If the certain sum of money becomes ₹810000 in 2 years and ₹2560000 in 6 years compounded annually. Find the principal invested.

## Answer

Correct Answer : a ) ₹455625

Explanation :**Quick Approach**

We know, that compound interest is the concept of interest on interest

For r rate of interest compounded annually, ratio of amount and principal for first year will be (100 + r) : 100, while this ratio changes to (100 + r)^{2} : 100^{2} in second year and so on

According to the question

The ratio between the amount obtained after sixth year and second year will be:

\(({100+r\over100})^4 = {Amount\space after\space 6th\space year\over Amount\space after\space 4th\space year} = {2560000\over810000}\)

\(({100+r\over100}) = \sqrt [4] {2560000 \over 810000} = {4\over3}\)

⇒r=\({10 \over 30}\)*100

⇒ 33.33%

The principal becomes ₹810000 in 2 years

We know that

⇒\(A=P(1+{r \over 100})^t\)

Where A, P, r and t is the amount, principal, rate and time respectively

Substituting the values

⇒\(810000=P(1+{33.33 \over 100})^2\)

⇒\(810000=P({4 \over 3})^2\)

⇒ P = ₹455625

Hence, (a) is the correct answer.

**Basic approach**

According to the question

Principal becomes ₹810000 in 2 years and ₹2560000 in 6 years compounded annually

Let the principal be P and rate of interest be r

We know that

⇒\(A=P(1+{r \over 100})^t\)

**For the time period of 6 years**

⇒\(2560000=P(1+{r \over 100})^4\).....(1)

**For time period of 2 years**

⇒\(810000=P(1+{r \over 100})^4\)....(2)

Dividing equation 1 by 2, we get

⇒\({2560000 \over 810000}= ( 1+{r \over 100})^4\)

⇒\((1+{r \over 100})=\sqrt [4] {256 \over 81}\)

⇒\({r \over 100}={4 \over 3}-1\)

⇒\(r=33{1 \over 3} \%\)

Hence, (a) is the correct answer.

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