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A person borrows rm 1000 from a bank at an interest rate of 10%. After some time, he pays the bank rm 1900 as full and final settlement of the loan. Estimate the duration of his loan.

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Answer to a math question A person borrows rm 1000 from a bank at an interest rate of 10%. After some time, he pays the bank rm 1900 as full and final settlement of the loan. Estimate the duration of his loan.

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Given: Total amount=A=1900. Principal amount=P=1000. annual interest rate=r=0.1%. Compound interest: A=P\left(1+\frac{r}{100}\right)^n Substitute the given values in above formula, we get 1900=1000\left(1+0.1\right)^n

1.1^n=\frac{1900}{1000}

1.1^n=1.9 apply natural logarithm on both sides, we get \ln\left(1.1^n\right)=\ln\left(1.9\right)

n\cdot\ln\left(1.1\right)=\ln\left(1.9\right)

n=\frac{\ln\left(1.9\right)}{\ln\left(1.1\right)} n=6.734. Therefore, The duration of the loan in compounding period is 6.734.

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A person borrows rm 1000 from a bank at an interest rate of 10%. After some time, he pays the bank rm 1900 as full and final settlement of the loan. Estimate the duration of his loan.

Answer to a math question A person borrows rm 1000 from a bank at an interest rate of 10%. After some time, he pays the bank rm 1900 as full and final settlement of the loan. Estimate the duration of his loan.

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