Question

Find the compound interest and the amount after 3 years on a principal of rupees 15,000 at 10% .​

Answer 1

$\sf :\implies Compound \: interest = P(1 + \dfrac{R}{100})^{n} \\ \\ \sf :\implies Compound \: interest = 15000(1 + \dfrac{10}{100})^{3} \\ \\ \sf :\implies Compound \: interest = 15000(1 + \dfrac{1}{10})^{3} \\ \\ \sf :\implies Compound \: interest = 15000(1 + 0.1)^{3} \\ \\ \sf :\implies Compound \: interest = 15000(1.1)^{3} \\ \\ \sf :\implies Compound \: interest = 15000(1.1)(1.1)(1.1) \\ \\ \sf :\implies Compound \: interest = 15000 \times 1.1 \times 1.1 \times 1.1\\ \\ \sf :\implies Compound \: interest = 15000 \times 1.21 \times 1.1\\ \\ \sf :\implies Compound \: interest = 15000 \times 1.331\\ \\ \sf :\implies Compound \: interest = 19965 \: rupees$

Answer 2

$\huge\tt{\underline\red{Answer}:-}$

• Pяiиϲiραℓ = Rѕ. 15,000

• Rατє = 10%

• Tiмє = 3 γяѕ.

• ϲοмρυи∂ iиτєяєѕτ =Aмουиτ-Pяiиϲiραℓ

• Aмουиτ = $P(1 + \frac{r}{100})^{n}$

⠀⠀⠀⠀ ⠀⠀= $15000(1 + \frac{10}{100})^{3}$

⠀⠀⠀⠀ ⠀⠀= $15000(\frac{110}{100})^{3}$

⠀⠀⠀⠀ ⠀⠀= $15000 \times \frac{110}{100} \times \frac{110}{100} \times \frac{110}{100}$

⠀⠀⠀⠀ ⠀⠀= $15\times11\times11\times11$

⠀⠀⠀⠀ ⠀⠀= 19965 ${Ans.}$

• ϲοмρυи∂ iиτєяєѕτ = 19965 - 15000

⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀ ⠀⠀= 4965 $\green{Ans.}$