M25-06

Question

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}

Bunuel · Accepted Answer

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

shasadou · Answer

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

nicomendez · Answer

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}

codeblue · Answer

I'm not following this last step... S(1-\frac{x}{100})= S*\sqrt{\frac{T}{S}}=\sqrt{ST} .

Any help?

Bunuel · Answer

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?

arunrnair · Answer

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.

xnthic · Answer

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)?

Bunuel · Answer

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} .

tommannanchery · Answer

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}

sashmat · Answer

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

Gnpth · Answer

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.

toludayo · Answer

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

Bunuel · Answer

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

Hope it helps.

PrayasT · Answer

I think this is a  high-quality  question and I agree with explanation.

SS.56 · Answer

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

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

Rebaz · Answer

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.

Palpable · Answer

No calculation is required.

We're supposed to find the geometric mean of s and t, which is (s*t)^1/2
Option E

agrasan · Answer

I like the solution - it’s helpful.

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