Question

a man invested ₹1000 for three years at 11% simple interest per annum and ₹1000 at 10% compounded interest per annum compounded annualy for three years Find which investment is better​

Answer 1

Answer:

The investment on compound interest is beneficial for the man.

Step-by-step explanation:

Given :

Rs. 1000 were invested for 3 years, 11% Rate p.a on Simple Interest.

Rs. 1000 were invested for 3 years, 10% rate p.a compounded annually.

To find :

Which investment was better.

Solution :

$\textsf{\underline{\underline{Investment I -}}}$

Here :

• Principal = Rs.1000
• Rate = 11%
• Time = 3 years
• SI = ??

$\bigstar \: {\boxed{\sf{SI = \frac{Principal \times Rate \times Time}{100}}}}$

$\sf{\implies} \: SI =\dfrac{Principal \times Rate \times Time}{100} \\ \\ \sf{\implies} \: SI = \dfrac{1000 \times 11 \times 3}{100} \\ \\ \sf{\implies} \: SI = \dfrac{10 \times 11 \times 3}{1} \\ \\ \sf{\implies} \: SI = 10 \times 33 \\ \\ \sf{\implies} \: SI = 330$

Interest = Rs. 330

$\rule{300}{1.5}$

$\textsf{\underline{\underline{Investment II -}}}$

Here :

• Principal = 1000
• Rate = 10%
• Number of times interest applied = 3

$\bigstar \: {\boxed{\sf{A= P\left( 1 + \frac{R}{100}\right)^{N}}}}$

$\sf{\implies} \:A= 1000\left( 1 + \dfrac{10}{100}\right)^{3} \\ \\ \sf{\implies} \:A= 1000\left(\dfrac{110}{100}\right)^{3} \\ \\ \sf{\implies} \:A=1000 \times \dfrac{110}{100} \times \dfrac{110}{100} \times \dfrac{110}{100} \\ \\ \sf{\implies} \:A=1 \times 11 \times 11 \times 11 \\ \\ \sf{\implies} \:A=1331$

Amount = Rs. 1331

$\rule{300}{1.5}$

$\bigstar \: \boxed{\sf{CI= Amount - Principal}}$

$\sf{\implies} \:CI=1331 - 1000 \\ \\ \sf{\implies} \:CI=331$

Interest = Rs. 331

$\rule{300}{1.5}$

331 > 330

Investment I < Investment II

$\therefore$ The investment on compound interest is beneficial for the man.