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

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

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12 Mar 2006, 09:14
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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
[Reveal] Spoiler: OA
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12 Mar 2006, 12:03
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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

m = k*(1+c)^2
m/k = (1+c)^2
(m/k)^(1/2) = 1+c
c= (m/k)^(1/2) - 1

the value of the watch in jan 1 1995 is:

m(1+c)
so m(1+(m/k)^(1/2) - 1)
= m(m/k)^(1/2))

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

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11 May 2012, 20:28

m = k*(1+c/100)^2 ? Since it says in the question, 'c percent more'..
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Re: At the end of each year, the value of a certain antique [#permalink]

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12 May 2012, 01:25
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Thiagaraj wrote:

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

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

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08 Jan 2013, 05:07
Bunuel wrote:
Thiagaraj wrote:

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

Shouldn't this year be raised by the third power? since its the third year.
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Re: At the end of each year, the value of a certain antique [#permalink]

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08 Jan 2013, 09:59
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fozzzy wrote:
Bunuel wrote:
Thiagaraj wrote:

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

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 $$m*(1+\frac{c}{100})$$.

Hope it's clear.
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Re: At the end of each year, the value of a certain antique [#permalink]

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09 Jan 2013, 23:55
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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

Only choice c gives desired result.

$$(m*sqrt(m))/sqrt(k)$$ = $$121 \sqrt{121} / \sqrt{100}$$ = 133.1
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11 Feb 2013, 18:40
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. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7119 Location: Pune, India Followers: 2132 Kudos [?]: 13632 [0], given: 222 Re: Arithematic [#permalink] ### Show Tags 11 Feb 2013, 20:45 4112019 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 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. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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06 Mar 2013, 05:41
VeritasPrepKarishma wrote:
4112019 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 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.
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Re: At the end of each year, the value of a certain antique [#permalink]

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13 Sep 2013, 03:46
Assume values, that is the fastest way to do this..

ex:

Value in 1992 - k - 100
percent increase - c - 10
=> Value in 1994 - m - 121

=> Answer should be 121 + 10% of 121 = 133.1

A quick check gives c as the only answer..
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Re: At the end of each year, the value of a certain antique [#permalink]

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23 Nov 2013, 17:47
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
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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, 20:43
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.
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Re: At the end of each year, the value of a certain antique [#permalink]

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25 Aug 2015, 20:34
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?

option d
(m^2)/2k

(100^2)/(2*25) = 200
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Re: At the end of each year, the value of a certain antique [#permalink]

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15 Oct 2015, 06:47
This is compounding rate formula. or compound interest formula.
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Re: At the end of each year, the value of a certain antique [#permalink]

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24 Dec 2015, 09:45
Bunuel wrote:
Thiagaraj wrote:

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

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

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06 Feb 2016, 14:34
Almost no calculations approach:

choose smart numbers.

c= 200 %
1992 => k = 1
1993 => 3
1994 => m = 9
So,
1995 => 27

Only choice c gives the desired result.
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At the end of each year, the value of a certain antique [#permalink]

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13 Apr 2016, 09:50
excuse me, c=100% K=10$, m=40$
so 92-> 10$93-> 20$
94-> 40$(m) 95->80$

so m^2/2k -> 40^2/2*10 or 1600/20=80

What is wrong here?
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Re: At the end of each year, the value of a certain antique [#permalink]

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22 Aug 2016, 15:44
iliavko wrote:
excuse me, c=100% K=10$, m=40$
so 92-> 10$93-> 20$
94-> 40$(m) 95->80$

so m^2/2k -> 40^2/2*10 or 1600/20=80

What is wrong here?

Let's try a different number to test only options (C) & (D)

Say c = 10% and K = 100

1st Jan 1992 = k = 100 (starting number)

1st Jan 1993 = 1.1 * 100 = 110

1st Jan 1994 = m = 1.1% * 110 = 121

1st Jan 1995 = 1.1 * 121 = 133.1

Option (C) m * √((m/k)) must equal 133.1 ===> 121 * √((121/100)) = 121 * 11/10 = 133.1 Yes

Option (D) m^2/2k must equal 133.1 ===> 121^2/(2 * 100) = (121 *121)/ (2 * 100) =121 * 60.5/100 =73.205 No

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Re: At the end of each year, the value of a certain antique   [#permalink] 22 Aug 2016, 15:44

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