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Let’s assume the speed of the boat in still water and the speed of the stream are ‘B’ and ‘C’ respectively. If the speed of the stream is 75% less than the speed of the boat. C = (100-75)% of B C = 25% of B C = (25/100) x B C = (1/4) x B B = 4C Eq. (i) The same boat can cover (d+20) km distance in still water in 11.75 hours. (d+20)/B = 11.75 Put the value of ‘B’ from Eq. (i) in the above equation. (d+20)/4C = 11.75 (d+20)/47 = C Eq. (ii) The time taken by the boat to cover (d+130) km distance in downstream is 1.9 hours less than the time taken by the same boat to cover (d-45) km distance in upstream. [(d+130)/(B+C)] = [(d-45)/(B-C)] - 1.9 Put the value of ‘B’ from Eq. (i) in the above equation. [(d+130)/(4C+C)] = [(d-45)/(4C-C)] - 1.9 [(d+130)/5C] = [(d-45)/3C] - 1.9 Put the value of ‘C’ from Eq. (ii) in the above equation. [47(d+130)]/[5(d+20)] = [47(d-45)]/[3(d+20)] - 1.9 By solving the above equation, the value of ‘d’ = 450
8(3/4) + 5(1/6) – 4(3/4) = ?
20 ×33 + 12 × 23 - 40 ÷ 15-1 + ? = 50
(750 / 15 × 15 + 152 + 20% of 125) = ?3
√ (12+√ (12+√ (12+ ⋯ ∞ ))
(60 × 8 ÷ 10) × 5 = ?
(√529 + 63 /8)% of 800 = ?% of 250
1 + 1 + 1/2+ 1/3 + 1/6 + 1/4 is equal to ____
68% of 450 – 1008 ÷ 14 + 516 ÷ 43 =?