Cost of an item is x. It's value increases by p% and
decreases by p% Now the new value is 1 rupee, what is the
actual value ?
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Answer / v venu gopal
original cost is 100^2 / (100^2 - p^2) Rs
| Is This Answer Correct ? | 21 Yes | 5 No |
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).
| Is This Answer Correct ? | 24 Yes | 9 No |
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)
| Is This Answer Correct ? | 7 Yes | 6 No |
Answer / kumar n
please given me the correct answer for this question.
| Is This Answer Correct ? | 3 Yes | 3 No |
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
| Is This Answer Correct ? | 2 Yes | 9 No |
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