Question

A car covers a distance of 400 km at a certain speed. had the speed been 12km/hr more,the time taken for the journey would have been 1 hour 40 minutes less. find the original speed of the car.

Answer 1

Let the original speed of the car be x kmph
case i:
Distance = 400 km
speed = x kmph
Time taken = 400/x hrs

case ii:
Distance = 400 km
speed = (x+12) kmph
Time taken = (400/x+12) 

But as per the question, the time taken is 1 hr 40 min lesser than the previous. 

that means,

400/x - (1 hr 40 min) = (400/x+12)

$\frac{400}{x} - \frac{5}{3} = \frac{400}{x+12}$

$\frac{1200-5x}{3x} = \frac{400}{x+12}$

[tex]1200x - 5x^2 + 14400 - 60x = 1200x [/tex]

x² + 12x - 2880 = 0

(x - 48)(x + 60) = 0

x = 48 as the negative value of x is ignored.

The original speed of the car = 48 kmph

Answer 2

Answer:

Step-by-step explanation:

Solution :-

Let the original speed of car be x km/h.

We know that,

Time = Distance/Time

t = 400/x

Increased speed = (x + 12) Km/h

According to the Question,

⇒ 400/x - 400/x + 12 = 1(40/60)

⇒ 400x + 4800 - 400x/x(x + 12) = 1(2/3)

⇒ 4800/x(x + 12) = 5/3

⇒ 5x² + 60x - 14400 = 0

⇒ x² + 12x - 2880 = 0

⇒ x² + 60x - 48x - 2880 = 0

⇒ x(x + 60) - 48(x + 60) = 0

⇒ (x + 60) (x - 48) = 0

⇒ x = - 60 or 48 (Neglecting negative sign)

⇒ x = 48 km/h

Hence, the original speed is 48 km/h.