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The value of an antique art-piece is x percent greater than its value

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The value of an antique art-piece is x percent greater than its value  [#permalink]

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New post 01 Oct 2018, 05:12
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The value of an antique art-piece is x percent greater than its value a year before, where x is same for each year. If the value of the art-piece was y dollars on January 1, 2011, and z dollars on January 1, 2013, then in terms of y and z, what was the value of the art-piece, in dollars, on January 1, 2014?


(A) \(z +\frac{1}{2}(z − y)\)

(B) \(z + \frac{1}{2}*\frac{(z − y)}{y}*z\)

(C) \(\frac{z\sqrt{z}}{\sqrt{y}}\)

(D) \(\frac{z^2}{2y}\)

(E) \(yz^2\)

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The value of an antique art-piece is x percent greater than its value  [#permalink]

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New post Updated on: 01 Oct 2018, 07:33
Bunuel wrote:
The value of an antique art-piece is x percent greater than its value a year before, where x is same for each year. If the value of the art-piece was y dollars on January 1, 2011, and z dollars on January 1, 2013, then in terms of y and z, what was the value of the art-piece, in dollars, on January 1, 2014?


(A) \(z +\frac{1}{2}(z − y)\)

(B) \(z + \frac{1}{2}*\frac{(z − y)}{y}*z\)

(C) \(\frac{z\sqrt{z}}{\sqrt{y}}\)

(D) \(\frac{z^2}{2y}\)

(E) \(yz^2\)


on January 1, 2011, Price \(= y\)
on January 1, 2012, Price \(= y[1+(x/100)]\)
on January 1, 2013, Price \(= y[1+(x/100)]^2 = z\)

i.e. \(1+(x/100) = √(z/y)\)

on January 1, 2014, Price \(= z[1+(x/100)] = z√(z/y)\)

Answer:Option C
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Originally posted by GMATinsight on 01 Oct 2018, 05:27.
Last edited by GMATinsight on 01 Oct 2018, 07:33, edited 1 time in total.
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The value of an antique art-piece is x percent greater than its value  [#permalink]

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New post 01 Oct 2018, 07:17
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This can be tested by putting values and checking answers.

Let y=100, x=10

2011= 100
2012= 100+10%= 110
2013= 110+10%=121 (now z)
2014= 121+10%= 133.1

Put the values in options to check the answers, C gives an exact match.

121x11/10= 1331/10= 133.1 (price in 2014)
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The value of an antique art-piece is x percent greater than its value  [#permalink]

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New post 01 Oct 2018, 08:10
We can substitute easy numbers to solve this problem

Since the value of the antique art-piece was x=10 percent greater every year,
if the value is z=1210 in 2013, it costs \(\frac{1210}{1.1} = 1100\)(10% less that 2013) in 2012.

In 2011, the value of the art-work must have been \(\frac{1100}{1.1} = 1000\) in 2011.

In 2014, the value, which is 1.1 times the value in 2013, is 1.1*1210 = 1331.

Therefore, \(\frac{z\sqrt{z}}{\sqrt{y}}\)(Option C) must be the value of the art-piece in 2014.

P.S We can substitute the values of y and z before we finally get an answer options that matches
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The value of an antique art-piece is x percent greater than its value   [#permalink] 01 Oct 2018, 08:10
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