Gold depreciated

At the end of 2002 gold costs (100 - x)*S/100

At the end of 2003 gold costs:

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

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

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

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

In 2002:

Cost = (100 - (100 - 100*sqrt(T/S)))*S/100
= (100*sqrt(T/S))S/100
= S*sqrt(T/S)
= sqrt (S*T) = Answer E

Thats not nice. I'm sure theres a better and quicker way but that does get you the answer.

let the cost of gold be 100 in 2000. let it decrease by 50% every year

2001: 100*1/2=50........S
2002: 50*1/2=25
2003: 25*1/2=12.50......T

substituting values of S and T in the answer choices, the one that gives the value of 25 will be the answer.

C. T sqrt (S) =12.50*(50)^1/2 >25

D. T S/sqrt (T) =(100/2 *25/2)/(25/2)^1/2=125 *2^1/2 >25

E . sqrt (S T)= (50*25/2)^1/2 =25

hence E

It is easier to pick up number and solve it

Let the cost of 1 lb of gold in year 2000 was 1000 and it depreciated 10 % annually

therefore, cost in 2001 = 900 (or S)
cost in 2002 = 810 ( the unkown)
cost in 2003 = 729 (or T)

A. T S/2 = 729*450 (this would be much greater than 810)
B. T sqrt (T/S) = 729 * sqrt(729/900) = 729 * 0.9 (less than 810)
C. T sqrt (S) = 729 * sqrt (900) = 729 *30 (much greater than 810)
D. T S/sqrt (T) = sqrt (T) * S = 81 * 900 ( much greater than 810)
E. sqrt (S T) = sqrt (729 * 900) = 27*30 = 810 Bingo!!

With these type of questions its always better to start with option E, and then try C

I have a question.

For this type of Q , how to understand picking which no will be ideal?

S ---> M ---> T where M is the price in 2002

depreciation is x% per year.

Therefore:

M = S*(100-x)/x
T = M*(100-x)/x

Lets say (100 - x)/x = y (I am doing this simplification because the answer choices are not in x)

So,

M = Sy and T=My

M = S*(T/M)
M^2 = ST
M = sqrt(ST) Choose E

-BM-