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

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Joined: 25 Dec 2018
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Location: India
GMAT 1: 490 Q47 V13
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The price of a pastry at 10AM was $p. Each hour after 11AM, the price  [#permalink]

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New post 12 Jan 2019, 13:36
3
4
00:00
A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

43% (02:46) correct 57% (02:41) wrong based on 62 sessions

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

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New post 12 Jan 2019, 15:34
1
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]

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New post 20 Jan 2019, 12:15
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] 20 Jan 2019, 12:15
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