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GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2506
Given Kudos: 1055
GPA: 3.39
At the end of each year, the value of a certain antique watch is "c"
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09 Mar 2022, 08:59
09 Mar 2022, 08:59
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At the end of each year, the value of a certain antique watch is "c" percent more than its value one year earlier, where "c" has the same value each year. If the value of the watch was "k" dollars on January 1, 1992, and "m" dollars on January 1, 1994, then in terms of "m" and "k", what was the value of the watch, in dollars, on January 1, 1995?
A. \(m+\frac{1}{2}(m-k)\)
B. \(m+\frac{1}{2}*\frac{m-k}{k}*m\)
C. \(\frac{m√m}{√k}\)
D. \(\frac{m^2}{2k}\)
E. \(km^2\)
A. \(m+\frac{1}{2}(m-k)\)
B. \(m+\frac{1}{2}*\frac{m-k}{k}*m\)
C. \(\frac{m√m}{√k}\)
D. \(\frac{m^2}{2k}\)
E. \(km^2\)
GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2506
Given Kudos: 1055
GPA: 3.39
Re: At the end of each year, the value of a certain antique watch is "c"
[#permalink]
10 Mar 2022, 10:50
10 Mar 2022, 10:50
Expert Reply
Price in 1992 - \(k\);
Price in 1993 - \(k*(1+\frac{c}{100})\);
Price in 1994 - \(k*(1+\frac{c}{100})^2=m\). From this we can get that \((1+\frac{c}{100})=\sqrt{\frac{m}{k}}\);
Price in 1995 - \(m*(1+\frac{c}{100})=m*\sqrt{\frac{m}{k}}\).
Answer: C.
Price in 1993 - \(k*(1+\frac{c}{100})\);
Price in 1994 - \(k*(1+\frac{c}{100})^2=m\). From this we can get that \((1+\frac{c}{100})=\sqrt{\frac{m}{k}}\);
Price in 1995 - \(m*(1+\frac{c}{100})=m*\sqrt{\frac{m}{k}}\).
Answer: C.
gmatclubot
Re: At the end of each year, the value of a certain antique watch is "c" [#permalink]
10 Mar 2022, 10:50
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