Question

at what rate percent per annum will 64 amount to 125 in 3 years a compound interest, interst being compounded annually​

Answer 1

use formual amount= p(1+r/100)^n

p =principal

n=no of years

r=rate of interest

125=64(1+r/100)^3r=25

Amount = P(1+r/100)^n where r is annual rate of interst in percent.125 = 64(1+r/100)^3125/64

=(1+r100)^3(1+r100)=5/4100+r=125r=125-100=25

so rate of interest = 25%

Answer 2

GIVEN :-

• Principal ( P ) = Rs. 64
• Amount ( A ) = Rs. 125
• Time ( T ) = 3 years.

TO FIND :-

• The rate ( R ).

SOLUTION :-

As we know that,

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

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

Now substitute the given values,

$\\ \\ : \implies \displaystyle \sf \: \dfrac{125}{64} = \bigg(1 + \dfrac{R}{100}\bigg)^{3}$

$: \implies \displaystyle \sf \: \bigg( \dfrac{5}{4} \bigg) ^{3} = \bigg(1 + \dfrac{R}{100}\bigg)^{3}$

On comparing both the sides we get,

$\\ \\ : \implies \displaystyle \sf \: \dfrac{5}{4} = 1 + \dfrac{R}{100}$

$: \implies \displaystyle \sf \: \dfrac{5}{4} - 1= \dfrac{R}{100}$

$: \implies \displaystyle \sf \: \dfrac{5 - 4}{4} = \dfrac{R}{100}$

$: \implies \displaystyle \sf \: \dfrac{1}{4} = \dfrac{R}{100}$

$: \implies \displaystyle \sf \: \dfrac{1}{4} \times 100 = R$

$: \implies \underline{ \boxed{ \displaystyle \sf \: Rate = 25\%}}$