Question

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

Answer 1

Answer:

The usual speed of the plane is 750 km/hr.

Step-by-step explanation:

Let the usual speed of the plane be x km/hr.

Increased speed of the plane = (x + 250) km/hr

Time taken to reach the destination at usual speed,  

$t_1=\dfrac{1500}{x}$

Time taken to reach the destination at increased speed,  

$t_2=\dfrac{1500}{x+250}$

Given,

• $t_1\minust_2$ = 30 min  = $\dfrac{30}{60}$hr

Therefore,

$\dfrac{1500}{x}-\dfrac{1500}{x+250}=\dfrac{30}{60}$

$\dfrac{1500x+375000-1500x}{x(x+250)}=\dfrac{1}{2}$

x² + 250x = 750000

x² + 250x - 750000 = 0

x² + 1000x - 750x - 750000 = 0

x(x + 1000) - 750(x + 1000) = 0

(x - 750)(x + 1000) = 0

x = 750,-1000

x = 750 (speed can never be in negative)

Answer 2

Answer:

Speed of plane is 750 km/hr.

Step-by-step explanation:

$\bold{\underline{\underline{Solution\::}}}$

• Time (t) = 30 min = 1/2 hr.
• Distance (d) = 1500 km
• Increased speed = 250 km/hr

$\bold{\underline{\underline{Assume\::}}}$

Let the -

• time be "M" sec.

Now,

Speed = Distance/Time

Speed = 1500/M

Now the plane increase it's speed by 250 km/hr.

So,

New speed = 1500/(M + 250)

New speed - Original speed = 1/2

Substitute the known values in above formula

→ $\dfrac{1500}{M}\:-\:\dfrac{1500}{M\:+\:250}\:=\:\dfrac{1}{2}$

→ $1500\bigg(\dfrac{1}{M}\:-\:\dfrac{1}{M\:+\:250}\bigg)\:=\:\dfrac{1}{2}$

→ $1500\bigg(\dfrac{M\:+\:250\:-\:M}{M(M+250)}\bigg)\:=\:\dfrac{1}{2}$

→ $\dfrac{375000}{M^2\:+\:250}\:=\:\dfrac{1}{2}$

Cross multiply them

→ $M^2\:+\:250M\:=\:750000$

→ $M^2\:+\:250M\:-\:750000\:=\:0$

→ $M^2\:+\:1000M\:-\:750M\:-\:750000\:=\:0$

→ $M(M+10000)\:-750(M+10000)\:=\:0$

→ $M\:=\:-10000,\:750$

(-10000 rejected because speed can't be negative)

∴ Speed of the plane is 750 km/hr.