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Re M2506 [#permalink]
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16 Sep 2014, 01:23
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1. Let´s say C=gold price in 2012
2. S(1X)/100=C, then (1x)/100=C/S 3. C(1X)/100=T, from statement 2 C*C/S=T.
Then C*C=S*T. C=\sqrt{S*T}

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Bunuel wrote: Official Solution:
Gold depreciated at a rate of \(X\%\) per year between 2000 and 2005. If 1 kg of gold cost \(S\) dollars in 2001 and \(T\) dollars in 2003, how much did it cost in 2002 in terms of \(S\) and \(T\)?
A. \(T\frac{S}{2}\) B. \(T\sqrt{\frac{T}{S}}\) C. \(T\sqrt{S}\) D. \(T\frac{S}{\sqrt{T}}\) E. \(\sqrt{ST}\)
Price of 1kg gold in 2001  \(S\) Price of 1kg gold in 2002  \(S(1\frac{x}{100})\) Price of 1kg gold in 2003  \(S(1\frac{x}{100})^2=T\). So, \((1\frac{x}{100})=\sqrt{\frac{T}{S}}\) Price of 1kg gold in 2002  \(S(1\frac{x}{100})=S*\sqrt{\frac{T}{S}}=\sqrt{ST}\).
Answer: E I'm not following this last step...\(S(1\frac{x}{100})= S*\sqrt{\frac{T}{S}}=\sqrt{ST}\). Any help?

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Re: M2506 [#permalink]
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01 Dec 2014, 06:54
codeblue wrote: Bunuel wrote: Official Solution:
Gold depreciated at a rate of \(X\%\) per year between 2000 and 2005. If 1 kg of gold cost \(S\) dollars in 2001 and \(T\) dollars in 2003, how much did it cost in 2002 in terms of \(S\) and \(T\)?
A. \(T\frac{S}{2}\) B. \(T\sqrt{\frac{T}{S}}\) C. \(T\sqrt{S}\) D. \(T\frac{S}{\sqrt{T}}\) E. \(\sqrt{ST}\)
Step 1: Price of 1kg gold in 2001  \(S\) Step 2: Price of 1kg gold in 2002  \(S(1\frac{x}{100})\) Step 3: Price of 1kg gold in 2003  \(S(1\frac{x}{100})^2=T\). So, \((1\frac{x}{100})=\sqrt{\frac{T}{S}}\) Step 4: Price of 1kg gold in 2002  \(S(1\frac{x}{100})=S*\sqrt{\frac{T}{S}}=\sqrt{ST}\).
Answer: E I'm not following this last step... S*\sqrt{\frac{T}{S}}=\sqrt{ST}[/m]. Any help? From step 2: Price of 1kg gold in 2002  \(S(1\frac{x}{100})\). From step 3: \((1\frac{x}{100})=\sqrt{\frac{T}{S}}\). Substitute this into the above equation: \(S(1\frac{x}{100})=S*\sqrt{\frac{T}{S}}=(\sqrt{S})^2*\sqrt{\frac{T}{S}}=\sqrt{ST}\). Does this make sense?
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Re: M2506 [#permalink]
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13 Feb 2015, 03:02
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Bunuel wrote: codeblue wrote: Bunuel wrote: Official Solution:
Gold depreciated at a rate of \(X\%\) per year between 2000 and 2005. If 1 kg of gold cost \(S\) dollars in 2001 and \(T\) dollars in 2003, how much did it cost in 2002 in terms of \(S\) and \(T\)?
A. \(T\frac{S}{2}\) B. \(T\sqrt{\frac{T}{S}}\) C. \(T\sqrt{S}\) D. \(T\frac{S}{\sqrt{T}}\) E. \(\sqrt{ST}\)
Step 2: Price of 1kg gold in 2002  \(S(1\frac{x}{100})\) Step 3: Price of 1kg gold in 2003  \(S(1\frac{x}{100})^2=T\). So, \((1\frac{x}{100})=\sqrt{\frac{T}{S}}\) Step 4: Price of 1kg gold in 2002  \(S(1\frac{x}{100})=S*\sqrt{\frac{T}{S}}=\sqrt{ST}\).
Answer: E I'm not following this last step... S*\sqrt{\frac{T}{S}}=\sqrt{ST}[/m]. Any help? From step 2: Price of 1kg gold in 2002  \(S(1\frac{x}{100})\). From step 3: \((1\frac{x}{100})=\sqrt{\frac{T}{S}}\). Substitute this into the above equation: \(S(1\frac{x}{100})=S*\sqrt{\frac{T}{S}}=(\sqrt{S})^2*\sqrt{\frac{T}{S}}=\sqrt{ST}\). Does this make sense? Step 1: Price of 1kg gold in 2001  \(S\) I take the substitution approach In 2001 , 1 Kg cost $100 . In 2002 cost is $ 90 . In 2003 cost is $81 S = 100, T = 81 Now i have subsituted the value in the options. I checked option a,b,c . All were not tallying. Option D : T * S/root T = 81 * 100/9 = 900 Option E : root of ST = root of 81*100 =90. If any better approach is there please highlight.

