Question of The Day14-06-2021

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

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 : 1002 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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