Question

find the sum on which the difference between the simple interest and and compound interest at the rate of 8% per annum compounded annually rupees 34 into 2 year ​

Answer 1

Given

Rate of interest = 8% p.a.

Time = 2 years.

Difference b/w CI & SI = Rs.34

To find

Sum of money

Solution

Formulas used :

➼ Simple interest = (P × r × n)/100

➼ Compound interest = P(1 + r)ⁿ - P

According to Question :

⇒ [P(1 + r)ⁿ - P] - [(P × r × n)/100] = 34

⇒ [P(1 + 8%)² - P] - [(P × 8 × 2)]/100 = 34

⇒ [P(1 + 0.08)² - P] - [(16P)/100] = 34

⇒ [P(1.08)² - P] - [16P/100] = 34

⇒ [P × 1.1664 - P] - [16P/100] = 34

⇒ [1.1664P - P] - [16P/100] = 34

⇒ [1664P/10000] - [16P/100] = 34

⇒ [1664P - 1600P]/10000 = 34

⇒ 64P = 34 × 10000

⇒ 64P = 340000

⇒ P = 340000/64

⇒ P = Rs.5312.5

Therefore,

Sum of money = Rs. 5312.5

Answer 2

Answer:

• Difference b/w CI & SI = Rs. 34
• Time = 2 years
• Rate = 8% p.a.
• Principal = ?

$\rule{100}{1}$

$\underline{\bigstar\:\textsf{Difference b/w CI \& SI for 2 years :}}$

$:\implies\sf Difference=Principal \times \Bigg\lgroup\dfrac{Rate}{100}\Bigg\rgroup^2\\\\\\:\implies\sf 34 = Principal \times \Bigg\lgroup\dfrac{8}{100}\Bigg\rgroup^2\\\\\\:\implies\sf 34 = Principal \times \dfrac{8}{100} \times \dfrac{8}{100}\\\\\\:\implies\sf \dfrac{34 \times 100 \times 100}{8 \times 8} = Principal\\\\\\:\implies\sf \dfrac{340000}{64} = Principal\\\\\\:\implies\underline{\boxed{\sf Principal= Rs.\:5312.50}}$

⠀

$\therefore\:\underline{\textsf{Hence, Principal will be \textbf{Rs. 5,312.50}}}.$