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Re: M2506 [#permalink]
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06 Aug 2015, 20:57
Hello Bunuel,
could you please clarify how we arrived at
Price of 1kg gold in 2003  S(1−x100)2
I always have issues in arriving at successive percentage change.
Thanks

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Re: M2506 [#permalink]
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Hi For this problem it would be easier to express percent change as simply "x": Step 12001=S 2002=S*x 2003=S*x*x=T Step 2We take 2003 and do some algebra (actually to get to this is the harderst thing about the problem) and express x in terms of "S" and "T": T=S*x^2 x^2=T/S x=root (T/S) Step 3We substitute the value of x into 2002: 2002=S*root (T/S) 2002=root S*root S*(root T/root S) 2002=root (ST) Hope this helps
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Hello WillGetIt2001  S 2002  S(1x/100) 2003  here the amount from 2002 is reduced by another x percent . so we multiply the amount from 2002 ie. s( 1x/100) by another (1x/100) Eg . if each year the reduction was by 40 percent. The steps would be 2001  S 2002  S(140/100) 2003  S(140/100)*(140/100) = S(140/100)^2 Similarly since here the percent reduction is x 2001  S 2002  S(1x/100) > This is the year we need but without the x ..we need it just in S and T. So we need to find a value for (1x/100) part in terms of S and T 2003 = S(1  X/100)^2 Now given 2003 is T .So S(1  X/100)^2 = T, (1  X/100)^2 = T/S (1  X/100) = (T/S)^1/2 > We found a value for (1x/100) in terms of S and T. Substitute this in the 2002 eqn  S(1x/100) = S((T/S)^1/2)) = (ST)^1/2

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Re: M2506 [#permalink]
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20 Jun 2016, 08:08
Bunuel im trying to refresh my math memory about rules again, how come the answer is root(ST) and not simply ST . Remember the rule was something about needing to place the answer back to its original form? please help thank you

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Re: M2506 [#permalink]
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20 Jun 2016, 08:13

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Re: M2506 [#permalink]
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21 Jun 2016, 05:42
thanks for such a quick reply ! im just wondering why the answer in the end is the root(ST), I follow you up until S x Root(T/S) , but then you square both terms which becomes S^2 x T/S which then equals ST, if i follow you right? but why is the final answer root(ST)?

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Re: M2506 [#permalink]
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Re: M2506 [#permalink]
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26 Jul 2016, 17:13
Let C be the cost of 1kg of gold in 2002.
Since, depreciation rate per year is X%, \(X = \frac{SC}{S}*100\), and \(X = \frac{CT}{C}*100\)
Hence, \(\frac{SC}{S} = \frac{CT}{C}\)
\(SC  C^2 = SC  ST\) \(C^2 = ST\)
Therefore, \(C = \sqrt{ST}\)

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Re: M2506 [#permalink]
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19 Mar 2017, 13:35
Let P equal price in 2002:
1) P = S(1+X%)
2) P(1+X%) = T >>> 3) P = \frac{T}{(1+X%)}
Multiply 1 and 3
P^2 = \frac{T}{(1+X%)} X S(1+X%)
Cancel out (1+X%)
P^2 = TS P = \sqrt{TS}

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Re: M2506 [#permalink]
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20 Mar 2017, 12:41
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I think, solving with real numbers is easier.
1) Let X=10, S=100 (cost of gold in 2001), T=81 (cost of gold in 2003). Then gold cost in 2002 will be 90. 2) Plug numbers in each answer option. Only E matches: \sqrt{100*90}=90

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Re: M2506 [#permalink]
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14 May 2017, 09:57
I did it by taking values
let x% is 10 year 2000  100  S year 2001  90 as 10% decrease year 2003  81 as 10% decrease  T
Now from option 3 T*S/sqrt(T) = 81*100/sqrt(81) = 81*100/9 = 90 satisfies. and also from option 4 sqrt(ST) = sqrt(100*81) = 90 satisfies
both options are same...T*S/sqrt(T) = sqrt(ST)
why is 4 preferred over 3

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Re: M2506 [#permalink]
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24 May 2017, 16:29
I wanted to post another way to think about setting up the equations for this.
Let \(G\) be the price of gold in 2002
\(1)\) \(G = S(1\frac{x}{100}) => S = \frac{G}{(1  x/100)}\)
\(2)\) \(T = G(1  \frac{x}{100}) => (1  \frac{x}{100}) = \frac{T}{G}\)
Now substitute \(2)\) into \(1)\)
\(S=\frac{G}{T/G} => ST = G^2\) \(G = \sqrt{ST}\)

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Re: M2506 [#permalink]
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19 Aug 2017, 18:34
Hi Bunuel, Can we assume numbers? For example I assumed 50% for the Percentage decrease and S= 100. And let the price in 2002 be A. A= \(100(1 \frac{50}{100})\) So Now A= 50. T= \(50(1 \frac{50}{100})\) = 25 Pulling in the values. Only option that gives 50 or A is \(\sqrt{ST}\) I thought this would be better option to avoid the confusion with the variables.
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