Question

A sum of money becomes Rs. 30,000 in 2 years and Rs. 32,500 in 3 years. find the principal and the rate of interest per year .

Answer 1

◀ Dear Friend!◀

◀ Question: ◀

→ A sum of money becomes Rs. 30,000 in 2 years and Rs. 32,500 in 3 years. find the principal and the rate of interest per year .

◀  Method of Solution:◀

→ Given: Amount = Rs 30,000

→  Given, Certain sum of money at compound interest becomes rupees 30,000 in 2 years.

So,→   Find the Compound Interest in 2 Years ?

Using Compound Interest Formula!

→ Amount = P(1+R/100)^t

30,000 = P(1+R/100)² -------(A)

Now,   →

Given Amounts = 32,500

Given, Certain sum of money at compound interest becomes rupees 32,500 in 3 years.

→ Amount = P(1+R/100)^t

⇒ 32,500 = p(1+R/100)³ -------(B)

→ Solving the Equation by Dividing!→

Dividing (B) by (A):

⇒   32,500= p(1+R/100)³ ÷ 30,000 = P(1+R/100)²

→  Dividing rule ! ( Divide LHS to LHS) and RHS to RHS →

⇒  32,500 = p(1+R/100)³ ÷ 30,000 = P(1+R/100)²

⇒ 32,500÷30,000= p(1+R/100)³÷P(1+R/100)²

⇒ 1.083 = (1+R/100)

⇒ 1.083-1 = R/100

⇒0.083 = R/100

•°• R= 100×0.083

•°•  ⇒ Rate = 8.3% →

◀  Hence, Rate of interest is 8.3%◀

Answer 2

Amount = P(1+R/100)^t

30000 = P(1+R/100)² --(1)


Given Amounts = 30000




Amount = P(1+R/100)^t

32500 = p(1+R/100)³ -(2)



Dividing (2) by (1)


32500 = p(1+R/100)^3/32500 = P(1+R/100)^2



32500 = p(1+R/100)^3 / 32500 = P(1+R/100)^2


32500/30000= p(1+R/100)^3/P(1+R/100)^2

1.083 = (1+R/100)

1.083-1 = R/100

0.083 = R/100



Rate = 8.3%