Question

in how many years will rupees 4000 amount to rupees 5324 @ 10% per annum compounded annually​

Answer 1

hello

CI=p {(1+r)^n-1}

5324-4000=4000 {(1+10/100)^n}-4000

5324=4000 (1+1/10)^n

5324/4000=(11/10)^n

1331/1000=(11/10)^n

(11/10)^3=(11/10)^n

n=3 

hence 3 years

Answer 2

$\mathfrak{Answer:}$

= 3 years.

$\mathfrak{Step-by-Step\:Explanation:}$

$\underline{\bold{Given\:in\:the\:Question:}}$

• Principle = P = Rs. 4000.
• Amount = A = Rs. 5324.
• Rate = r = 10%.
• Interest is compounded annually.

$\bold{Solution:}$

Let the time be n years.

We know that ,

$\boxed{\bold{Amount=P\left(1+\dfrac{r}{100}\right)^n}}\\\\\\\mathfrak{According\:to\:question:}\\\\\\\implies\tt{5324=4000\left(1+\dfrac{10}{100}\right)^n}\\\\\\\implies\tt{\dfrac{5324}{4000}=\left(\dfrac{11}{10}\right)^n}\\\\\\\implies\tt{\dfrac{1331}{1000}=\left(\dfrac{11}{10}\right)^n}\\\\\\\implies\tt{\left(\dfrac{11}{10}\right)^3=\left(\dfrac{11}{10}\right)^n}\\\\\\\implies\tt{n=3.}\\\\\\\\\boxed{\boxed{\bold{Time=3\:years.}}}$