Latest TCS Aptitude Question SOLUTION: Cost of an item is x. It's value increases by p% and then decreases by p%, now the new value is 1 rupee. What is the actual value ?

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Cost of an item is x. It's value increases by p% and then decreases by p%, now the new value is 1 rupee. What is the actual value ?

Solution

After increment of p%, value = x+(xp/100)
then after p% decrement, value = {x+(xp/100)}-[{x+(xp/100)}p/100]
= {x+(xp/100)}{1-(p/100)}
= x{1+(p/100)}{1-(p/100)}
= x[1-{(p^2)/10000}] = 1 (given)
so x = 1/[1-{(p^2)/10000}] = 10000/{10000-(p^2)}

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Asked In TCS SUHAS SUTARIYA (12 years ago)

Solved Read Solution (2)

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CHANDAN KR SINGH  (12 years ago)

ANSWER : 10000/(10000-p^2)

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naresh kumar n  (12 years ago)

solution s (10000/(10000-p^2))

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