A sum of ₹ 8000 is divided into two parts. Th | vedclass

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(A) Let the first part be $₹ x$ and the second part be $₹ (8000 - x)$.Since the time period is the same (let it be $T = 1$ year),the simple interest f

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$1050$

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(A) Let the first part be $₹ x$ and the second part be $₹ (8000 - x)$.Since the time period is the same (let it be $T = 1$ year),the simple interest formula is $SI = \frac{P 	imes R 	imes T}{100}$.According to the problem,the simple interest on both parts is equal:$\frac{x 	imes 21 	imes 1}{100} = \frac{(8000 - x) 	imes 35 	imes 1}{100}$$21x = 35(8000 - x)$$21x = 280000 - 35x$$21x + 35x = 280000$$56x = 280000$$x = \frac{280000}{56} = 5000$.So,the first part is $₹ 5000$ and the second part is $₹ (8000 - 5000) = ₹ 3000$.The interest on the first part is $\frac{5000 	imes 21 	imes 1}{100} = ₹ 1050$.The interest on the second part is $\frac{3000 	imes 35 	imes 1}{100} = ₹ 1050$.Thus,the interest of each part is $₹ 1050$.

$A$ sum of ₹ $8000$ is divided into two parts. The simple interest on the first part at the rate of $21 \%$ per annum is equal to the simple interest on the second part at the rate of $35 \%$ per annum. What is the interest (in ₹) of each part?

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