If Rs. (y-200) was invested in scheme C at the rate of 25% per annum on compound interest compounded annually, then after two years Rs. 2812.5 will be obtained as an interest. (y-200) of (100+25)% of (100+25)% - (y-200) = 2812.5 (y-200) of 125% of 125% - (y-200) = 2812.5 (y-200)[1.5625-1] = 2812.5 0.5625x(y-200) = 2812.5 (y-200) = 5000 y = 5000+200 = 5200 Rs. ‘y’ was invested in scheme A at the rate of 21% per annum for (t+2) years. Rs. (y+2400) was invested in scheme B at the rate of 18% per annum for ‘t’ years. The interest obtained from both of the schemes together is Rs. 14484. y x 21% x (t+2) + (y+2400) x 18% x t = 14484 Put the value of ‘y’ in the above equation. 5200 x 21% x (t+2) + (5200+2400) x 18% x t = 14484 5200 x 21% x (t+2) + 7600 x 18% x t = 14484 1092 x (t+2) + 1368 x t = 14484 1092t+2184 + 1368t = 14484 2460t = 14484-2184 2460t = 12300 Value of ‘t’ = 5
√2401 × (√2116 ÷ 23) × 21 ÷ 3 = ?
(3984 ÷ 24) x (5862 ÷ 40) = ?
40% are the passing marks. A student gets 250 marks yet fails by 38 marks. What is the maximum marks?
The value of ((0.27)2-(0.13)2) / (0.27 + 0.13) is:
Find the HCF of 15x2 + 8x – 12, 3x² + x – 2, 3x² - 2x, 9x² - 12x + 4
7(3/6) of 534 + 262 = 61800 - ?
Find the simplified value of the given expression:
224 ÷ 4 + 7 X 36 - 8 of 15 + 162 - 300
?% of 24% of 1200 + 12 × 11 = 204
( 12.43×12.43×12.43 +19.57×19.57×19.57) / (12.43×12.43-12.43×19.57 +19.57×19.57)
30% of 600 –