Question
A boat travels from dock βMβ to dock βNβ 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 βMβ and dock βNβ is:Solution
According to the question, Let the original speed of the boat be βxβ km/h and the original time taken to reach the destination be βtβ hours. So, the distance between dock βMβ and βNβ = x Γ t = xt km So, xt/(x + 8) = t β 4..........................(1) And, xt/(x + 16) = t β 4 β 1 Or, xt/(x + 16) = t β 5........................(2) Dividing equation (1) by equation (2), we get (x + 16)/(x + 8) = (t β 4)/(t β 5) So, xt β 48 β 8x + 16t = xt + 8t β x β 40 Or, 8t β 8x = 8 Or, x = t β 1 Putting the value of βxβ in equation (1), we get 8tΒ² β 32t = 8tΒ² β 40t + 40 Or, t = 5 So, x = 5 β 1 = 4 Desired distance = 5 Γ 4 = 20 km
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