If a sum of money at compound interest amount | vedclass

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(B) Let the principal be $P$. The amount $A$ after $t$ years is given by $A = P(1 + r/100)^t$.According to the problem,the sum becomes $3P$ in $3$ yea

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$6$

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(B) Let the principal be $P$. The amount $A$ after $t$ years is given by $A = P(1 + r/100)^t$.According to the problem,the sum becomes $3P$ in $3$ years,so $3P = P(1 + r/100)^3$,which implies $(1 + r/100)^3 = 3$.We want to find the time $T$ such that the sum becomes $9P$,so $9P = P(1 + r/100)^T$,which implies $(1 + r/100)^T = 9$.Since $9 = 3^2$,we can write $(1 + r/100)^T = (3^2)$.Substituting $(1 + r/100)^3 = 3$,we get $(1 + r/100)^T = ((1 + r/100)^3)^2 = (1 + r/100)^6$.Comparing the exponents,we get $T = 6$ years.

If a sum of money at compound interest amounts to thrice itself in $3$ years,then in how many years will it be $9$ times itself?

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