Question

find the difference between the compound interest and simple interest on a sum of rs 16000 for 3 years if the rate of interest is 2%per annum

Answer 1

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Here is your answer
P = Rs 16000
R = 2%p.a
T = 3 years
SI = PRT/100
$= \frac{16000 \times 2 \times 3}{100} \\ = 960$
Simple interest is Rs 960

Now
For Compound Interest
A = P(1+ R/100)^n
$= 16000 \times (1 + \frac{2}{100} ) {}^{3} \\ = 16000 \times (1 + \frac{1}{50} ) {}^{3} \\ = 16000 \times ( \frac{51}{50} ) {}^{3} \\ = \frac{16000 \times 51 \times 51 \times 51}{50 \times 50 \times 50} \\ = 16979.328$
Therefore Compound Interest is
= A - P
= 16,979.328 - 16000
= Rs 979.328

Now The difference between Compound Interest and simple interest is
= 979.328 - 960
= Rs 19.328


NOW FOR YOUR SECOND QUESTION PLEASE
CHECK

P = 3000
R = 10% p.a
= 10/4 % for Compound quarterly
T = 1 years
n = 1× 4 Compounded quarterly

Now
A = P (1+ R/100)^n
$= 3000 \times (1 + \frac{10}{4 \times 100} ) {}^{4} \\ = 3000 \times (1 + \frac{1}{40} ) {}^{4} \\ = 3000 \times ( \frac{41}{40} ) {}^{4} \\ = \frac{3000 \times 41 \times 41 \times 41 \times 41}{40 \times 40 \times 40 \times 40} \\ = 3311.43$
So Compound Interest is
= A - P
= 3311.43 - 3000
= Rs 311.43

HOPE YOU CHECK BOTH ANSWER


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Answer 2

Hi friends

Here is your answer
SI = PRT/100
$= \frac{16000 \times 3 \times 2}{100} \\ = 960$
Now
A = P(1+R/100)^n
$= 16000 \times (1 + \frac{2}{100} ) {}^{3} \\ = 16000 \times ( \frac{51}{50} ) {}^{3} \\ = 16979.3$
CI = 16979.3 - 16000
= 979.3

Difference is
= 979.3 - 960
= 19.3

Hope it helps

@MSD