Question

If the difference between the simple interest and the compound interest on a sum of money for two years at 4% per annum is Rs. 800, find the sum.​

Answer 1

Answer:

• When Time is 2 years and Principal, Rate is Same. Then Use this Formula.

$\bigstar\:\boxed{\sf Difference = P\bigg(\dfrac{r}{100}\bigg)^2}$

$\rule{150}{1}$

$\underline{\bigstar\:\:\textsf{According to the Question :}}$

$:\implies\sf Difference = P\bigg(\dfrac{r}{100}\bigg)^2\\\\\\:\implies\sf 800 = P\bigg(\dfrac{4}{100}\bigg)^2\\\\\\:\implies\sf 800 = P\bigg(\dfrac{1}{25}\bigg)^2\\\\\\:\implies\sf 800 = P \times\dfrac{1}{625}\\\\\\:\implies\sf 800 \times 625= P\\\\\\:\implies\sf\underline{\boxed{\textsf{\textbf{P = Rs. 500,000}}}}$

⠀

$\therefore\:\underline{\textsf{Therefore, Sum will be \textbf{Rs. 500,000}}}.$

Answer 2

Solution:-

$\frak{We \: have }\begin{cases}\sf{Time (n) = 2 \: years}\\\sf{Rate \: of \: interest(r) = 4\% \: p.a.}\\\sf{Difference \: b/w \: CI \: \& \: SI = Rs.800}\\\sf\red{Sum \: of \: money = ?}\end{cases}$

Now we know formula:-

☛ Simple interest = Prn

→ Simple interest = P × 4% × 2

→ Simple interest = P × (4/100) × 2

→ Simple interest = P × (1/25) × 2

→ Simple interest = 2P/25

→ Simple interest = 0.08P

$\rule{200}{1}$

We also know,

☛ CI = P(1 + r/100)ⁿ - P

→ CI = P(1 + 4/100)² - P

→ CI = P(1 + 1/25)² - P

→ CI = P [(25 + 1)/25 ]² - P

→ CI = P(26/25)² - P

→ CI = P(1.04)² - P

→ CI = P × 1.0816 - P

→ CI = 1.0816P - P

→ CI = 0.0816P

$\rule{200}{1}$

A/q,

→ 0.0816P - 0.08P = 800

→ 0.0016P = 800

→ P = 800/0.0016

→ P = 500,000

Therefore,

$\therefore\underline{\boxed{\textsf{Sum of money = {\textbf{Rs.500,000 }}}}}$