Question
The distance of the School and house of Suresh is 80 km.
One day he was late by 1 hour than the normal time to leave for the college, so he increased his speed by 4km/h and thus he reached to college at the normal time. What is the changed speed of Suresh?Solution
ATQ, we can say that Let the actual speed of Suresh be x km/hr According to the question 80/xβ 80/(x + 4) = 1 80 [(x + 4 β x) / (x2 + 4x)] = 1 320 = x2 + 4x x2 + 4x β 320 = 0 x2 + 20x β 16x β 320 = 0 x (x + 20) β 16 (x + 20) = 0 (x + 20) (x β 16) = 0 Taking, (x + 20) = 0 x = -20 (not possible) Taking, (x β 16) = 0 x = 16 Speed of Suresh = 16 km/hr β΄ Changed speed of Suresh = 16 + 4 = 20 km/hr
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