A sum of money placed at compound interest do | vedclass

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(C) The formula for compound interest is $A = P(1 + r/100)^n$,where $A$ is the amount,$P$ is the principal,$r$ is the rate,and $n$ is the time in year

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$8$

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(C) The formula for compound interest is $A = P(1 + r/100)^n$,where $A$ is the amount,$P$ is the principal,$r$ is the rate,and $n$ is the time in years.Given that the sum doubles in $4$ years,we have $2P = P(1 + r/100)^4$,which simplifies to $(1 + r/100)^4 = 2$.We want to find the time $n$ such that the sum becomes four times itself,i.e.,$4P = P(1 + r/100)^n$.This simplifies to $(1 + r/100)^n = 4$.Since $4 = 2^2$,we can substitute the value of $2$ from the first equation: $(1 + r/100)^n = ((1 + r/100)^4)^2$.Therefore,$(1 + r/100)^n = (1 + r/100)^8$.Comparing the exponents,we get $n = 8$ years.

$A$ sum of money placed at compound interest doubles itself in $4$ $years$. In how many $years$ will it amount to four times itself?

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