Question

solve it fast plzz
it's my paper tomorrow

Answer 1

Important Formulas:

(i) Speed = Distance/Time

(ii) Time = Distance/Speed.

According to the Question,

Let the usual speed be 'x'.

Given that it has increased its speed by 100 km/hr.

So, the speed of the plane = (x + 100) km/hr.

Given that it left 30 minutes later = (30/60) = 1/2.

Given Distance = 1500 km.

Now, We have Distance and Speed. Apply the 2nd formula:

⇒ (1500/x) - (1500/x + 100) = 1/2

⇒ (1500x + 150000 - 1500x)/x(x + 100) = 1/2

⇒ 150000 * 2 = x(x + 100)

⇒ 300000 = x^2 + 100x

⇒ x^2 + 100x - 300000 = 0

⇒ x^2 + 600x - 500x - 300000 = 0

⇒ x(x + 600) - 500(x + 600) = 0

⇒ (x - 500)(x + 600) = 0

⇒ x = 500,-600 {Speed cannot be negative}

⇒ x = 500.

Therefore, the usual speed of the plane = 500 km/hr.

Hope this helps!

Answer 2

Let to be usual speed of plane X km/hr

Increased speed of the plane = (x + 100) km/hr

Time taken to reach the destination at usual speed,

T1=1500/x hr

Time taken to reach the destination at increased speed,

T2=1500/100+x

Given,

 t 1 – t 2 = 30 minutes

$\bold{ \frac{1500}{x} - \frac{1500}{x + 100} = \frac{30}{60}} \\ \\ \bold{ \frac{1500}{x} - \frac{1500}{x + 100} = \frac{ \cancel30}{ \cancel60}} \\ \\ \bold{ \frac{1500}{x} - \frac{1500}{x + 100} = \frac{1}{2}} \\ \\ \bold{ 1500( \frac{ \cancel x + 100 - \cancel x}{x(x + 100)}} ) = \frac{1}{ 2} \\ \\ \bold{ \frac{3000 }{ {x}^{2} + 100x } = 1} \\ \\ \bold{3000 = {x}^{2} + 100x} \\ \\ \bold{{x}^{2} + 100x - 3000}$

$\bold{ {x}^{2} + 600x - 500x - 3000} \\ \\ \bold{ x(x + 600)} - 500(x + 600) \\ \\ \bold{x - 500 = 0} \\ \\ \bold{x = 500km/hr}$

Note:- Speed Never be Negative so I taken positive value.

Thanks!!!