Answer / v venu gopal

original cost is 100^2 / (100^2 - p^2) Rs

Answer / phaniram

Let the item cost be I then,

Item cost by increasing p% will be:
I(100+p)/100

Item Cost by decreasing p% will be:
I(100+p)(100-p)/100*100

But the above is equal to 1.

So,

I(100^2-p^2)/100^2 = 1 => I = 100^2/(100^2 - p^2).

Answer / sekhar

it depends on p value

if any item is increased and decresed by p percent actually
the value of item is decreased by 0.P^2%
so 100%-.p^2%=1rupee
100%=x



x=1oo*1/(100-.p^2)

Answer / chandradukpa

1=0
0=1
So, 1+1=0.

Answer / kumar n

please given me the correct answer for this question.

Answer / sandeep

consider '1' rupee as 100 and p=20
now it is increased by 20% then the cost is 120
and decreased by 20% now the cost is 96
since it is 1 rupee we get 0.96 ps

If 1/x = 3.5. then 1/(x+2) =

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