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

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Joined: 02 Sep 2009
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The value of an antique art-piece is x percent greater than its value  [#permalink]

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01 Oct 2018, 04:12
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Difficulty:

45% (medium)

Question Stats:

67% (02:48) correct 33% (02:52) wrong based on 30 sessions

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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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Updated on: 01 Oct 2018, 06:33
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, 04:27.
Last edited by GMATinsight on 01 Oct 2018, 06:33, 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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01 Oct 2018, 06:17
1
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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01 Oct 2018, 07:10
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, 07:10
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# The value of an antique art-piece is x percent greater than its value

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