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At the end of each year, the value of a certain antique [#permalink]

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12 Mar 2006, 10:14

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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+1/2(m-k) B. m+1/2((m-k)/k)m C. (m*sqrt(m))/sqrt(k) D. m^2/2k; E. km^2

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+1/2(m-k) B) m+1/2((m-k)/k)m C) (m*sqrt(m))/sqrt(k) D)m^2/2k; E) km^2

please show any work and explanations please. OA to follow.

m = k*(1+c/100)^2 ? Since it says in the question, 'c percent more'..

Yes, it should.

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+1/2(m-k) B. m+1/2((m-k)/k)m C. (m*sqrt(m))/sqrt(k) D. m^2/2k; E. km^2

Price in 1992 - \(k\); Price in 1993 - \(k*(1+\frac{c}{100})\); Price in 1994 - \(k*(1+\frac{c}{100})^2=m\) --> \((1+\frac{c}{100})=\sqrt{\frac{m}{k}}\); Price in 1995 - \(m*(1+\frac{c}{100})=m*\sqrt{\frac{m}{k}\).

Re: At the end of each year, the value of a certain antique [#permalink]

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08 Jan 2013, 06:07

Bunuel wrote:

Thiagaraj wrote:

@Conocieur: Shouldn't the equation read

m = k*(1+c/100)^2 ? Since it says in the question, 'c percent more'..

Yes, it should.

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+1/2(m-k) B. m+1/2((m-k)/k)m C. (m*sqrt(m))/sqrt(k) D. m^2/2k; E. km^2

Price in 1992 - \(k\); Price in 1993 - \(k*(1+\frac{c}{100})\); Price in 1994 - \(k*(1+\frac{c}{100})^2=m\) --> \((1+\frac{c}{100})=\sqrt{\frac{m}{k}}\); Price in 1995 - \(m*(1+\frac{c}{100})=m*\sqrt{\frac{m}{k}\).

Answer: C.

Shouldn't this year be raised by the third power? since its the third year. _________________

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m = k*(1+c/100)^2 ? Since it says in the question, 'c percent more'..

Yes, it should.

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+1/2(m-k) B. m+1/2((m-k)/k)m C. (m*sqrt(m))/sqrt(k) D. m^2/2k; E. km^2

Price in 1992 - \(k\); Price in 1993 - \(k*(1+\frac{c}{100})\); Price in 1994 - \(k*(1+\frac{c}{100})^2=m\) --> \((1+\frac{c}{100})=\sqrt{\frac{m}{k}}\); Price in 1995 - \(m*(1+\frac{c}{100})=m*\sqrt{\frac{m}{k}\).

Answer: C.

Shouldn't this year be raised by the third power? since its the third year.

It is actually.

Price in 1994 is \(k*(1+\frac{c}{100})^2\) which is \(m\), so the price in 1995 is \(k*(1+\frac{c}{100})^2*(1+\frac{c}{100})\) or \(m*(1+\frac{c}{100})\).

Re: At the end of each year, the value of a certain antique [#permalink]

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10 Jan 2013, 00:55

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buckkitty wrote:

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+1/2(m-k) B. m+1/2((m-k)/k)m C. (m*sqrt(m))/sqrt(k) D. m^2/2k; E. km^2

I went with smart numbers.

c= 10 % 1992 => k = 100 1994 => m = 121 obviously, 1995 => 133.1

I find using real numbers helps the most with these types of q's. Let's say the value in 1992 is $100 (this would be k). Let's use a 10% growth rate for c. This means the value in '93 is $110, and '94 is $121 (this would be m). In 1995 the value of the antique would be 133.1 based on the 10% growth rate. So we have k=100, m=121. By substituting m and k into the answer choices we must arrive at 133.1. By estimating it a little you can start eliminating answer choices fairly quickly. Only answer choice C gives you the desired result.

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 January1, 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 +1/2(m–k) B. m +1/2(m - k)m Cm square root m /square root k D.\(m^2\)/2k E. k\(m2\)

Value on Jan 1, 1992 = k Value on Jan 1, 1993 = k(1+c/100) Value on Jan 1, 1994 = \(k(1 + c/100)^2 = m\) So, \((1 + c/100) = \sqrt{\frac{m}{k}}\) Value on Jan 1 1995 = \(k(1+c/100)^3 = k(1 + c/100)^2 * (1 + c/100)\) = \(m*\sqrt{\frac{m}{k}}\)

Yes, I generally prefer plugging in numbers but the calculations here are a little painful (with squares and roots) so using algebra is not a bad idea. _________________

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 January1, 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 +1/2(m–k) B. m +1/2(m - k)m Cm square root m /square root k D.\(m^2\)/2k E. k\(m2\)

Value on Jan 1, 1992 = k Value on Jan 1, 1993 = k(1+c/100) Value on Jan 1, 1994 = \(k(1 + c/100)^2 = m\) So, \((1 + c/100) = \sqrt{\frac{m}{k}}\) Value on Jan 1 1995 = \(k(1+c/100)^3 = k(1 + c/100)^2 * (1 + c/100)\) = \(m*\sqrt{\frac{m}{k}}\)

Yes, I generally prefer plugging in numbers but the calculations here are a little painful (with squares and roots) so using algebra is not a bad idea.

frankly , it would take more than 10 mins if we plug in the numbers. GMAT Writers know tat folks would use pluggin in and hence they create crazy algebra.

Aside, this question appeared in question pack1 and this thread was created in 2006.. I wonder how this question had leaked back then. _________________

hope is a good thing, maybe the best of things. And no good thing ever dies.

Re: At the end of each year, the value of a certain antique [#permalink]

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23 Nov 2013, 18:47

I solved without algebra, at least after the first step I noticed there would be a quadractic in there, and the only answer that had a sqrt in it was C

Re: At the end of each year, the value of a certain antique [#permalink]

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05 Dec 2014, 06:50

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Re: At the end of each year, the value of a certain antique [#permalink]

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01 Jul 2015, 21:43

Sample numbers is a great strategy here. Looking at the answer choices, we should choose numbers that are perfect squares. In this way, we won't have to waste time solving for c, since we know that it is the same each year. c=100 k=25 m = 25(2)^2=100 P_1995=200

Going through the answer choices, we see that only C produces 200.

Re: At the end of each year, the value of a certain antique [#permalink]

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25 Aug 2015, 21:34

TooLong150 wrote:

Sample numbers is a great strategy here. Looking at the answer choices, we should choose numbers that are perfect squares. In this way, we won't have to waste time solving for c, since we know that it is the same each year. c=100 k=25 m = 25(2)^2=100 P_1995=200

Going through the answer choices, we see that only C produces 200.

both c and d give 200 when we consider k=25 and m =100..isnt it?

m = k*(1+c/100)^2 ? Since it says in the question, 'c percent more'..

Yes, it should.

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+1/2(m-k) B. m+1/2((m-k)/k)m C. (m*sqrt(m))/sqrt(k) D. m^2/2k; E. km^2

Price in 1992 - \(k\); Price in 1993 - \(k*(1+\frac{c}{100})\); Price in 1994 - \(k*(1+\frac{c}{100})^2=m\) --> \((1+\frac{c}{100})=\sqrt{\frac{m}{k}}\); Price in 1995 - \(m*(1+\frac{c}{100})=m*\sqrt{\frac{m}{k}\).

Answer: C.

Price in 1995 - \(m*(1+\frac{c}{100})=m*\sqrt{\frac{m}{k}\).

How come in 1995 you don't do k * (1 + c/100) ^ 3 ?

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