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# m25. q.6

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21 Feb 2009, 16:56
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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 ?

(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
E

Source: GMAT Club Tests - hardest GMAT questions

The price was changing at a constant rate:
$$\begin{eqnarray*} P_{2001}&=&S\\ P_{2002}&=&kS\\ P_{2003}&=&k^2S=T\\ \end{eqnarray*}$$

From the last equation:
$$k=\sqrt{\frac{T}{S}}$$

Then $$P_{2002}=kS=\sqrt{TS}$$ .
------------------------------------------------

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22 Feb 2009, 05:08
Assume initial price is P

S= P(1 - x/100)

T= P (1 -x/100)^3

T/S = (1 -x/100)^2

(1 -x/100) = sqrt (T/S)

Price in 2002 will be S(1-x/100)

= S ( sqrt (T/s))

Hope this helps
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24 Feb 2009, 07:16
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It is a concept of geometric series..where you have constant (r) between two successive terms in series.

2001 S
2002 Sr
2003 S * r Square = T (r is ratio between two successive terms).

so r = root (t/s)
so rs = root (t/s) * s = root (t * s).
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02 Sep 2009, 13:51
I cannot understand the following:

- Gold depreciated at a rate of X% per year and then in answer explanation factor k is brought in. Is there any specific reason for that?

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13 Nov 2009, 13:06
patedhav wrote:
It is a concept of geometric series..where you have constant (r) between two successive terms in series.

2001 S
2002 Sr
2003 S * r Square = T (r is ratio between two successive terms).

so r = root (t/s)
so rs = root (t/s) * s = root (t * s).

I obviously need root practice but in the last step how is root (t/s) * s = root (t*s)??
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07 Jan 2010, 17:50
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ajonas wrote:
patedhav wrote:
It is a concept of geometric series..where you have constant (r) between two successive terms in series.

2001 S
2002 Sr
2003 S * r Square = T (r is ratio between two successive terms).

so r = root (t/s)
so rs = root (t/s) * s = root (t * s).

I obviously need root practice but in the last step how is root (t/s) * s = root (t*s)??

Does anyone have an answer to the question above? I do not understand how root (t/s)*s = root (t*s). Please help... my test is on Saturday!
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23 Jan 2010, 00:18
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i plugged in numbers

in 2000, the price of gold was 100. the depreciation rate is constant. so 10%.
so, in 2001 it was 90
and in 2002 it was 81.

then, i tested the options, starting with C, to know which one fits.

E's the one!
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09 Feb 2010, 05:31
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@ajonas and @hatethegmat... guys it's so simple, just a basic math multiplication.
it's as folllows

sqrt(T/S)*S= (sqrt T)/(sqrt S) * (sqrt S) (sqrt S)=(sqrt T)*1/(sqrt S)*(sqrt S) (sqrt S)=(sqrt T)* (sqrt S)
==> = sqrt(T*S)

Actually S=(sqrt S)^2= (sqrt S)*(sqrt S)

And ones of the (sqrt S) will be canceled by 1/(sqrt S)

Hope it helps
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09 Feb 2010, 11:26
Ans is E
cost in 2001 is S
cost in 2002 is S-XS/100
cost in 2003 is (S-XS/100) -X/100(S-XS/100) =T
solving the equation of 2003 S((100-X)/100)^2=T
or S-SX/100 = Square root of [TS] so ans is E
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11 Feb 2010, 22:52
I solved it by picking numbers
let's say x%=50%
and at 2000 gold value is 200
so 2001 after 50% dec.. = 100 T
2002 = 50 S
2003 = 25
by sub.ing values we get for ans choice E
sqrt.(TS) = sqrt.(100 *25) = 25
so my choice is E

Posted from my mobile device
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12 Feb 2010, 01:50
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hatethegmat wrote:
ajonas wrote:
patedhav wrote:
It is a concept of geometric series..where you have constant (r) between two successive terms in series.

2001 S
2002 Sr
2003 S * r Square = T (r is ratio between two successive terms).

so r = root (t/s)
so rs = root (t/s) * s = root (t * s).

I obviously need root practice but in the last step how is root (t/s) * s = root (t*s)??

Does anyone have an answer to the question above? I do not understand how root (t/s)*s = root (t*s). Please help... my test is on Saturday!

$$rS=\sqrt{T}*S/\sqrt{S}$$
Multiply both numerator and denominator $$\sqrt{S}$$
Now you have $$rS=\sqrt{T}*S*\sqrt{S}/\sqrt{S}*\sqrt{S}$$
Simplifying gives us $$\sqrt{S}*\sqrt{T}*S/S$$ because $$\sqrt{S}*\sqrt{S}=S$$
Cancel out S and you get $$\sqrt{ST}$$
Hope it's clear now and good luck with your test.
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12 Feb 2010, 19:52
iwillmakeit wrote:
I cannot understand the following:

- Gold depreciated at a rate of X% per year and then in answer explanation factor k is brought in. Is there any specific reason for that?

