Question

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

Answer 1

distance=300km
let it's original speed₁ be y km/h
speed₂=y+10
as per condition
t₁=distance/speed₁
t₁ =300/y.........(1)
t₂=distance/speed₂
t₂=300/(y+10)
as per second condition
t₁-t₂=1
300/y - 300(y+10)=1
300{1/y - 1/(y+10)}=1
$300( \frac{1}{y} - \frac{1}{y + 10} ) = 1 \\ 300( \frac{y + 10 - y}{y \times (y + 10)} ) = 1 \\ \frac{300 \times 10}{ {y}^{2} + 10y} = 1 \\ 3000 = {y}^{2} + 10y$
y²+10y-3000=0
y²+60y-50y-3000=0
y(y+60)-50(y+60)=0
(y+60)(y-50)=0
∴either y+60=0
gives y=-60
or
y-50=0
gives y=50
∴y=50km/h[∵length can never be negetive]
∴the original speed of the car is 50km/h

Answer 2

Let original speed be x km/hr
Distance=300km
Original time = 300/x hrs
Increased speed = (x + 10) km/hr
New time = 300/(x+10)
According to the question,
$\frac{300}{x} - \frac{300}{x + 10} = 1 \\ \frac{300(x + 10) - 300x}{ {x}^{2} + 10x } = 1 \\ {x}^{2} + 10x - 3000 = 0$

D=b^2-4ac
D=100-4(1)(-3000)
D=100+12000
D=12100
Roots of equation are
$x = \frac{ - b + \sqrt{d} }{2a} and \: \frac{ - b - \sqrt{d} }{2a} \\ x = - \frac{ - 10 + 110}{2} and \: \frac{ - 10 - 110}{2} \\ x = 50 \: and \: - 60$

x=-60 rejected (because speed cannot be negative)
Thus speed=50km/hr