Question

if the speed of a car is increased by 10 km per hour it takes 1 hour less to cover a distance of 300 km find the original speed of a car

Answer 1

Answer:

Step-by-step explanation:

Distance=300km

Speed=distance/time

According to question:

  Original speed = x

  Increased speed = (x+10)

  (300/x+10)-(300/x)=1

      Cross multiply;

   300(x)-300(x+10)=x^2+10x

   X^2+10x-3000=0 ........(1)

   On solving the eqs with the help of quadratic formula(-b+-√b^2-4ac/2a)

 We get :

          x=50

 This speed is 50km/hr

Answer 2

Answer:

Let the original speed be x Km/h.

Total time travel be t Hour.

Case 1

300÷x = t .................... eq.(1)

Case 2

Speed is increased by 10 Km/h.

And time is 1 hour less

Now new equation is

300÷(x+10)= t-1

put the value of t from equation 1 to above equation

300÷(x+10)=(300÷x)-1

⇒(300÷x)-[300÷(x+10)]=1

taking LCM

[300(x+10)-300x]÷($x^{2} +10x$)=1

300x+3000-300x=$x^{2} +10x$

$x^{2}+10x-3000=0$

$x^{2} +60x-50x-3000=0$

$x(x+60)-50(x+60)=0$

$(x+60)(x-50)=0\\x+60=0  or  x-50=0\\x=-60 or x=50$

x≠-60

so value of x is 50

the original value of speed of the car is 50 Km/h