Question

5. Find the difference between simple interest and compound interest for the sum of
25700 at the rate of 10% p.a. for 1 year.​

Answer 1

Given :-

• Principal = ₹25700
• Rate = 10% per annum
• Time = 1 year

Aim :-

• To find the difference between simple interest and compound interest

Simple interest :-

Formula to use :-

$\longrightarrow \sf{simple \: interest \: = \dfrac{principal \times \: rate \times time}{100} }$

Substituting the values,

$\implies \: \sf{simple \: interest = \cfrac{25700 \times 10 \times 1}{100} }$

$\implies \sf{simple \: interest = \dfrac{257 \not0 \not0 \times 10 \times 1}{1 \not0 \not0} }$

$\implies \sf{simple \: interest = rs.2570}$

Compound interest :-

In order to find the compound interest, we first have to find the Amount.

Formula to use :-

$\longrightarrow \sf{amount = principal \bigg(1 + \dfrac{rate}{100} \bigg)^{time} }$

$\longrightarrow \sf{compound \: interest = amount \: - principal}$

Substituting the values,

$\implies \sf{amount = 25700 \bigg(1 + \dfrac{10}{100} \bigg) ^{1} }$

Taking LCM as 100, and adding the terms inside the brackets,

$\implies \sf{25700 \bigg( \dfrac{100 + 10}{100} \bigg) ^{1} }$

Adding,

$\implies \sf{25700 \bigg( \dfrac{110}{100} \bigg)^{1} }$

Reducing to the lowest terms,

$\implies \sf{25700 \bigg(\dfrac{11 \not0}{1 0 \not0} \bigg)^{1} }$

$\implies \sf{25700 \bigg( \dfrac{11}{10} \bigg) ^{1} }$

$\implies \sf{25700 \times \dfrac{11}{10} }$

Cancelling,

$\implies \sf{2570 \not0 \times \dfrac{11}{1 \not0} }$

$\implies \sf{2570 \times 11}$

$\implies \sf rs.{ 28,270}$

Therefore amount = ₹28270.

Compound interest :-

$\implies \sf{28270 - 25700}$

$\implies \sf rs.{2570}$

Difference :-

$\implies \sf{2570 - 2570}$

$\implies \sf{rs.0}$

Some more formulas :-

• When interest is compounded half-yearly

$\longrightarrow \sf{amount = principal \bigg(1 + \dfrac{rate}{200} \bigg)^{2 \times time} }$

• When interest is compounded quarterly

$\longrightarrow \sf{amount = principal \bigg(1 + \dfrac{rate}{400} \bigg)^{4 \times time}}$