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The price of a pastry at 10AM was $p. Each hour after 11AM, the price

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akurathi12
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The price of a pastry at 10AM was $p. Each hour after 11AM, the price [#permalink] New post   12 Jan 2019, 13:36
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The price of a pastry at 10 AM was $p. Each hour after 11 AM, the price decreases by 20%. If Julie bought the pastry at 11:15 AM whereas her sister bought the same 3 hours later, then the price paid by Julie's sister was what percent less than that paid by Julie?(There is no price change apart from those mentioned above)

A. 33.33%
B. 37.5%
C. 48.8%
D. 51.2%
E. None of these
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C
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MahmoudFawzy
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Re: The price of a pastry at 10AM was $p. Each hour after 11AM, the price [#permalink] New post   12 Jan 2019, 15:34
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visualizing the time line:

10:00 <---(p)---> 11:00 <---(p)---> 12:00 <---(\(\frac{4}{5}\)p)---> 13:00 <---(\((\frac{4}{5})^2\)p)---> 14:00 <---(\((\frac{4}{5})^3\)p)---> 15:00

Julie's price = p
sister's price = \((\frac{4}{5})^3\)p

\((\frac{4}{5})^3\)p = p(1- \(\frac{x}{100}\))

\((\frac{4}{5})^3\) = 1- \(\frac{x}{100}\)

\(\frac{x}{100}\) = 1 - \(\frac{64}{125}\)

x = \(\frac{61}{125} * 100\) = \(\frac{488}{1000} * 100\) = 48.8 %

So C IMO
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Re: The price of a pastry at 10AM was $p. Each hour after 11AM, the price [#permalink] New post   20 Jan 2019, 12:15
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at 11:15 price =\(4/5*p\) Let this price be p1
after three hours of 11:15 it will be \((4/5)^3 * (4/5*p)\)= \((256/625)*p\) let this price be p2

percentage decrease in price will be
\((p1-p2)/p1 *100\)
substitute the values

\(4/5*p\) - \((256/625)*p\) / \(4/5*p\) *100
= 48.88 %
= answer C
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Re: The price of a pastry at 10AM was $p. Each hour after 11AM, the price [#permalink] New post   27 Nov 2019, 21:12
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can be done by assuming the initial price i.e. price of the pastry 10 AM - > 10

10 AM - > $10 (price paid by Julie)
11 - > $10 *.8 -> $ 8 (price paid by Julie)
12 - > $8 * .8 - > $ 6.4
1 -> $6.4 * .8 -> $ 5.12
@ 2 -> $5.12*.8 -> 4.096

price paid of Julie's sis - > 4.096

% difference= > (4.096/8) -> here numerator is greater than 50% of 8, thus the %value will be slightly lower than 50%.

from the given choices - > only C works out as an option. thus ANS - > C
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Re: The price of a pastry at 10AM was $p. Each hour after 11AM, the price [#permalink] New post   27 Nov 2019, 22:30
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akurathi12 wrote:
The price of a pastry at 10 AM was $p. Each hour after 11 AM, the price decreases by 20%. If Julie bought the pastry at 11:15 AM whereas her sister bought the same 3 hours later, then the price paid by Julie's sister was what percent less than that paid by Julie?(There is no price change apart from those mentioned above)

A. 33.33%
B. 37.5%
C. 48.8%
D. 51.2%
E. None of these


There is a reduction of 20% every hour so the price becomes (4/5) of what it was in the previous hour.

At 11:15, Julie bought the pastry for p*(4/5)
3 hrs later, at 2:15, her sister bought the pastry for p*(4/5)*(4/5)^3 = p*(4/5)^4 = p*(256/625)

% Diff = [p*(4/5) - p*(256/625)] / p*(4/5) = 1 - (256 * 5)/(625 * 4) = 61/125 = 48.8% ( because 61 is a bit less than 62.5 which is half of 125)

Answer (C)
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Re: The price of a pastry at 10AM was $p. Each hour after 11AM, the price [#permalink] New post   12 Dec 2022, 09:02
Julie will by it at $p. Her sister will buy it at

0.8x0.8x0.8p = 0.512p.

Percentage decrease =


p-0.512p

= 0.488p

0.488p/p x 100 = 48.8%

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Re: The price of a pastry at 10AM was $p. Each hour after 11AM, the price [#permalink]
12 Dec 2022, 09:02
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