Question

find the compound interest and amount.
please answer these questions

Answer 1

$\bold{\large{\underline{\underline{Answer\: 5 \: \colon}}}}$

P = ₹ 62,500

R = 12% p.a.

T = 2 years 6 month

t = 2 $\dfrac{1}{2}$ year

$\bold{\large{\underline{\underline{Formulae\: \colon}}}}$

$\implies \bold{ A \: = \: P\: \Big(1\: + \: \dfrac{R}{100}\Big)^{n} \times \Big(1\: +\: \dfrac{\dfrac{1}{2}R}{100}\Big) }$

$\implies \bold{ A \: = \: 62500\: \Big(1\: + \: \dfrac{12}{100}\Big)^{2} \times \Big(1\: +\: \dfrac{\dfrac{1}{2} \times 12}{100}\Big) }$

$\implies \bold{ A \: = \: 62500 \: \times \: (1\: + \: 0.12)^{2} \: \times \: (1+0.06)}$

$\implies \bold{ A \: = \: Rs\: 83,104}$

Interest amount ₹ 20,604

$\bold{\large{\underline{\underline{Answer\: 6 \: \colon}}}}$

P = ₹ 9,000

R = 10% p.a.

T = 2 years 4 month

t = 2 $\dfrac{1}{3}$ year

$\bold{\large{\underline{\underline{Formulae\: \colon}}}}$

$\implies \bold{ A \: = \: P\: \Big(1\: + \: \dfrac{R}{100}\Big)^{n} \times \Big(1\: +\: \dfrac{\dfrac{1}{3}R}{100}\Big) }$

$\implies \bold{ A \: = \: 9,000\: \Big(1\: + \: \dfrac{10}{100}\Big)^{2} \times \Big(1\: +\: \dfrac{\dfrac{1}{3} \times 10}{100}\Big) }$

$\implies \bold{ A \: = \: 9,000 \: \times \: (1\: + \: 0.1)^{2} \: \times \: (1+0.033)}$

$\implies \bold{ A \: = \: Rs\:11249.37 }$

$\bold{\large{Thanks}}$

Interest amount ₹ 2,249.37