Question
A bus travels from point βXβ to βYβ at a
constant speed. If its speed was increased by 15 km/h, it would have taken 2 hours less to cover the distance. It would have taken further 30 minutes less if the speed was further increased by 10 km/h. The distance between point βXβ and point βYβ is:Solution
According to the question, Let the original speed of the bus be βxβ km/h and the original time taken to reach the destination be βtβ hours. So, the distance between point βXβ and βYβ = x Γ t = xt km So, xt/(x + 15) = t β 2..........................(1) And, xt/(x + 25) = t β 2 β 1/2 Or, xt/(x + 25) = t β 2.5........................(2) Dividing equation (1) by equation (2), we get (x + 25)/(x + 15) = (t β 2)/(t β 2.5) So, xt β 50 β 5x/2 + 25t = xt + 15t β x β 30 Or, 15t β 5x/2 = 80 Or, 5x/2 = 15t β 80 Or, x = 2/5 Γ (15t β 80) Putting the value of βxβ in equation (1), we get 30tΒ² β 60t = 30tΒ² β 75t + 120 Or, t = 8 So, x = 2/5 Γ (120 β 80) = 32 Desired distance = 8 Γ 32 = 256 km
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