Aptitude - Compound Interest - Discussion

Can any one provide shortcut for difference (CI-SI) questions for more than two years?

Apply below simple formula and get the answer easily.

Difference = P(R)^2/(100)^2.

C.P - S.I
[p(1-r/100)^n-p ]-[(P*R*N)/100] = 96.

Where,
p=1500, n=2, R=?, C.P and S.I difference is 96.

[15000(1-r/100)^2-15000 ]-[(1500*R*2)/100] = 96 ...(1).

Now, 15000 common in above so take it out,

15000[1(1-r/100)^2 - 1 ]-[(1*R*2)/100] = 96 ...(2).

Take L.C.M. we get,

15000[(100 + R)^2 - 10000 - (200 x R)/10000] = 96 ...(3).

Where (100 + R)^2 is like (a + b)^2 = (a^2 + b^2 + 2*a*b).

(100^2 + R^2 + 2*100*R) = (10000 + R^2 + 200R).

Sub (10000 + R^2 + 200R) in eqn(3) we get,

15000[ (10000 + R^2 + 200R) - 10000 - (200 x R)/10000] = 96.

15000[ (10000 + R^2 + 200R - 10000 - 200R) /10000] = 96.

Now 10000 and 200R is cancelled, remaining.

15000[ R^2 /10000] = 96.

[15000R^2/10000] = 96.

[15R^2/10] = 96.

divide by 5 we get,

[3R^2/2] = 96.

Now keep the R^2 in left hand side and move 3/2 to right side

R^2 = 96(2/3) // 96/3 = 32.

R^2 = 32*2.

R^2 = 64 // square root of 64 is 8.

R = 8%.

In 4th step how they done R^2=64 ?

If we solve it there should be R^2-200R=64 and this is a quadratic equation. Can anyone explain me how to solve this equation ?

It's simply p(r/100)^2
p=1500,r=x,t=2yrs
Thus
15000(x/100)^2 it finally comes out like
x^2=64
x^2=8^2
x=8
Therefore rate=8%

Thanks prawin.

shortcut:
p = d (100/r)^2

15000 = 96*100*100/r^2

r^2 = 96*100*100/15000

r = 8.

SIMPLE FORMULA FOR TWO YEARS
C.I.-S.I.=P(R/100)*(R/100)
96=15000(R/100)(R/100)
R*R=960/15=64
R=8

Kindly explain in 3rd step howcum -1 would be showing...is this problem solve throuGh the LCM?

Please tell something more about compound interest.