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M12-35

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Bunuel
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M12-35 [#permalink] New post   16 Sep 2014, 00:48
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65% (01:51) correct 35% (02:06) wrong based on 187 sessions
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If the price of a certain item increased by \(x\%\) from 2001 to 2002, and then increased by \(y\%\) from 2002 to 2003, where both \(x\) and \(y\) are positive numbers, what is the overall percentage increase in price from 2001 to 2003?



(1) \(xy = 30\)

(2) \(100x + 100y + xy = 1330\)
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Re M12-35 [#permalink] New post   16 Sep 2014, 00:48
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If the price of a certain item increased by \(x\%\) from 2001 to 2002, and then increased by \(y\%\) from 2002 to 2003, where both \(x\) and \(y\) are positive numbers, what is the overall percentage increase in price from 2001 to 2003?

The price in 2001: \(p\);

The price in 2002: \(p*(1+\frac{x}{100})\);

The price in 2003: \(p*(1+\frac{x}{100})(1+\frac{y}{100})=p*(\frac{y}{100}+\frac{x}{100}+\frac{xy}{10,000})*100\);

The percentage increase from 2001 to 2003 is \(\frac{2003-2001}{2001}*100=\)

\(=\frac{p(1+\frac{y}{100}+\frac{x}{100}+\frac{xy}{10,000})-p}{p}*100=\)

\(=((1+\frac{y}{100}+\frac{x}{100}+\frac{xy}{10,000})-1)*100=\)

\(=x+y+ \frac{xy}{100}\).

(1) \(xy=30\). Not sufficient.

(2) \(100x+100y+xy=1330\).

Dividing both sides by 100, we get \(x+y+\frac{xy}{100}=13.3\), which directly gives us the percentage increase in price. Therefore, statement (2) is sufficient to answer the question.


Answer: B
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Re: M12-35 [#permalink] New post   22 Mar 2023, 04:21
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M12-35 [#permalink] New post   12 Aug 2023, 10:47
I think this is a high-quality question and I agree with explanation.
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Re: M12-35 [#permalink] New post   04 Jan 2024, 04:42
Bunuel

Question number 1

The price in 2003: P ∗ (1+x/100) (1+y/100) = (y/100+x/100+xy/10,000) ∗ 100

How do you get rid of P in the second part of the equation?

Question number 2

((1+y/100+x/100+xy/10,000) −1)∗100= x+y+xy/100

How do you get rid of *100 at the end of the first part?


Thanks in advance!
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Re: M12-35 [#permalink] New post   04 Jan 2024, 05:48
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Rebaz wrote:
Bunuel

Question number 1

The price in 2003: P ∗ (1+x/100) (1+y/100) = (y/100+x/100+xy/10,000) ∗ 100

How do you get rid of P in the second part of the equation?

Question number 2

((1+y/100+x/100+xy/10,000) −1)∗100= x+y+xy/100

How do you get rid of *100 at the end of the first part?

Thanks in advance!



1. p was just missing there. Edited.

2. \(\frac{p(1+\frac{y}{100}+\frac{x}{100}+\frac{xy}{10,000})-p}{p}*100=\)

Reduce by p:

\(=((1+\frac{y}{100}+\frac{x}{100}+\frac{xy}{10,000})-1)*100=\)

\(=(\frac{y}{100}+\frac{x}{100}+\frac{xy}{10,000})*100=\)

\(=x+y+ \frac{xy}{100}\).

Hope it's clear.
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Re: M12-35 [#permalink]
04 Jan 2024, 05:48
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