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# M25-06

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Intern
Joined: 05 Feb 2018
Posts: 1
Concentration: Finance, Real Estate
WE: Project Management (Real Estate)

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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 - $$S(1-\frac{x}{100})=S*\sqrt{\frac{T}{S}}=\sqrt{ST}$$.

I get this but I just dont understand how you simplified S*\sqrt{\frac{T}{S}}=\sqrt{ST}[/m].
Math Expert
Joined: 02 Sep 2009
Posts: 60566

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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 - $$S(1-\frac{x}{100})=S*\sqrt{\frac{T}{S}}=\sqrt{ST}$$.

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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Joined: 26 Feb 2018
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GMAT 1: 640 Q45 V34
GPA: 3.9
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19 Jun 2018, 13:19
I think this is a high-quality question and I agree with explanation.
Director
Joined: 08 Jun 2013
Posts: 538
Location: France
GMAT 1: 200 Q1 V1
GPA: 3.82
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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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Joined: 27 Nov 2017
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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.
VP
Joined: 14 Feb 2017
Posts: 1364
Location: Australia
Concentration: Technology, Strategy
GMAT 1: 560 Q41 V26
GMAT 2: 550 Q43 V23
GMAT 3: 650 Q47 V33
GMAT 4: 650 Q44 V36
GMAT 5: 650 Q48 V31
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GMAT 7: 710 Q47 V41
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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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Here's how I went from 430 to 710, and how you can do it yourself:
VP
Joined: 14 Feb 2017
Posts: 1364
Location: Australia
Concentration: Technology, Strategy
GMAT 1: 560 Q41 V26
GMAT 2: 550 Q43 V23
GMAT 3: 650 Q47 V33
GMAT 4: 650 Q44 V36
GMAT 5: 650 Q48 V31
GMAT 6: 600 Q38 V35
GMAT 7: 710 Q47 V41
GPA: 3
WE: Management Consulting (Consulting)

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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 - $$S(1-\frac{x}{100})=S*\sqrt{\frac{T}{S}}=\sqrt{ST}$$.

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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Here's how I went from 430 to 710, and how you can do it yourself:
Math Expert
Joined: 02 Sep 2009
Posts: 60566

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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 - $$S(1-\frac{x}{100})=S*\sqrt{\frac{T}{S}}=\sqrt{ST}$$.

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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11 Aug 2019, 10:47
Bunuel
Is it just coincidence that is the geometric mean?
Re: M25-06   [#permalink] 11 Aug 2019, 10:47

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# M25-06

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