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M25-06

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Re: M25-06  [#permalink]

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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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New post 19 Jun 2018, 10:18
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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New post 19 Jun 2018, 13:19
I think this is a high-quality question and I agree with explanation.
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New post 21 Jun 2018, 10:27
Alternative approach: As we are dealing with all variable problem

Let's say a gold cost Rs 100 in the year 2001 so S = 100

Assume X = 10 %

So the value in 2002 - Rs. 90 and in the year 2003 Rs. 81

Hence T = 81

Option E - (8100)^0.5 = 90 = (TS)^0.5
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M25-06  [#permalink]

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New post 10 Feb 2019, 13:35
In the question,we are given that X% will be depreciated for every year.In 2001,1 kg of gold cost s dollar. Is this S dollars depreciated cost or cost at start of 2001?
I think the question should have this information.Correct me If I'm wrong.
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New post 11 Jun 2019, 23:27
Took me so long to actually solve this. Genius explanation! Really simplifies things. Cheers mate

The plug-in method is significantly quicker though.

Use the following
x= 10, s = 100

2001 S = 100
2002 = 100*9/10 = 90
2003 =90*9/10 = 81 = T
Plug-in to each Answer Choice quickly to find that (e) is the only answer choice that results in our desired number

\(\sqrt{ST} = \sqrt{100*81} = 90\)
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M25-06  [#permalink]

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New post 11 Jun 2019, 23:44
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


Bunuel - just for clarity. To get from \(S*\sqrt{\frac{T}{S}}\) to \(\sqrt{ST}\) you rationalized the denominator of the fraction yea?
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New post 12 Jun 2019, 00:05
dcummins 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


Bunuel - just for clarity. To get from \(S*\sqrt{\frac{T}{S}}\) to \(\sqrt{ST}\) you rationalized the denominator of the fraction yea?


Right.

\(S*\sqrt{\frac{T}{S}}=(\sqrt{S})^2*\sqrt{\frac{T}{S}}=\sqrt{ST}\)
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New post 11 Aug 2019, 10:47
Bunuel
Is it just coincidence that is the geometric mean?
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Re: M25-06   [#permalink] 11 Aug 2019, 10:47

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