Question

find the difference between simple interest and compound interest for 3 year on 30,000 at 15 ℅ per annum when the interest is compounded annually​

Answer 1

Answer:

si =13,500

Explanation:

p= 30,000

R=15%

T=3years

SI = P×R×T/100

= 30,000×15×3/100

=13,500

Now, compound Interesrt

Answer 2

S O L U T I O N :

$\underline{\bf{Given\::}}$

• Principal, (P) = Rs.30000
• Rate, (R) = 15% p.a
• Time, (T), (n) = 3 years

$\underline{\bf{Explanation\::}}$

$\underbrace{\bf{1^{st}\:Case\::}}$

We get simple Interest according to the question.

As we know that formula of the S.I;

$\boxed{\bf{Simple\:Interest=\frac{P\times R \times T}{100} }}$

$\mapsto\tt{S.I. = \dfrac{P \times R \times T}{100} }$

$\mapsto\tt{S.I. = \dfrac{30000 \times 15 \times 3}{100} }$

$\mapsto\tt{S.I. = \dfrac{300\cancel{00} \times 15 \times 3}{\cancel{100}} }$

$\mapsto\tt{S.I. = Rs.(300 \times 15 \times 3)}$

$\mapsto\bf{S.I. = Rs.13500}$

$\underbrace{\bf{2^{nd}\:Case\::}}$

As we know that formula of the compounded annually;

$\boxed{\bf{Amount = Principal\bigg(1+\frac{R}{100} \bigg)^{n}}}$

$\mapsto\tt{Amount = 30000\bigg(1+\dfrac{15}{100} \bigg)^{3}}$

$\mapsto\tt{Amount = 30000\bigg(1+\cancel{\dfrac{15}{100}} \bigg)^{3}}$

$\mapsto\tt{Amount = 30000\bigg(1+\dfrac{3}{20} \bigg)^{3}}$

$\mapsto\tt{Amount = 30000\bigg(\dfrac{20+3}{20} \bigg)^{3}}$

$\mapsto\tt{Amount = 30000\bigg(\dfrac{23}{20} \bigg)^{3}}$

$\mapsto\tt{Amount = 30000 \times \dfrac{23}{20} \times \dfrac{23}{20} \times \dfrac{23}{20} }$

$\mapsto\tt{Amount = \cancel{30000 }\times \dfrac{23}{\cancel{20}} \times \dfrac{23}{\cancel{20}} \times \dfrac{23}{\cancel{20}} }$

$\mapsto\tt{Amount = Rs.(3.75 \times 23 \times 23 \times 23)}$

$\mapsto\bf{Amount = Rs.45626.25}$

Now, as we know that Compound Interest;

→ C.I. = Amount - Principal

→ C.I. = Rs.45626.25 - Rs.30000

→ C.I. = Rs.15626.25

Now,

→ Difference = Compound Interest - Simple Interest

→ Difference = Rs.15626.25 - Rs.13500

→ Difference = Rs.2126.25