A boat travels from dock ‘M’ to dock ‘N’ at a constant speed. If its speed was increased by 8 km/h, it would have taken 4 hours less to cover the distance. It would have taken further 1 hour less if the speed was further increased by 8 km/h. The distance between dock ‘M’ and dock ‘N’ is:

According to the question, Let the original speed of the boat be ‘x’ km/h and the original time taken to reach the destination be ‘t’ hours. So, the distance between dock ‘M’ and ‘N’ = x × t = xt km So, xt/(x + 8) = t – 4..........................(1) And, xt/(x + 16) = t – 4 – 1 Or, xt/(x + 16) = t – 5........................(2) Dividing equation (1) by equation (2), we get (x + 16)/(x + 8) = (t – 4)/(t – 5) So, xt – 48 – 8x + 16t = xt + 8t – x – 40 Or, 8t – 8x = 8 Or, x = t – 1 Putting the value of ‘x’ in equation (1), we get 8t² – 32t = 8t² – 40t + 40 Or, t = 5 So, x = 5 – 1 = 4 Desired distance = 5 × 4 = 20 km