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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