A sum of money doubles in 3 years at compound | vedclass

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(D) Let the principal amount be $P$.According to the compound interest formula,$A = P(1 + r/100)^n$.Given that the sum doubles in $3$ years: $2P = P(1

Vedclass · Accepted Answer

$6$

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(D) Let the principal amount be $P$.According to the compound interest formula,$A = P(1 + r/100)^n$.Given that the sum doubles in $3$ years: $2P = P(1 + r/100)^3$,which implies $(1 + r/100)^3 = 2$.We want to find the time $t$ in which the sum becomes $4$ times: $4P = P(1 + r/100)^t$.This simplifies to $4 = (1 + r/100)^t$.Since $4 = 2^2$,we can substitute $2 = (1 + r/100)^3$ into the equation:$4 = ((1 + r/100)^3)^2 = (1 + r/100)^6$.Comparing the exponents,we get $t = 6$ years.

$A$ sum of money doubles in $3$ years at compound interest,compounded annually. It will become $4$ times of itself in: (in $years$)

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