Question

On a sum of money, the simple interest for 2 years is Rs.660, while compound interest is Rs.696.30, the rate of interest being the same in both the cases. The rate of interest is?

Answer 1

Heya friend,

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Let ₹P be the principal and R be rate of interest. Then,

S.I. = P×R×T/100

=> 660 = P×R×2/100

=> 660 = P×R/50

=> 660×50 = P×R

=> 33,000 = P×R

=> P = 33,000


C.I. = P{(1+R/100)^n - 1}

=> 696.30 = 33,000/R{(100+R/100)^2 - 1}

=> 69,630R/100×33,000 = {(100)^2 + (R)^2 + 2×100×R/(100)^2 - 1}

=> 211R/10,000 = {10,000 + (R)^2 + 200R - 10,000/10,000}

=> 211R×10,000/10,0000 = (R)^2 + 200R

=> 211R = R(R+200)

=> 211R/R = R+200

=> 211 = R+200

=> 211-200 = R

=> R = 11 % per annum

Hence, the rate of interest is 11 % per annum.


Thanks

With regards@

Tanisha

Answer 2

Given:

SI of two years $=Rs\ 660$

Formula for SI is

$S.I.=\frac{P\times R\times T}{100}$

where P is principle, T is time and R is rate of interest

So,   $S.I.=\frac{P\times R\times 2}{100}$

$660=\frac{P\times R\times2}{100}$

$P\times R=33000$

$P=\frac{33000}{R}$

Now, Compound Interest for two years

$C.I.=P[({\frac{1+R}{100} )^n-1]$

$696.30=P[(\frac{1+R}{100})^2-1]$

$696.30=\frac{33000}{R} [(\frac{1+R}{100})^2-1]$

Solving the above equation, we get

$R=11\%$

So the rate of interest will be $11\%$

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