Question

Anshu covers a distance of 200km traveling with a uniform speed of x km per hour.The distance could have been covered 2hours less,had the speed been(x+5)km per hour .Calculate the value of x.
​

Answer 1

$\huge{\blue{\bold{\underline{\underline{Answer :}}}}}$

$\:\:$

$\large{\green{\underline \bold{\tt{Given :-}}}}$

$\:\:$

• Total distance = 200 km

$\:\:$

• Original speed = x km/hour

$\:\:$

• Time taken will be 2 hours less, if the speed is (x+5) km/hour

$\:\:$

$\large{\red{\underline \bold{\tt{To \: Find :-}}}}$

$\:\:$

Case I [ Speed = x ]

$\:\:$

• Let the time taken be t1

$\:\:$

$\underline{\bold{\texttt{We know that :}}}$

$\:\:$

➠ $\sf Time = \dfrac { Distance } { Speed }$ ---- (1)

$\:\:$

• Time = t1

• Distance = 200 km

• Speed = x

$\:\:$

$\underline{\bold{\texttt{Putting these values in (1) }}}$

$\:\:$

➜ $\sf t1 = \dfrac { 200} { x}$ ------ (2)

$\:\:$

Case II [ Speed = x +5 ]

$\:\:$

• Let the time taken be t2

$\:\:$

• Time = t2

• Distance = 200 km

• Speed = x + 5

$\:\:$

$\underline{\bold{\texttt{Putting these values in (1) }}}$

$\:\:$

➜ $\sf t2 = \dfrac { 200} { x + 5}$ ------ (3)

$\:\:$

$\purple{\underline \bold{According \: to \: the \ question :}}$

$\:\:$

Time taken will be 2 hours less, if the speed is (x+5) km/hour

$\:\:$

➠ t1 = t2 + 2

$\:\:$

➜ $\sf \dfrac { 200} { x} = \dfrac { 200} { x + 5} + 2$

$\:\:$

➜ $\sf \dfrac { 200} { x} = \dfrac { 200 + 2x + 10} { x + 5}$

$\:\:$

Cross multiplying

$\:\:$

➜ 200x + 1000 = 200x + 2x² + 10x

$\:\:$

➜ 2x² + 10x - 1000 = 0

$\:\:$

Dividing the above equation by "2"

$\:\:$

➜ x² + 5x - 500 = 0

$\:\:$

➜ x² + 25x - 20x - 500 = 0

$\:\:$

➜ x(x + 25) -20(x + 25) = 0

$\:\:$

➜ (x + 25)(x - 20) = 0

$\:\:$

• x = -25

• x = 20

$\:\:$

As speed can't be negative hence x = 20

$\:\:$

• Hence the original speed is 20 km/hr

Answer 2

$\huge{\blue{\bold{\underline{\underline{Answer :}}}}}$

$\:\:$

$\large{\green{\underline \bold{\tt{Given :-}}}}$

$\:\:$

• Total distance = 200 km

$\:\:$

• Original speed = x km/hour

$\:\:$

• Time taken will be 2 hours less, if the speed is (x+5) km/hour

$\:\:$

$\large{\red{\underline \bold{\tt{To \: Find :-}}}}$

$\:\:$

Case I [ Speed = x ]

$\:\:$

Let the time taken be t1

$\:\:$

$\underline{\bold{\texttt{We know that :}}}$

$\:\:$

➠ $\sf Time = \dfrac { Distance } { Speed }$ ---- (1)

$\:\:$

Time = t1

Distance = 200 km

Speed = x

$\:\:$

$\underline{\bold{\texttt{Putting these values in (1) }}}$

$\:\:$

➜ $\sf t1 = \dfrac { 200} { x}$ ------ (2)

$\:\:$

Case II [ Speed = x +5 ]

$\:\:$

Let the time taken be t2

$\:\:$

Time = t2

Distance = 200 km

Speed = x + 5

$\:\:$

$\underline{\bold{\texttt{Putting these values in (1) }}}$

$\:\:$

➜ $\sf t2 = \dfrac { 200} { x + 5}$ ------ (3)

$\:\:$

$\purple{\underline \bold{According \: to \: the \ question :}}$

$\:\:$

Time taken will be 2 hours less, if the speed is (x+5) km/hour

$\:\:$

➠ t1 = t2 + 2

$\:\:$

➜ $\sf \dfrac { 200} { x} = \dfrac { 200} { x + 5} + 2$

$\:\:$

➜ $\sf \dfrac { 200} { x} = \dfrac { 200 + 2x + 10} { x + 5}$

$\:\:$

Cross multiplying

$\:\:$

➜ 200x + 1000 = 200x + 2x² + 10x

$\:\:$

➜ 2x² + 10x - 1000 = 0

$\:\:$

Dividing the above equation by "2"

$\:\:$

➜ x² + 5x - 500 = 0

$\:\:$

➜ x² + 25x - 20x - 500 = 0

$\:\:$

➜ x(x + 25) -20(x + 25) = 0

$\:\:$

➜ (x + 25)(x - 20) = 0

$\:\:$

x = -25

x = 20

$\:\:$

• As speed can't be negative hence x = 20

$\:\:$

• Hence the original speed is 20 km/hr