Question

in how many years will a sum of Rs4000 amount to Rs5290 at 15% per annum if the interest is compounded annually?​

Answer 1

Given: A sum of Rs. 4000 amount to Rs. 5290 at 15% per annum.

To find: In how many years will a sum of Rs. 4000 amount to Rs. 5290 at 15% per annum if the interest is compounded annually?

Solution:-

According to the given information, we have been given that,

• Principal amount, P = Rs. 4000
• Final amount, A = Rs. 5290
• Rate of interest, R = 15%
• Time period, T = ?

We can find calculate the time period by using the final amount formula. The final amount formula is given by,

• A = P(1 + R)ᵀ

By substituting the given values in the formula, we get the following results:

⇒ 5290 = 4000(1 + 15/100)ᵀ

⇒ 5290 = 4000(1 + 0.15)ᵀ

⇒ 5290 = 4000(1.15)ᵀ

⇒ 5290/4000 = 1.15ᵀ

⇒ 1.3225 = 1.15ᵀ

⇒ 1.15² = 1.15ᵀ

⇒ T = 2

Hence, the time period is 2 years.

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Answer 2

Answer:

Given :-

• A sum of Rs 4000 amount to Rs 5290 at 15% per annum if the interest is compounded annually.

To Find :-

• What is the time period.

Formula Used :-

$\clubsuit$ Amount Formula :

$\longrightarrow \sf\boxed{\bold{\pink{A =\: P\bigg(1 + \dfrac{r}{100}\bigg)^n}}}$

Solution :-

Given :

• Principal = Rs 4000
• Amount = Rs 5290
• Rate of Interest = 15% per annum

According to the question by using the formula we get,

$\dashrightarrow \sf\bold{\purple{A =\: P\bigg(1 + \dfrac{r}{100}\bigg)^n}}$

By putting :

• A = Rs 5290
• P = Rs 4000
• r = 15% per annum

$\implies \sf 5290 =\: 4000\bigg(1 + \dfrac{15}{100}\bigg)^n$

$\implies \sf 5290 =\: 4000\bigg(\dfrac{100 + 15}{100}\bigg)^n$

$\implies \sf 5290 =\: 4000\bigg(\dfrac{115}{100}\bigg)^n$

$\implies \sf \dfrac{529\cancel{0}}{400\cancel{0}} =\: \bigg(\dfrac{115}{100}\bigg)^n$

$\implies \sf \dfrac{529}{400} =\: \bigg(\dfrac{\cancel{115}}{\cancel{100}}\bigg)^n$

$\implies \sf \bigg(\cancel{\dfrac{23}{20}}\bigg)^2 =\: \bigg(\cancel{\dfrac{23}{20}}\bigg)^n$

$\implies \sf 2 =\: n$

$\implies \sf\bold{\red{n =\: 2\: years}}$

$\therefore$ The time period is 2 years .