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Let the speed of boat and stream be x km/hr and y km/hr respectively. Speed in upstream = (x – y) km/hr Speed in downstream = (x + y) km/hr ATQ: => 253/(x – y) = 3/2 × 340/(x + y) => {253/(x – y)}/{340/(x + y)} = 3/2 [ratio of time taken] Now, given that the total time taken to cover the distance is 5 hrs 42 sec or 5 hrs approx. So, 253/(x – y) = 3 and 340/(x + y) = 2 x - y = 84-----(i) x + y = 170-----(ii) solving equation (i) and (ii) we get, x = 127 y = 43 Therefore, approximate distance covered in upstream in 3 hrs 10 sec = (127 – 43) × 3 = 252 km
If log 32 = 3.816, then the value of log 16 is:
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