Question

A plane left 30 minutes late than the schedule time and in order to reach its destination 1500 km away in time,it has to increase its speed by 250 km/h from its usual speed.find its usual speed.

Answer 1

Given
Distance =1500km
Left 30min late
Speed Increase by 250km/h

Let usual speed=x

Time =distance /speed

=1500/x

Now speed in increased by 250km/hr

New speed =(x+250)

Then
Time =1500/(x+250)


So we can write

$\frac{1500}{x} - \frac{1}{2} = \frac{1500}{x + 250}$

$\frac{1500}{x} - \frac{1500}{x - 250} = \frac{1}{2}$

$\frac{375000}{x(x + 250)} = \frac{1}{2}$

X²+250x-750000=0

Now find two numbers who's sum=+250 and multiple =-750000

1000,-750

X²+1000x-750x-750000

X(x+1000)-750(x+1000)

(x+1000)(x-750)

X+1000=0

X=-1000

Speed can't be in negative

X-750=0

X=750km/h

Usual speed =750km/h

This is ur ans hope it will help you

Answer 2

hello

time taken by the aeroplane = x km/hr

Distance to destination = 1500 km

Case (i)

Speed = Distance / Time

= (1500 / x) Hrs
 
Case (ii)

Time taken by aeroplane = (x - 1/2) Hrs

Distance to destination = 1500 km

Speed = Distance / Time = 1500 / (x - 1/2) Hrs
 
Increased speed = 250 km/hr
 
⇒ [1500 / (x - 1/2)] - [1500 / x] = 250

⇒ 1/(2x2 - x) = 1/6

⇒ 2x2 - x = 6

⇒ (x - 2)(2x + 3) = 0

⇒ x = 2 or -3/2

time can not be negative,

time taken by the aeroplane

= 2 hrs

usual speed

= (1500 / 2) = 750 km/hr.