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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 ?

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.

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}.

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

Answer: 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.

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

Hope it helps _________________

"Don't be afraid of the space between your dreams and reality. If you can dream it, you can make it so." Target=780 http://challengemba.blogspot.com Kudos??

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

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

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.