According to the question, Let the original speed of the bus be ‘x’ km/h and the original time taken to reach the destination be ‘t’ hours. So, the distance between point ‘X’ and ‘Y’ = x × t = xt km So, xt/(x + 15) = t – 2..........................(1) And, xt/(x + 25) = t – 2 – 1/2 Or, xt/(x + 25) = t – 2.5........................(2) Dividing equation (1) by equation (2), we get (x + 25)/(x + 15) = (t – 2)/(t – 2.5) So, xt – 50 – 5x/2 + 25t = xt + 15t – x – 30 Or, 15t – 5x/2 = 80 Or, 5x/2 = 15t – 80 Or, x = 2/5 × (15t – 80) Putting the value of ‘x’ in equation (1), we get 30t² – 60t = 30t² – 75t + 120 Or, t = 8 So, x = 2/5 × (120 – 80) = 32 Desired distance = 8 × 32 = 256 km
I’ve got abusydaytomorrow, so IthinkI’ll hit thesack.
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