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 βxβ km/h and βyβ km/h. Therefore, (x + y) = 70β¦. (I) And (x β y) = 30β¦β¦ (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 β 50 = 20 km/h Let the racing distance be βdβ km. (d/20) β (d/50) = (6/60) the Or {(50d β 20d)/1000} = (6/60) Or (30d/1000) = (6/60) Or d = (6/60) Γ (1/0.03) = (10/3) km
Eructation
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