On a certain sum, the compound interest accrued in the first two years is 5520 and that in the first three years is 9576. What is the rate of interest?

Question

On a certain sum, the compound interest accrued in the   first two years is 5520 and that in the first three years is  9576. What is the rate of interest?

hartnn · Answer

@neta Hi :)
$\huge \color{red}{\text{Welcome to Open Study}}\ddot\smile$

Do you know this formula for compound interest ?
$\large F = P (1+r )^n $
you can use this, by putting n =3 first; then n=2, and just divide.

anonymous · Answer

f means

hartnn · Answer

oh sorry, i forgot to explain the terms ,
"F" means final amount after 'n' years,
'P' means principal amount, and 'r' means rate of interest in decimal.

hartnn · Answer

since you are given compound interest,  say I,
you can use, 
$\large I = F-P = P (1+r )^n-P=P[(1+r^n)-1]$

anonymous · Answer

k i will try

hartnn · Answer

good :) let me know if you couldn't do it.

anonymous · Answer

but here we dont know two things p and r

anonymous · Answer

i have consider two equations

hartnn · Answer

sorry for my typo, it should be 
$\large I = F-P = P (1+r )^n-P=P[(1+r)^n-1]$
also, you'll have 2 equations, once when you put n=2 and then n=3
if you divide those ,P gets eliminated

anonymous · Answer

p(1+r)2-p(1+r)3=4056

hartnn · Answer

you have subtracted .
try dividing the equations, P will get cancelled.

hartnn · Answer

and you'll just be left with an equation in r

anonymous · Answer

k

anonymous · Answer

but we have -p

hartnn · Answer

$\huge \dfrac{I_1}{I_2}=\dfrac{P [(1+r)^{n_1}-1]}{P[(1+r)^{n_2}-1]}=\dfrac{ [(1+r)^{n_1}-1]}{[(1+r)^{n_2}-1]}$

anonymous · Answer

r=1.734%

anonymous · Answer

i think i made mistake in divison

hartnn · Answer

yeah, try again...

anonymous · Answer

k thanks a lot

hartnn · Answer

welcome ^_^