Question

.In how many years Rs.700 will amount to Rs.847 at a compound interest rate 10 p.c.p.a

Answer 1

Answer:

• In 2 years ₹ 700 will amount to ₹ 847 at a rate of 10%.

Step-by-step explanation:

Given that:

• Principal = ₹ 700
• Amount = ₹ 847
• Rate = 10%

To Find:

• Time

As, we know that:

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

Where,

• A = Amount
• P = Principal
• R = Rate
• n = Time

Substituting the values,

$\sf{847 = 700\bigg(1+\dfrac{10}{100}\bigg) ^n}\\\\\sf{\longrightarrow \dfrac{847}{700}= \bigg(\dfrac{100 + 10}{100}\bigg)^n}\\\\\sf{\longrightarrow \dfrac{847}{700}= \bigg(\dfrac{110}{100}\bigg)^n}\\\\\sf{\longrightarrow \dfrac{121}{100}= \bigg(\dfrac{11}{10}\bigg)^n}\\\\\sf{\longrightarrow \dfrac{11\times11}{10\times10}= \bigg(\dfrac{11}{10}\bigg)^n}\\\sf{\longrightarrow \bigg(\dfrac{11}{10}\bigg)^2 = \bigg(\dfrac{11}{10}\bigg)^n}\\\\\sf{\longrightarrow \; 2= n}$

Hence, time = 2 years

More Formulas:

• CI = A - P

When the interest is compounded annually but rates are different for different years:

$\sf{P \times\bigg(1+\dfrac{a}{100}\bigg)\times\bigg(1+\dfrac{b}{100}\bigg)}$

Where,

• a is rate for 1st year.
• b is rate for 2nd year.

This formula can be extended for any numbers of years.

Eg:-

$\sf{P \times\bigg(1+\dfrac{a}{100}\bigg)\times\bigg(1+\dfrac{b}{100}\bigg)...\bigg(1+\dfrac{\infty}{100}\bigg)}$

When time is given in fraction:

Suppose time is 1 ½ years. Then,

$\sf{A = P\times \bigg(1+\dfrac{R}{100}\bigg)^1\times\Bigg(1+\dfrac{\dfrac{1}{2}\times R}{100} \Bigg)}$

Answer 2

Answer:

Please see the attached file for the answer