Aptitude - Compound Interest - Discussion

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?

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

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.

SIMPLE STEP.

FORMULA :
Diff b/w SI and CI when T = 2yrs.

P = (D * 100^2)/R^2.
D =RS 1.
R = 4%.
SO, Answer is 625.


FOR DIFF B/W CI AND SI FOR 3 YRS
FORMULA

P = (D * 100^3)/R^2(300 + R).

Can anyone please explain to me how we get 51/625 x in the first answer?

Interest of interest of amount is difference between SI and CI at rate given,

4% of x = 0.04x,
4% of 0.04x = (0.04x)4/100 = 1,
x = 625.

We know that,

(C.I - S.I) = P * R * R/100 * 100.
HERE, 1 = P * 4 * 4/100 * 100.

P = 25 * 25 = 625.

Well said, Thank you @Muthuu.

Hi, one thing I could not understand. In question, it is given that the difference between simple and compound. Not difference between compound and simple. These two questions are distinct in the first, the answer will be in minus. So how can it possible? So please help me please.

Assume principal 100.

Then, CI=8.16 & SI=8.00,
1÷ 0.16 * 100 = 625.