one of nice sums I've come across

ok here we go

2001 -> s
2002 -> s - (x*s)/100 => s(1-x/100) => s(100-x/100) (depreciates at x%)
2003 -> s - (x*s)/100 - (x/100) * (s - (x*s)/100) [basically x% further depreciation in next year]

now 2003 = T

so T = s - (x*s)/100 - (x/100) * (s - (x*s)/100)

the right hand turns out following way

T = s - x*s/100 - x*s/100 - x^2 * s/10000
T = s(10000 - 200x - x^2)/10000
T = s *(100-x/100)^2
sqr.rt(T) = sqr.rt(s) * (100-x/100)
sqr.rt(T) * sqr.rt(s) = sqr.rt(s) * sqr.rt(s) * (100-x/100)
sqr.rt(T * s) = s*(100-x/100) which is basically price in 2002 from above equations

HTH

bhumika.k.shah wrote: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*S/2

B. T * sq rt T/S

C.T sq rt S

D.T *S/sq rt T

E. sq rt T *S

OA E

I dint even know how to start solving this question

i find that plugging in numbers is the best way to approach such problematic problems.
let's assume that the price of gold was 1000 per kg in 2000 and depreciated at a constant rate of 10% after 2000.

so in 2001, the price was 900 = S
in 2002, it was 810.
and in 2003, it was 729 = T

now, use the answer options to back solve. we need to find the option which gives us the price in 2002, i.e. 810

i always start with C.

you'll notice that only E gives us 810.

sqrt729 * sqrt 900 = 27 * 30 = 810

plugging numbers is a good solution but sometimes I feel it becomes matter of luck because on exam due to pressure you might panick considering 5 options to try and one doesn't know which one could be correct- but it depends on person to person . using equations I might take few seconds extra but I'm sure I'll land on correct answer

yea i agree with linkinpark...
it depends from person to person...
i might be comfortable by plugging in #s and backsolving...
but who knows on D day i might get a right answer by going the equation way