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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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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?

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

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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?

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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30 Aug 2017, 21:57

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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%

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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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.