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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} 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} . The correct answer is E. ------------------------------------------------ Once again can someone please help with this question I dont understand the concept here.
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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!  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}=SCancel out S and you get \sqrt{ST}Hope it's clear now and good luck with your test. 
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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#p871864PLUG-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=90Answer: E. 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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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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@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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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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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}. Answer: E.
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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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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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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?
Thanks in advance
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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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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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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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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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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?
Thanks in advance 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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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) So the answer is E. Good Luck for your test. Regards, Subhash
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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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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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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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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.
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