a trick question...

[vishal007win]

R=(3065-2965)/(3064+2964)

find the range of R?

can any please help me out..
what approach i need to attack this problem?

[tiny-tim]

R=(3065-2965)/(3064+2964)

find the range of R?

can any please help me out..
what approach i need to attack this problem?

Hi vishal007win! :smile:

I don't understand :redface: …

(3065-2965)/(3064+2964) is a constant, so how can there be a range? :confused:

[vishal007win]

sorry the question was..
what is the range in which R lies?
is it
1. 0< R <0.1
2. 0.1< R < 0.5
3. 0.5< R <0.7
4 0.7< R <1

[retracell]

I think the question is asking for something else.

If you find the value of R, it ends up being approximately 24 and doesn't lie in any of those choices.

[vishal007win]

@retracell
how did you approached to this solution??

[hamster143]

yep, approximately 24.

The easiest way to see that it's on the order of 30 is to divide the numerator and the denominator by 30^65 and to use the approximation (1+x)^y \approx e^{xy}, which holds with good accuracy if x is small and y is large.

[vishal007win]

yup after dividin by 30^65 both num. and den.
i get something like this
R=(1-y65)/30*(1+y64)

where y=29/30

now writing y=(1-x)
where x=1/30


(1-(1-x)^{65})/30*(1+(1-x)^{64})

now using this

(1+x)^y \approx e^{xy}

i got


(e^{65x} -1)/30*e^x*(e^{64x}+1)


which finally gives the answer R=0.024

now please check the solution...n point out my mistakes

[hamster143]

No, it should be

30 * (1-e^{-65x}) / (1+e^{-64x}) \approx 30 * 0.9 / 1.1 \approx 24

[vishal007win]

:cry: calculation mistake..

thnx
now i got it...