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Find the simple interest on 540000/= invested for 18 months at rate of 10% pa |
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Answer» P = Rs. 540000 R = 10% T = 18 months = \(1 \frac{1}{2}\) = \(\frac{3}{2}\) years S.I = \(\frac{P \times R \times T}{100}\) = \(\frac{540000 \times 10 \times 3}{100 \times 2}\) = Rs.5400 x 5 x 3 = Rs.81000 |
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