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M25-06

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Bunuel
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M25-06 [#permalink] New post   16 Sep 2014, 01:22
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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. \(\sqrt{st}\)
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Bunuel
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Re M25-06 [#permalink] New post   16 Sep 2014, 01:23
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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. \(\sqrt{st}\)


In 2001, the price for 1kg of gold was \(s\).

In 2002, the price became \(s(1-\frac{x}{100})\).

It 2003, the price became \(s(1-\frac{x}{100})^2 = t\). This implies that \((1-\frac{x}{100})=\sqrt{\frac{t}{s} }\).

Substituting \((1-\frac{x}{100})=\sqrt{\frac{t}{s} }\) into the 2002 price formula, we find that \(s(1-\frac{x}{100})=s*\sqrt{\frac{t}{s} }=\sqrt{st}\)


Answer: E
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Re: M25-06 [#permalink] New post   12 Aug 2015, 08:18
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Hi

For this problem it would be easier to express percent change as simply "x":

Step 1
2001=S
2002=S*x
2003=S*x*x=T

Step 2
We 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 3
We 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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Re: M25-06 [#permalink] New post   06 Oct 2014, 12:25
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1. Let´s say C=gold price in 2012

2. S(1-X)/100=C, then (1-x)/100=C/S
3. C(1-X)/100=T, from statement 2 C*C/S=T.

Then C*C=S*T. C=\sqrt{S*T}
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Re: M25-06 [#permalink] New post   01 Dec 2014, 06:43
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: M25-06 [#permalink] New post   01 Dec 2014, 06:54
Expert Reply
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: M25-06 [#permalink] New post   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: M25-06 [#permalink] New post   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: M25-06 [#permalink] New post   21 Jun 2016, 08:00
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lydennis8 wrote:
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)?


No, we don't square. We just write S as \((\sqrt{S})^2\):

\(S*\sqrt{\frac{T}{S}}=(\sqrt{S})^2*\sqrt{\frac{T}{S}}=\sqrt{ST}\).
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Re: M25-06 [#permalink] New post   26 Jul 2016, 17:13
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Let C be the cost of 1kg of gold in 2002.

Since, depreciation rate per year is X%,
\(X = \frac{S-C}{S}*100\), and \(X = \frac{C-T}{C}*100\)

Hence, \(\frac{S-C}{S} = \frac{C-T}{C}\)

\(SC - C^2 = SC - ST\)
\(C^2 = ST\)

Therefore, \(C = \sqrt{ST}\)
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Re: M25-06 [#permalink] New post   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: M25-06 [#permalink] New post   19 Aug 2017, 18:34
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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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Re: M25-06 [#permalink] New post   19 Jun 2018, 07:07
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 get this but I just dont understand how you simplified S*\sqrt{\frac{T}{S}}=\sqrt{ST}[/m].
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Re: M25-06 [#permalink] New post   19 Jun 2018, 10:18
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toludayo 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. \(\sqrt{ST}\)


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 get this but I just dont understand how you simplified S*\sqrt{\frac{T}{S}}=\sqrt{ST}[/m].


\(S*\sqrt{\frac{T}{S}}=(\sqrt{S})^2*\sqrt{\frac{T}{S}}=\sqrt{ST}\).

Hope it helps.
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Re: M25-06 [#permalink] New post   19 Jun 2018, 13:19
I think this is a high-quality question and I agree with explanation.
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Re: M25-06 [#permalink] New post   14 Aug 2023, 13:19
Bunuel wrote:
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. \(\sqrt{ST}\)


A plug-in method is a bit easier and less confusing

let x=100%

2001-32 (S)
2002-16
2003-8 (T)
2002-4
2001-2
2000-1

Option E- \(\sqrt{ST}\)= \(\sqrt{32*8}\) =\(\sqrt{256}\)=16

Ans E
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Re: M25-06 [#permalink] New post   15 Aug 2023, 04:22
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M25-06 [#permalink] New post   20 Dec 2023, 01:06
Bunuel did a great job by simplifying the question. But where i and i think others go wrong is his last step:

To go from S∗√S/T to (√S)^2 * √S/T. You may wondering how the hell is that possible? Well here is the answer.

S= (√S)^2 just like 2=(√2)^2 or 3=(√3)^2 or 8=(√8)^2. to extend this to variables, it will look like this:

V=(√V)^2 or T=(√T)^2 or Z=(√Z)^2 etc...

This is not the only question that is see that this simple but yet important knowledge/step is crucial to solve such questions. If donot know this, you will be stuck and lose a lot of time during the exam.
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Re: M25-06 [#permalink] New post   26 Dec 2023, 05:36
No calculation is required.

We're supposed to find the geometric mean of s and t, which is (s*t)^1/2
Option E
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Re: M25-06 [#permalink]
26 Dec 2023, 05:36
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