Question

A man drives his car on a highway where the speed limit is 60 km/hr. He has to cover a distance of 240 km at a uniform speed on this road. If he increases his speed by 20
km/hr, he can reach his destination 1 hr earlier. What is his original speed at which he travels actually.​

Answer 1

speed =60km/hr

distance=240km

Time =d/s

T=240/60=4hours

if he reach increases in 20km/hr he get 1 hour earlier

so,Time Is=4-1=3hours

Speed=d/t

s=240/3

s=80km/hr

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Answer 2

$\fbox{\texttt{\green{Answer:}}}$

Original speed = 60 km/hr

$\fbox{\texttt{\pink{Given:}}}$

Speed limit of the highway = 60 km/hr

Total distance to he covered = 240 km

$\fbox{\texttt{\orange{To\:find:}}}$

Original speed at which he travels.

$\fbox{\texttt{\red{How\:to\:Find:}}}$

$\dagger$Total distance = 240 km

$\dagger$Let the uniform speed = x km/h

$\dagger$Time taken, $\bold{t_1=\frac{240}{x}hr}$

$\dagger$Increased speed = x + 20 km/h

$\dagger$Time taken, $\bold{t_2=\frac{240}{x+20}hr}$

$\hrulefill$

According to the question,

⇝ $\bold{\dfrac{240}{x}-\dfrac{240}{x+20}=1}$

⇝ $\bold{240\left(\dfrac{1}{x}-\dfrac{1}{x+20}\right)=1}$

⇝ $\bold{\dfrac{x+20-x}{x(x+20)}=\dfrac{1}{240}}$

⇝ $\bold{\dfrac{20}{x^2+20x}=\dfrac{1}{240}}$

Now, we get,

⇝ x² + 20x - 4800 = 0

By splitting the middle term,

⇝ x² + 80x - 60x - 4800 = 0

⇝ x(x + 80) - 60(x + 80) = 0

⇝ (x + 80)(x - 60) = 0

∴ x = -80 or x = 60

Since speed cannot be negative, his speed = 60 km/hr.