Question

A plane left 30 min later than its scheduled time to reach its destination 1500 km away. In order to reach in time, it increases its speed by 250 km/hr. What is its original speed

Answer 1

answers ☺️☺️ by an easy method.

hope it will help you

Answer 2

Given :-

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

To find :-

• What is its original speed ?

Solution :-

Let usual speed of plane = x km / hr

then usual time taken $(t_{1})$ ${\implies}\frac{Distance}{speed}$

${\implies}\frac{1500}{x} hrs$

New speed

${\implies} ( {x + 250} )km /hr$

New time $t_{2}$

${\implies}$Distance / speed

${\implies}$1500 / x + 250 hrs

So,

${\implies}t_{1} - t_{2}$ = $\frac{1}{2}$

${\implies} \frac{1500}{x} - \frac{1500}{x \: + \: 250} = \frac{1}{2} \\ \\ {\implies} 1500 \: ( \: \frac{1}{x} - \frac{1}{x \: + 250} ) \: = \frac{1}{2} \\ \\ {\implies} 1500 \: ( \frac{ x \: + 250 \: - \: x}{x \: ( \: x \: + \: 250 \: ) \: } ) = \frac{1}{2} \\ \\ {\implies} \frac{1500 \: \times \: 250}{ {x}^{2} + \: 250 \: x} = \frac{1}{2} \\ \\ {\implies} {x}^{2} + 250x \: = \: 1500 \: \times 250 \: \times 2 \: \\ \\ {\implies} {x}^{2} + 250x \: = 750000 \: = 0 \\ \\{\implies} {x}^{2} + 250x \: - 750000 \: = 0 \\ \\ {\implies} {x}^{2} + (1000 - 750)x - 750000 = 0 \\ \\{\implies} { x }^{2} + 1000x - 750x - 750000 = 0 \\ \\ {\implies}x(x + 1000) - 750(x + 1000) = 0 \\ \\{\implies} (x + 1000) \: (x - 750) = 0$

${\implies} x = - 1000 \: or \: x = 750$

Speed cannot be negative .

So, the original speed of plane is 750 km/hr.

_____________________