Question

Naresh puts rs 10000 at interested compounded annually for 2 years . Had he invested it at 1 % more rate of intrest compounded annually . He would have earned rs 209 .what is the intrest rate​

Answer 1

» Question :

Naresh puts ₹ 10000 at a interest, compounded annually for 2 years .If he had invested it at 1 % more than the rate of intrest compounded annually . He would have earned rs 209 .What is the interest rate.

» To Find :

The Rate of the Interest for the whole transaction .

» Given :

• Principal = ₹ 10000

• Time = 2 years

• Simple Interest = ₹ 209

» We Know :

Amount Formula :

$\sf{\underline{\boxed{A = P\left(1 + \dfrac{R}{100}\right)^{n}}}}$

Amount :

$\sf{\underline{\boxed{A = P + SI}}}$

Where ,

• A = Amount
• P = Principal
• R = Rate of interest
• n = Time Period
• SI = Simple Interest

» Concept :

According to the question ,it said that the when the Rate was increased by 1% then the interest was ₹ 209 ,so we can take the Rate as (1 + R%) p.a.

For Finding the Amount : (When Rate is (1 + R%)p.a.

• Principal = ₹ 10000
• Interest = ₹ 209

By Substituting the values in the formula ,we get :

$\sf{A = P + SI}$

$\sf{\Rightarrow A = 10000 + 209}$

$\sf{\Rightarrow A = 10209}$

Hence , the Amount is ₹ 10209.

Now ,by using the Amount Formula, we can find the Rate of interest.

» Solution :

The values which are given in the question :

• Principal = ₹ 10000
• Amount = ₹ 10209
• Time = 2 years

Taken Rate by ATQ :

• R = (1 + R)% p.a.

Formula For finding the Amount :

$\sf{A = P\left(1 + \dfrac{R}{100}\right)^{n}}$

By substituting the values in it ,we get :

$\sf{\Rightarrow 10209 = 10000\left(1 + \dfrac{(1 + R)}{100}^{2}\right)}$

$\sf{\Rightarrow \dfrac{10209}{10000} = \left(1 + \dfrac{(1 + R)}{100}\right)^{2}}$

$\sf{\Rightarrow \dfrac{10209}{10000} = \left(\dfrac{100 + (1 + R)}{100}\right)^{2}}$

$\sf{\Rightarrow \dfrac{10209}{10000} = \left(\dfrac{101 + R}{100}\right)^{2}}$

$\sf{\Rightarrow \sqrt{\dfrac{10209}{10000}} = \left(\dfrac{101 + R}{100}\right)}$

$\sf{\Rightarrow \dfrac{101.04(approx.)}{100} = \left(\dfrac{101 + R}{100}\right)}$

$\sf{\Rightarrow \dfrac{101.04}{100} \times 100 = 101 + R}$

$\sf{\Rightarrow \dfrac{101.04}{\cancel{100}} \times \cancel{100} = 101 + R}$

$\sf{\Rightarrow 101.04 = 101 + R}$

$\sf{\Rightarrow 101.04 - 101 = R}$

$\sf{\Rightarrow 0.04\% = R}$

Hence ,the rate of interest is 0.04 %.

» Additional information :

• Simple Interest :

$\sf{SI = \dfrac{P \times R \times t}{100}}$

• Rate :

$\sf{R = \dfrac{SI \times 100}{P \times t}}$

• Time :

$\sf{t = \dfrac{SI \times 100}{P \times R}}$

• Principal :

$\sf{P = \dfrac{SI \times 100}{R \times t}}$

Where ,

• P = Principal
• SI = Simple Interest
• t = Time
• R = Rate of interest