Find the least number of complete years in wh | vedclass

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(B) Let the principal sum be $P$.The formula for the amount $A$ under compound interest is $A = P(1 + \frac{R}{100})^n$,where $R$ is the rate of inter

Vedclass · Accepted Answer

$4$

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(B) Let the principal sum be $P$.The formula for the amount $A$ under compound interest is $A = P(1 + \frac{R}{100})^n$,where $R$ is the rate of interest and $n$ is the number of years.According to the question,we want the amount to be more than double the principal,so $A > 2P$.Substituting the values: $P(1 + \frac{20}{100})^n > 2P$.Dividing both sides by $P$: $(1.2)^n > 2$.Now,test values for $n$:For $n = 3$: $(1.2)^3 = 1.728 For $n = 4$: $(1.2)^4 = 2.0736 > 2$.Thus,the least number of complete years required is $4$.

Find the least number of complete years in which a sum of money put out at $20 \%$ compound interest will be more than double?

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