Question

If a sum of money becomes 256/81 times of itself im 4 years compounded annually, find the rate of interest.​

Answer 1

Answer:

Let P = Rs x. , A =Rs.(625/256).x. , T = 1year = 4 quarter. , rate of interest = r%

p.a. = r/4 % p.quarter.

A = P .(1 + r/100)^T.

or, (625/256).x = x .(1+ r/400)^4.

or, (5/4)^4 = (1+ r/400)^4.

or, 5/4 = 1 + r/400.

or, 5/4 - 1 = r/400.

or, 1/4 = r/400.

or, 4.r. = 400

or, r = 400/4 = 100 % p.a. Answer.

Answer 2

Answer:

Question:- A sum of money becomes 256/81 times of itself in 4 years compounded annually. Find rate of interest. ?

Solution :

Let us assume that, the given sum of money

is Rs.81x.

So,

→ Principal = Rs * 0.81x

→ Rate = Let R% per annum compounded annually.

→ Time = 4 years.

,Amount=(256/81) times of Principal = (256/81) * 81x = Rs.256x

we know that, when rate is compounded annually,

Amount= Principa l^ * [1+(Rate/ 100) ]^(Time)

Step-by-step explanation:

Putting all values we get :

256x=81x^ * [1 + (R / 100)] ^ 4

→ (256x / 81 * x) = [1 + (R / 100)] ^ 4

-> (256/81) = [1 + (R / 100)] ^ 4 → (4 ^ 4 / (3 ^ 4)) = [1 + (R / 100)] ^ 4

-

-> (4/3) ^ 4 = [1 + (R / 100)] ^ 4

Now we know that, if m = n in a ^ ^ m=b^ ^ n .

than, a = b .

Therefore,

(4/3) = [1 + (R / 100)]

(4/3) - 1 = (R / 100)

→ (1/3) = (R / 100) -

- R = (100/3)

→ R=33.33\%.( Ans .)

Hence, Required rate of interest will

be 33.33%.