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# M25-06

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 01:22
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Difficulty:

75% (hard)

Question Stats:

64% (02:44) correct 36% (02:11) wrong based on 322 sessions

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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. $$\sqrt{ST}$$
[Reveal] Spoiler: OA

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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. $$\sqrt{ST}$$

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

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

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

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

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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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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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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18 Aug 2015, 23:32
Hello WillGetIt

2001 - S
2002 - S(1-x/100)
2003 - here the amount from 2002 is reduced by another x percent . so we multiply the amount from 2002 ie. s( 1-x/100) by another (1-x/100)
Eg . if each year the reduction was by 40 percent. The steps would be
2001 - S
2002 - S(1-40/100)
2003 - S(1-40/100)*(1-40/100) = S(1-40/100)^2

Similarly since here the percent reduction is x

2001 - S
2002 - S(1-x/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 (1-x/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 (1-x/100) in terms of S and T.

Substitute this in the 2002 eqn - S(1-x/100) = S((T/S)^1/2)) = (ST)^1/2

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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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20 Jun 2016, 08:13
lydennis8 wrote:
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

Don't know which rule are you referring to... Can you please tell me which step from the solution is unclear?
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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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21 Jun 2016, 08:00
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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26 Jul 2016, 17:13
Let C be the cost of 1kg of gold in 2002.

Since, depreciation rate per year is X%,
$$X = \frac{S-C}{S}*100$$, 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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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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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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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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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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19 Aug 2017, 18:34
Top Contributor
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] 19 Aug 2017, 18:34
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# M25-06

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