Aptitude - Compound Interest - Discussion

Take as, C.I - S.I = Rs.1.

Then, for C.I ->use the Formula[x + y + {(xy)/100}].

For x&y ->put R% value( i.e.,) X=4 &y=4.

By this ,we get [4+4+{4*4/100}] = 8.15.
For S.I -> R * T = 4 * 2 = 8.
So , 8.15 - 8 = RS.1 => 0.16=1.

By this, 0.16% = 1 means,then 100%=?
So, cross multiply it
i.e., 100 * 1/0.16 = 625.

I am just trying to explain briefly, hence the sum looks like long . But you all just write the numerical step and look it .
It just takes few secs only try it.

(SI -CI) for 2 year = (r/100)^2 * principal.

Can anybody tell me the formula for finding rate and principle when compound interest and simple interest are given with time 2 years and SI = 800 and CI = 850?

Solution for CI on certain amount for 2 years is 2200 and 3 years is 3640. Find the rate?

Guys please help me.

CI for 2 years - SI for 2 years = PR^2/100^2.
1 = P16/10000,
10000/16 = P,
P = 625.

How we derive the shortcut trick for 2 or 3 years for the sum of money if the difference given b/w CI and SI? Anyone help me to clear these shortcut to improve myself. Please.

We can use the direct formula for this one.

The difference after 2 years = PR^2/100^2.

Here, P - Principal, R - Interest of Rate.

@Gagan

SI = P * (n * r)/100.
CI = P((1 + r/100)^n - 1).

Now simplify both equations such that P comes to the left and rest to the right equate both to get r.

Very simple formula is;

p.(R/100)^2 = (compound interest - Simple interest) time duration 2 years.

Why not R/2?