6 The price of an article increased by X and then fell by X such that the latest price of

As the x% varies,the original price also varies.
So the solution comes in equation form
let original price = y
when x% price increases,then new price = y*(100+x)/100
now x% price fell down,so latest price = y*((100+x)/100)*((100-x)/100) = 2
y*((100)^2 -x^2) = 2*10^4
y = (2*10^4)/(10^4 - x^2)
taking common 10^4
y = 2/(1-(x/100)^2)
so, y = 2/(1-(x%)^2)
this is the final equation ......
now putting the differnt values 'y' you get the original price as written in the options(choice) of question
10 years agoHelpfull: Yes(5) No(2)
1 RS as per given conditions
10 years agoHelpfull: Yes(1) No(8)
in succesion we have a+b+ab/100=> (x-x-x^2/100)% total => x^2/100 loss total
=>loss= L = ((x^2/100)*P)/100 => P*x^2/10000
=>p=2+L => P(1-x^2/10000)=2 = > P=20000/(10000-x^2) Ans
10 years agoHelpfull: Yes(1) No(0)
ans = 2.02
10 years agoHelpfull: Yes(1) No(0)
we know if x% is increased and x% decreased ,then overall decreased (x^2/100)%
if y is original price then
y=2*100/(x^2)
10 years agoHelpfull: Yes(0) No(1)
let original price be rs p
then the price increases by x%
which is =(1+x/100)p
now
ATQ it decreases by x%
then (1-x/100)p(1+x/100)=2
by solving we get
p= 20000/(10000-x^2)
thats it
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original price= 20000/(10000-x^2)
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p[[1-[x/100]^2]]=2 ---its highly impossible to figure out the answer..with out assumptions
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6. The price of an article increased by X% and then fell by X% such that the latest price of the article is Rs. 2. What was the original price ?