Question : A motorcyclist left $6\frac{6}{9}$ minutes later than the scheduled time but to reach its destination 21 km away on time, he had to increase his speed by 12 km/hr from the usual speed. What is the usual speed (in km/h) of the motorcyclist?
Option 1: 28
Option 2: 35
Option 3: 42
Option 4: 64
Correct Answer: 42
Solution : Distance = 21 km Let the usual speed of a motorcyclist be $x$ km/hr. Time taken, t 1 = $\frac{\text{distance}}{\text{speed}}$ = $\frac{21}{x}$ When left late by $6\frac{6}{9}$ min = $\frac{1}{9}$hr, Time, t 2 = travel time + waiting time = $\frac{21}{(x+12)}+\frac{1}{9}$ According to the question, $\frac{21}{x}$ = $\frac{21}{(x+12)}$ + $\frac{1}{9}$ $⇒\frac{21x+252-21x}{x(x+12)}$ = $\frac{1}{9}$ $⇒\frac{252}{x^2+12x}$ = $\frac{1}{9}$ $⇒x^2+12x =2268$ $⇒x^2+54x-42x-2268=0$ $⇒(x+54)(x-42)=0$ $⇒x=-54$ or $x = 42$ Since speed can't be negative, $x$ = 42 km/hr Hence, the correct answer is 42.
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