Aptitude - Compound Interest - Discussion

Hai I'm prudhvi. Can any one help me out?

Why are we using 676/625x in the step one?

Please help me.

Hello friends!

The question is very easy . First know the diff b/w S.I. and C.I.

S.I. = P*T*R/100
Amount = S.I. + P

C.I. = Amount - P
Amount = P*((1 +(R/100))^ n) where (n= no of years)

Given that S.I. and C.I diff is 1rupee

and also given T=time =2 yrs , R=rate = 4%,
and he asked P ie Principal Amount.


So S.I. = P*2*4/100; = P*2/25;
C.I. = Amount - P

= P*(1+ (4/100))^2 - P

= P*(1 + (1/25))*2 - P
= P*((26/25)^2) - p
= P*(676/625) - p

= (P*676-P*625)/625
= (P*51)/625

WE KNOW S.I. = (P*2)/25

AND S.I.-C.I. = 1
HENCE
(P*2)/25 - (P*51)/625 = 1
BY SOLVING U GET
P = 625 RS

How to solve very simple?

Please any one tel me how to solve it in simple method.

shotcut method:
If you give the difference between si and ci for 2 years means,
4*4/100
=.16
1/.16*100
100/16*100
100/4*25
625

For beginners those are asking why they used -x in step one.

Let me explain you clearly.

First take Amount=Rs.5000 and Rate=2% Years=1

Then calculate SI, SI=PNR/100

So, SI=5000*1*2/100=RS100

[Here you are getting only interest]

Now calculate CI, CI=P(1+r)^n

So, CI=5000*(1+2/100)^1=RS5100

[Here we are getting interest+Principal amount]

That's why there are using -x in the formula.

If you use -x(for our example its P) we will get the interest only.

Shortcut method

Difference = (D*R^2)/100^2 for 2 year

Difference = {(D*R^2)(300+R)}/100^3 for 3 year.

@sudharshan.

Rightly said, cleared because of you.

Shortcut Method

sum = difference*(100/R)^2 for 2 year

sum = {difference*(100^3)}/{R^2*(300+R)} for 3 year

A=P( 1+R/100)
C.I=A-P
DIFFERENCE OF CI AND S.I FOR X YEAR= P*(R/100)^X