Gold deprecated by X% => New Gold Rate = Old Gold Rate * (1-X/100)
and in the explanation, 1-X/100 was substituted by k
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14 Feb 2011, 06:22
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If we assume A was the price of Gold in 2002 then following equations can be written:
A= S * (1+x%)
T= A * (1+x%) ==> T/A = (1+x%)

substitute (1+x%) in first equation
A= S*T/A ==> A^2 = S*T ==> A = (S*T)^0,5!
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15 Feb 2011, 06:20
Let y be the rate of Gold in 2002, so we have the following:

y = (1-x/100) S

T = (1-x/100)y => y = T/(1-x/100)

Multiply both sides, and we have :

=> y * y = (1-x/100) S * T/(1-x/100)

=> y = sq root(ST)

Regards,
Subhash
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15 Feb 2011, 06:59
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This question was also posted in PS subforum. Below is my solution from there.

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?

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 - $$S(1-\frac{x}{100})^2=T$$ --> $$(1-\frac{x}{100})=\sqrt{\frac{T}{S}}$$;

Price of 1kg gold in 2002 - $$S(1-\frac{x}{100})=S*\sqrt{\frac{T}{S}}=\sqrt{ST}$$.

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15 Feb 2011, 22:18
Picking numbers makes this pretty easy!

Let rate be 10%
(i)2001 Price = 10
(ii)2002 Price = 10*110/100 = 11
(iii)2003 Price = 11*110/100 = 12.1

Try out the answers. Only E works i.e √(12.1*10) = 11 which is 2002 Price
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30 Mar 2012, 19:23
rahulms wrote:
i plugged in numbers

in 2000, the price of gold was 100. the depreciation rate is constant. so 10%.
so, in 2001 it was 90
and in 2002 it was 81.

then, i tested the options, starting with C, to know which one fits.

E's the one!

Can someone explain how the number plugging method works?

According to rahulms, in 2002, the value was 81. How would you go about testing this with the answers?
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31 Mar 2012, 01:56
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restore wrote:
rahulms wrote:
i plugged in numbers

in 2000, the price of gold was 100. the depreciation rate is constant. so 10%.
so, in 2001 it was 90
and in 2002 it was 81.

then, i tested the options, starting with C, to know which one fits.

E's the one!

Can someone explain how the number plugging method works?

According to rahulms, in 2002, the value was 81. How would you go about testing this with the answers?

Algebraic solution is offered here: m25-q-75939.html#p871864

PLUG-IN METHOD:
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?

A. $$T\frac {S}{2}$$

B. $$T\sqrt{\frac{T}{S}}$$

C. $$T\sqrt{S}$$

D. $$T\frac{S}{\sqrt {T}}$$

E. $$\sqrt{ST}$$

Say the price of gold in 2001 was $100 and depreciated rate 10%, then: Price of 1kg gold in 2001 - S=$100;
Price of 1kg gold in 2002 - $90; Price of 1kg gold in 2003 - T=$81;

Now, plug S=$100 and T=$81 in the asnwer choices to see which one gives \$90 (the price of 1kg gold in 2002). Only option E works: $$\sqrt{ST}=\sqrt{100*81}=10*9=90$$

Note that for plug-in method it might happen that for some particular number(s) more than one option may give "correct" answer. In this case just pick some other numbers and check again these "correct" options only.

Hope it helps.
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02 Apr 2012, 05:45
E

2001 = s
2002 = s * [1-x/100] = z -------- I
2003 = z * [1-x/100] = T ------------ II

T = z * [1-x/100]
T = s * [1-x/100] * [1-x/100]
T = s * [1-x/100]^2
T/S = [1-x/100]^2
SQRT T/S = [1-x/100]

THEREFORE

Z = s * [1-x/100]

Z = S * SQRT T/S
Z = SQRT TS
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26 May 2012, 10:37
I used numbers to plug into the solution. used simple nos assumed x% = 10% and s=100 hence 2002=110 and 2003 or t= 121. though must say it took time.. slightly over 3 mins. maybe from time perspective algebraic method will be quicker.
Re: m25. q.6   [#permalink] 26 May 2012, 10:37
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# m25. q.6

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