Question

find the difference between compound interest and simple interest on rs.15000 for 3years at 5% per annum​

Answer 1

AnswEr :

• Principal = Rs. 15,000
• Rate = 5% p.a.
• Time = 3 Years

$\sf{Difference = Compound \: Interest - Simple \: Interest}$

$\sf{Diff. = \bigg[P \bigg(1 + \dfrac{r}{100} \bigg)^{t} - 1 \bigg]- \bigg[\dfrac{PRT}{100} } \bigg]$

$\sf{Diff. = \bigg[15000 \bigg(1 + \cancel\dfrac{5}{100} \bigg)^{3} - 1 \bigg]- \bigg[\dfrac{150 \cancel{00} \times 5 \times 3} {\cancel{100}} } \bigg]$

$\sf{Diff. = \bigg[15000 \bigg(1 + \dfrac{1}{20} \bigg)^{3} - 1 \bigg]- \bigg[150 \times 5 \times 3 } \bigg]$

$\sf{Diff. = \bigg[15000 \bigg( \dfrac{21}{20} \bigg)^{3} - 1 \bigg]- 2250}$

$\sf{Diff. = \bigg[15000 \bigg( \dfrac{9261}{8000} - 1\bigg) \bigg]- 2250}$

$\sf{Diff. = \bigg(15000 \times \dfrac{1261}{8000} \bigg) - 2250}$

$\sf{Diff. = Rs.(2364.375 - 2250)}$

$\sf{Diff. = Rs. \: 114.375}$

⠀

$\therefore$ Difference of CI and SI is Rs. 114.375

Answer 2

Answer:

114,375

Explanation:

→ ( P ) = Principal amount ( 15,000 )

→ ( R ) = Rate ( 5% )

→ ( N ) = Time ( 3 years )

→ ( N) = number of years (Time)

Hence

→Let D be the Difference.

→ D = (P (1 + r / 100 )^n - 1) - (PRT) / 100

→ (15000(1+5/100)^3)-(15000×5×3/10 0)

→ D = (15000 (21 / 20)^3 - 1) - 2250

→ D = (15000 (9261 / 8000 - 1) - 2250

→ D = (15000 × 1261 / 8000) - 2250

→ D = (2364.375 - 2250)

= 114.35

Therefore , 114.35 is the answer.