Question
The sum and difference of the speeds of the two cars are
70km/h and 30km/h respectively. In a race, if the car with more speed finishes 6 minutes earlier than the other. The racing distance was equal to:Solution
Let the speeds of the faster car and the slower car be ‘x’ km/h and ‘y’ km/h. Therefore, (x + y) = 70…. (I) And (x – y) = 30…… (II) Adding equation (I) and (II), we get 2x = 100 Or x = (100/2) = 50 Therefore, the speed of the faster car = 50 km/h From equation (I), the speed of the slower car = 70 – 50 = 20 km/h Let the racing distance be ‘d’ km. (d/20) – (d/50) = (6/60) the Or {(50d – 20d)/1000} = (6/60) Or (30d/1000) = (6/60) Or d = (6/60) × (1/0.03) = (10/3) km
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