Question

a plane left 30 min. later than the schedule time and in order to reach its destination 1500km away in time in has to increase its speed by 250 km /h from its usual speeds .find usual speed ?​

Answer 1

Answer:

750km/hr

Step-by-step explanation:

Given :

A plane left the air port 30 minutes late then the scheduled time,. When it is 1500km away from the destination , The pilot increased the speed by 250km/hr than usual speed to reach the destination on time.

To Find :

The average/usual speed ,.

Solution :

Statement 1 :

The plane left the air port 30 minutes late then the scheduled time,.

⇒ Let the original speed be x,

Statement 2 :

When it is 1500km away from the destination,The pilot increased the speed by 250km/hr than usual speed to reach the destination on time.

Then,

We know that,

$30 \ minutes = \frac{1}{2} \ hour$

$Time =\frac{Distance}{Speed}$

So, $Change \ in \ Time =\frac{Distance}{Original \ Speed} - \frac{Distance}{change \ in \ Speed}$

$Original \ time = \frac{Distance}{Original \ Speed} = \frac{1500}{x}$ ...(1)

$Original \ Time = \frac{Distance}{increased \ Speed} +Decrease \ in \ time$

⇒ $Original \ time = \frac{1500}{x+250} + \frac{1}{2}$

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

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

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

⇒ $\frac{1}{2} = 1500(\frac{250{x^2 + 250x})$

⇒ $\frac{1}{2} = \frac{375000}{x^2 + 250x}$

⇒ $750000 = x^2 + 250x$

⇒ $x^2 + 250x - 750000 = 0$

⇒ $x^2 + 1000x - 750x - 750000 = 0$

⇒ $x(x+1000) - 750( x + 1000) = 0$

⇒ $(x + 1000) (x - 750) = 0$

For this equation to be zero,.

⇒ x + 1000 = 0 (or) x - 750 = 0

⇒ x ≠ -1000 (as speed can't be in negative)

⇒ x - 750 = 0 ⇒ x = 750

∴ The original speed of the plane is 750 km/hr

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\therefore{\text{ Original\:speed=750\:km/h}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

• In the given question informatio given about a plane left 30 min. later than the schedule time and in order to reach its destination 1500km away in time in has to increase its speed by 250 km /h from its usual speeds.

• We have to find Original speed.

$\green{\underline \bold{Given : }} \\ :\implies \text{Time(t) = 30 \: min} \\ \\ :\implies \text{Distance(d) = 1500 \: km} \\ \\ :\implies \text{Increased \: speed = 250 \: km/h} \\ \\ \red{\underline \bold{To \: Find : }} \\ :\implies \text{Original \: speed = ?}$

• According to given question :

$\bold{Let \: original \: speed = x \: km/h} \\ :\implies Original \: time = \frac{Distance}{Speed} \\ \\ :\implies Original \: time = \frac{1500}{x} - - - - - (1) \\ \\ \bold{For \: Increased \: speed : } \\ :\implies Original\: time - 30 \: min = \frac{distance}{original \: speed + 250} \\ \\ \bold{Putting \: value \: of \: Original \: time \: from \: (1)} \\ :\implies \frac{1500}{x} - \frac{1}{2} = \frac{1500}{x + 250} \\ \\ :\implies \frac{1}{2} = \frac{1500}{x} - \frac{1500}{x + 250} \\ \\ :\implies \frac{1}{2} = \frac{375000}{ {x}^{2} + 250x } \\ \\ :\implies {x}^{2} + 250x = 750000 \\ \\ :\implies {x}^{2} + 250x - 750000 = 0 \\ \\ :\implies {x}^{2} + 1000x - 750x - 750000 = 0 \\ \\ :\implies x(x + 1000) - 750(x + 1000) = 0 \\ \\ :\implies (x - 750)(x + 1000) = 0 \\ \\ \bold{Note : Value \: of \: speed \: never \: be \: in \: negative} \\ \bold{:\implies \text{x = 750 \: km/h}}$