Interest for long years

A=P(1+r)^t

6000=4800(1+r)^4

5=4(1+r)^4

Cubing both sides,

125=64(1+r)^12

Multiplying both sides by 75,

9375= 4800(1+r)^12

The compound interest for 4 years is 6000/4800= 5/4

Money in bank: -After 8 years: 6000× 5/4 = 7500 -After 12years: 7500× 5/4 = 9375

*After 12 years Manish will get Rs9375

just now Edit comment 1 1 Farhan Rasool

let M be the money at any time t. at t=0 M=4800 at t=4 M=6000 at t=12 M=? By given condition derivative with respect to "t" is D(M)=k where k is a constant Now after integration we get M=C(exp kt) at t=0 M=4800 we get C=4800 at t=4 M=6000 we get k=0.055 now M=4800(exp 0.055t) Now at t=12 we get M=9375 Ans.

Though can be solved with Compound Int. formula, it can be solved without using formula.

Since 4800 becomes 6000 in 4 years, in next 4 years 6000 would become 6000*6000/4800=7500

Since 4800 becomes 6000 in 4 years, in next 4 years 7500 would become 7500*6000/4800=9375

r^4= 6000/4800 =1.25

r^12=(r^4)^3=1.25^3

4800*(1.25^3)= 9375

Let rate of interest be some percentage, so after four years we multiply the principle by the interest rate scale factor 'x', 48 times for each month.

So after twelve years we multiply by the scale factor 'x^48' 3 times.

Simple algebra from here:

x^48 = 6000/4800 = 5/4

( x^48)^3 = 125/64

So final amount = 125/64*4800 = 9375.

It is much simpler. 4800 became 6000 in 4 years thus (1+r)^4 is 1.25. Thus every 4 years the Compund interest is calculated as 1.25

4800 x 1.25 x 1.25 x 1.25. = 9375.

I=4800, N=6000 t=5, k=0.0559 N=Ie^-kt if t=12 N=4800e^-(0.0559)(12) N=9375

just apply the formula..A=P(1+r)^t