Question

The difference between the compound interest and the simple interest on an amount of ₹18,000 in 2 years was ₹405. What is the rate of interest per annum?

Answer 1

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Correct Question:-

• The difference between the compound interest and the simple interest accrued on an amount of Rs. 18,000 in 2 years was Rs. 405. What was the rate of interest p.c.p.a. ?

$\text{\large\underline{\red{Required\:Answer:-}}}$

• Required interest per annum is 15%.

$\text{\large\underline{\orange{Step-by-step explanation: :-}}}$

• Let the rate of interest be R% p.a.

Here,

• Principal ( P ) is Rs 18,000 and time ( T ) is 2 years.

According to the question:-

• Compound Interest - Simple interest = Rs 405

$\begin{gathered}\boxed{\mathsf{ Simple\:Interest =\dfrac{Principal \times Rate \times Time}{100}}}\\\\\boxed{\mathsf{Compound\:Interest=P\bigg[ \bigg( 1+\dfrac{r}{100} \bigg)^n - 1 \bigg] }}\end{gathered}$

[ Rs$\:18,000\: (1 +$$\frac{R}{100}$ )² –$18,000$ ] – [ $\frac{18000×R×2}{100}$ ] = 405

• $⟹$ [ 18000( 1 + R / 100 )^2 - 18000 ] - [ 180 x R x 2 ] = 405

• $⟹$[ 18000( 1 + R² / 10000 + R / 50 ) ] - 18000 - 360R = 405

• $⟹$18000 + 9R² / 5 + 360R - 18000 - 360R = 405
• $⟹$9R² / 5 = 405

• $⟹$ R² = 405 x 5 / 9

• $⟹$ R² = 225

• R = 15

Hence,

The required interest per annum is 15%.

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Answer 2

Rs\:18,000\: (1 +18,000(1+ \frac{R}{100}

100R)² –18,00018,000 ] – [ \frac{18000×R×2}{100}

100

18000×R×2 ] = 405

⟹⟹ [ 18000( 1 + R / 100 )^2 - 18000 ] - [ 180 x R x 2 ] = 405

⟹⟹ [ 18000( 1 + R² / 10000 + R / 50 ) ] - 18000 - 360R = 405

⟹⟹ 18000 + 9R² / 5 + 360R - 18000 - 360R = 405

⟹⟹ 9R² / 5 = 405

⟹⟹ R² = 405 x 5 / 9

⟹⟹ R² = 225

R = 15

Hence,

The required interest per annum is 15%.