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18 Jan 2010, 01:29
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B
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63% (01:37) correct
37%
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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) \(100x + 100y + xy = 1330\)
(1) \(xy = 30\)
(2) \(100x + 100y + xy = 1330\)
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20 Sep 2013, 11:58
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Official Solution:
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: \(p*(1+\frac{x}{100})(1+\frac{y}{100})=p*(1+\frac{y}{100}+\frac{x}{100}+\frac{xy}{10,000})\);
The percentage increase from 2001 to 2003 is \(\frac{2003-2001}{2001}*100=\)
(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
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: \(p*(1+\frac{x}{100})(1+\frac{y}{100})=p*(1+\frac{y}{100}+\frac{x}{100}+\frac{xy}{10,000})\);
The percentage increase from 2001 to 2003 is \(\frac{2003-2001}{2001}*100=\)
\(=\frac{p(1+\frac{y}{100}+\frac{x}{100}+\frac{xy}{10,000})-p}{p}*100=\)
\(=((1+\frac{y}{100}+\frac{x}{100}+\frac{xy}{10,000})-1)*100=\)
\(=x+y+ \frac{xy}{100}\).
\(=((1+\frac{y}{100}+\frac{x}{100}+\frac{xy}{10,000})-1)*100=\)
\(=x+y+ \frac{xy}{100}\).
(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
18 Jan 2010, 05:54
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Price in 2002=(1+x)*Price-in-2001
Price in 2003=(1+y)*Price-in-2002=(1+x)(1+y)*Price-in-2001=(1+x+y+xy)*Price-in-2001
So basically, we need to determine (1+x+y+xy), where x=X/100, y=Y/100. 1+X/100+Y/100+XY/10,000
St1: XY=30 - > Clearly Insufficient
St2: 100X+100Y+XY=800; => X+Y+XY/100 = 8 => X/100+Y/100+XY/10,000 = 0.08 That's what we wanted. Sufficient
(B)
Price in 2003=(1+y)*Price-in-2002=(1+x)(1+y)*Price-in-2001=(1+x+y+xy)*Price-in-2001
So basically, we need to determine (1+x+y+xy), where x=X/100, y=Y/100. 1+X/100+Y/100+XY/10,000
St1: XY=30 - > Clearly Insufficient
St2: 100X+100Y+XY=800; => X+Y+XY/100 = 8 => X/100+Y/100+XY/10,000 = 0.08 That's what we wanted. Sufficient
(B)
