Question

Calculate the compound interest on Rs 6000 at 10% per annum for 3 years.​

Answer 1

Step-by-step explanation:

Given Rate of interest= 10% per annum

Principal for the first year = Rs 6000

Interest for the first year = Rs (6000 × 10 × 1) / 100

= Rs 600

Amount at the end of first year = Rs 6000 + Rs 600

= Rs 6600

Principal for the second year = Rs 6600

Interest for the second year = Rs (6600 × 10 × 1) / 100

= Rs 660

Amount for the second year = Rs 6600 + Rs 660

= Rs 7260

Principal for the third year = Rs 7260

Interest for the third year = Rs (7260 × 10 × 1) / 100

= Rs 726

Amount for the Third year = Rs 7260 + Rs 726

= Rs 7986

Therefore, compound interest for 3 years = final amount – (original) Principal

= Rs 7986 – Rs 6000

We get,

= Rs 1986

Answer 2

Given :

• Principal (P) = Rs.6000
• Rate (R) = 10%
• Time Period (t) = 3years

To Find :

• Compound Interest (CI)

Formula Applied :

$\displaystyle{\underline{\boxed{\sf Compound\:Interest = P \Bigg[\Big(1+\dfrac{R}{100}\Big)^t -1\Bigg]}}}$

Solution :

$\implies\sf P\Bigg[\Big(1+\dfrac{R}{100}\Big)^t -1\Bigg]$

$\implies \sf 6000 \Bigg[\Big(1+\dfrac{10}{100}\Big)^3 -1\Bigg]$

$\implies \sf 6000 \Bigg[\Big(\dfrac{100+10}{100}\Big)^3 -1\Bigg]$

$\implies \sf 6000 \Bigg[\Big(\dfrac{110}{100}\Big)^3 -1\Bigg]$

$\implies \sf 6000\Bigg [\Big (\dfrac {110\times110\times 110}{100\times 100\times 100}\Big) -1\Bigg]$

$\implies \sf 6000 \Big(\dfrac{1331000}{1000000}-1\Big)$

$\implies \sf 6000 \Big(\dfrac{1331000-1000000}{1000000}\Big)$

$\implies \sf 6{\cancel{0}} {\cancel{0}}{\cancel{0}}\Big(\dfrac{331{\cancel{0}}{\cancel{0}}{\cancel{0}}}{1{\cancel{0}}{\cancel{0}}{\cancel{0}}{\cancel{0}}{\cancel{0}}{\cancel{0}}}\Big)$

$\implies \sf 6 \times 331$

$\implies \sf Rs. 1986$

Required Answer :

Compound interest for the given condition is $\underline{ \sf Rs. 1986}$