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The price of a pair of earrings is x percent more than the price of a

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The price of a pair of earrings is x percent more than the price of a  [#permalink]

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New post 19 Nov 2019, 01:48
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The price of a pair of earrings is x percent more than the price of a bracelet, and the price of a necklace is y percent less than the price of the earrings. Is the price of the bracelet less than the price of the necklace?


(1) \(\frac{xy}{100} < x-y\)

(2) \(x-y > 0\)


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Re: The price of a pair of earrings is x percent more than the price of a  [#permalink]

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New post 19 Nov 2019, 07:26
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The price of a pair of earrings is x percent more than the price of a bracelet,----\(p*(1+\frac{x}{100})\) and
the price of a necklace is y percent less than the price of the earrings.----\(p*(1+\frac{x}{100})(1-\frac{y}{100})\)
Is the price of the bracelet less than the price of the necklace? Is \(p<p*(1+\frac{x}{100})(1-\frac{y}{100})....1<(\frac{100+x}{100})(\frac{100-y}{100})............10000<100x-100y-xy+10000.......xy<100(x-y)\)


(1) \(\frac{xy}{100} < x-y.........xy<100(x-y)\)..Yes..Suff

(2) \(x-y > 0\).....
xy<100(x-y).....say x-y=1
x=3 and y=2.....3*2<100(3-2)
x=100 and y=99...100*99>100(100-99)
Insuff

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Re: The price of a pair of earrings is x percent more than the price of a  [#permalink]

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New post 21 Nov 2019, 06:33
Bunuel wrote:
The price of a pair of earrings is x percent more than the price of a bracelet, and the price of a necklace is y percent less than the price of the earrings. Is the price of the bracelet less than the price of the necklace?


(1) \(\frac{xy}{100} < x-y\)

(2) \(x-y > 0\)


\(e=(1+x/100)b…b=100e/(100+x)… n=(1-y/100)e\)
\(b<n…100e/(100+x)<(1-y/100)e…100/(100+x)<(1-y/100);\)
\(10^4<(100-y)(100+x)…10^4-10^4<100x-100y-xy…xy<100(x-y)\)


(1) \(\frac{xy}{100} < x-y\) sufic

\(xy<100(x-y)…xy/100<(x-y)\)

(2) \(x-y > 0\) insufic

\(x-y>0…x>y:(x,y)=(5,4)…xy<100(x-y)…20/100<1=yes\)
\(x-y>0…x>y:(x,y)=(5,4.99)…xy<100(x-y)…24.95/100<(0.01)=no\)

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Re: The price of a pair of earrings is x percent more than the price of a   [#permalink] 21 Nov 2019, 06:33
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