best approach. Always try to use 1 if possible. Makes lifes easier
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A particular parking garage is increasing its rates by 15 pe 27 Jul 2015, 10:30
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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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A particular parking garage is increasing its rates by 15 pe 02 Sep 2017, 07:23
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tejal777
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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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

Answer: D
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A particular parking garage is increasing its rates by 15 pe 11 Sep 2017, 03:51
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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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A particular parking garage is increasing its rates by 15 pe 25 Jan 2022, 12:11
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hgp2k
\(R*D=E\)
\(1.15R*(1-x)D=E\)
\(R*D=1.15R*(1-x)D\)
\(1-\frac{1}{1.15}=x\)

Solving, \(x=13%\)
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What do your variables stand for?
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A particular parking garage is increasing its rates by 15 pe 25 Jan 2022, 12:42
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Sushen
hgp2k
\(R*D=E\)
\(1.15R*(1-x)D=E\)
\(R*D=1.15R*(1-x)D\)
\(1-\frac{1}{1.15}=x\)

Solving, \(x=13%\)


What do your variables stand for?
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Total Cost = Price x Day = constant

if we increase price by 15%, then new price will be (1 +15/100) = 115/100
In order to keep total Cost same i.e. constant we will have to reduce day by a factor of 100/115.

115/100 Price X 100/115 Days = Price X Day = constant

Thus Days will reduce by (1-100/115) = 15/115
i.e. 15/115x100 = 13 %
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A particular parking garage is increasing its rates by 15 pe 25 Jan 2022, 12:44
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Sushen
hgp2k
\(R*D=E\)
\(1.15R*(1-x)D=E\)
\(R*D=1.15R*(1-x)D\)
\(1-\frac{1}{1.15}=x\)

Solving, \(x=13%\)


What do your variables stand for?
Show more

R stands for Rate
D stands for Days
E stands for Expenditure
x stands for fraction dreacrese in Days
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A particular parking garage is increasing its rates by 15 pe 07 Jun 2025, 01:03
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