Author 
Message 
TAGS:

Hide Tags

Senior Manager
Joined: 17 Aug 2005
Posts: 279
Location: Boston, MA

At the end of each year, the value of a certain antique
[#permalink]
Show Tags
12 Mar 2006, 09:14
Question Stats:
73% (02:52) correct 27% (03:03) wrong based on 569 sessions
HideShow timer Statistics
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(mk) B. m+1/2((mk)/k)m C. (m*sqrt(m))/sqrt(k) D. m^2/2k; E. km^2
Official Answer and Stats are available only to registered users. Register/ Login.




Math Expert
Joined: 02 Sep 2009
Posts: 61549

Re: At the end of each year, the value of a certain antique
[#permalink]
Show Tags
12 May 2012, 01:25
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(mk) B. m+1/2((mk)/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.
_________________




Manager
Joined: 04 Oct 2011
Posts: 165
Location: India
Concentration: Entrepreneurship, International Business
GPA: 3

Re: At the end of each year, the value of a certain antique
[#permalink]
Show Tags
09 Jan 2013, 23:55
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(mk) B. m+1/2((mk)/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




Manager
Joined: 20 Mar 2005
Posts: 165
Location: Colombia, South America

Re: PSvalue of watch
[#permalink]
Show Tags
12 Mar 2006, 12:03
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(mk) B) m+1/2((mk)/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)^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



Intern
Joined: 02 Apr 2012
Posts: 2

Re: At the end of each year, the value of a certain antique
[#permalink]
Show Tags
11 May 2012, 20:28
@Conocieur: Shouldn't the equation read
m = k*(1+c/100)^2 ? Since it says in the question, 'c percent more'..



Director
Joined: 29 Nov 2012
Posts: 677

Re: At the end of each year, the value of a certain antique
[#permalink]
Show Tags
08 Jan 2013, 05:07
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(mk) B. m+1/2((mk)/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.



Math Expert
Joined: 02 Sep 2009
Posts: 61549

Re: At the end of each year, the value of a certain antique
[#permalink]
Show Tags
08 Jan 2013, 09:59
fozzzy wrote: 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(mk) B. m+1/2((mk)/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 \(m*(1+\frac{c}{100})\). Hope it's clear.
_________________



Manager
Joined: 18 Oct 2011
Posts: 76
Location: United States
Concentration: Entrepreneurship, Marketing
GMAT Date: 01302013
GPA: 3.3

Re: Arithematic
[#permalink]
Show Tags
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: 10130
Location: Pune, India

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
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Senior Manager
Status: Gonna rock this time!!!
Joined: 22 Jul 2012
Posts: 416
Location: India
GMAT 1: 640 Q43 V34 GMAT 2: 630 Q47 V29
WE: Information Technology (Computer Software)

Re: Arithematic
[#permalink]
Show Tags
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.
_________________
hope is a good thing, maybe the best of things. And no good thing ever dies.Who says you need a 700 ?Check this out : http://gmatclub.com/forum/whosaysyouneeda149706.html#p1201595My GMAT Journey : http://gmatclub.com/forum/endofmygmatjourney149328.html#p1197992



Intern
Joined: 11 Sep 2013
Posts: 3
Location: India
Concentration: Marketing, Entrepreneurship
GPA: 3.65
WE: Editorial and Writing (Journalism and Publishing)

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



Manager
Joined: 26 Sep 2013
Posts: 182
Concentration: Finance, Economics
GMAT 1: 670 Q39 V41 GMAT 2: 730 Q49 V41

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



Manager
Joined: 10 Mar 2013
Posts: 166
GMAT 1: 620 Q44 V31 GMAT 2: 690 Q47 V37 GMAT 3: 610 Q47 V28 GMAT 4: 700 Q50 V34 GMAT 5: 700 Q49 V36 GMAT 6: 690 Q48 V35 GMAT 7: 750 Q49 V42 GMAT 8: 730 Q50 V39
GPA: 3

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



Manager
Joined: 24 Nov 2013
Posts: 55

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



Manager
Joined: 28 Dec 2013
Posts: 65

Re: At the end of each year, the value of a certain antique
[#permalink]
Show Tags
24 Dec 2015, 09:45
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(mk) B. m+1/2((mk)/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 ?



Retired Moderator
Joined: 29 Oct 2013
Posts: 248
Concentration: Finance
GPA: 3.7
WE: Corporate Finance (Retail Banking)

At the end of each year, the value of a certain antique
[#permalink]
Show Tags
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.
_________________
Please contact me for super inexpensive quality private tutoring
My journey V46 and 750 > http://gmatclub.com/forum/myjourneyto46onverbal750overall171722.html#p1367876



Senior Manager
Joined: 08 Dec 2015
Posts: 282

At the end of each year, the value of a certain antique
[#permalink]
Show Tags
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
Answer D
What is wrong here?



Manager
Joined: 02 Jun 2015
Posts: 169
Location: Ghana

Re: At the end of each year, the value of a certain antique
[#permalink]
Show Tags
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
Answer D
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 So Answer is C



Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 9542
Location: United States (CA)

Re: At the end of each year, the value of a certain antique
[#permalink]
Show Tags
05 Nov 2017, 07:30
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(mk) B. m+1/2((mk)/k)m C. (m*sqrt(m))/sqrt(k) D. m^2/2k; E. km^2 We can make the following expressions: Value of the watch in 1992 = k Value of the watch in 1993 = k*(1 + c/100) Value of the watch in 1994 = k*(1 + c/100)^2 = m Value of the watch in 1995 = k*(1 + c/100)^3 = m(1 + c/100) Since k*(1 + c/100)^2 = m (the value of the value in 1994), we have: (1 + c/100)^2 = m/k 1 + c/100 = √(m/k) Thus, the value of the watch in 1995 is: m(1 + c/100) = m*√(m/k) = m*√m/√k Answer: C
_________________
5star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button.



NonHuman User
Joined: 09 Sep 2013
Posts: 14159

Re: At the end of each year, the value of a certain antique
[#permalink]
Show Tags
02 Jan 2020, 23:43
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________




Re: At the end of each year, the value of a certain antique
[#permalink]
02 Jan 2020, 23:43






