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Answer 1

$\huge \mathsf{ \color{purple}{✦Solution :-}}$

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

Increased speed = 100 km/hr

Time taken to cover 1500 km = 1500/x hr.

Time taken to cover 1500 km with increased speed = 1500/x + 100 hr

According to the Question,

$⇒ 1500/x - 1500(x + 100) = 30/60$

$⇒ x² + 100x - 300000 = 0$

$⇒ x² + 600x - 500x - 300000 = 0$

$⇒ (x + 600)(x - 500) = 0$

$⇒ x = -600 or 500$

(Neglecting negative sign as speed cannot be negative)

$⇒ x = 500$

$\small \mathsf{ \color{purple}{Hence, \: the \: usual \: speed \: of \: plane \: is \: 500 \: km/hr.}}$

Answer 2

let usual speed = x km/hr

speed= (x+100) km/hr

distance = 1500km

$time = \frac{distance}{usual \: speed} = \frac{1500}{x} hrs$

$time \: taken = \frac{1500}{x + 100}hrs$

$time \: taken \: = usual \: time - 0.5 \: hrs$

$= \frac{1500}{x} - \frac{1500}{x + 100} = \frac{1}{2} hrs$

$= {x}^{2} + 100x = 1500 \times 100 \times 2$

$= {x}^{2} + 100x - 300000 = 0$

$x = \frac{ - 100 + \frac{ + }{ - } \sqrt{ {100}^{2} + 4} \times 300000 }{2}$

$= \frac{ - 100 \frac{ + }{ - } 1100}{2} = \frac{ - 100}{2} + \frac{1100}{2}$

=500,600