Question

An aeroplane started $30$ minutes later than the schedule time from a place $1500km$ away from its destination to reach the destination at the schedule time the pilot had to increase the speed by $250$km/hr what was the speed of the aeroplane per hour during the journey?

• A. $100$km/hr
• B. $1000$km/hr
• C. $1050$km/hr
• D. $900$km/hr

Answer 1

The correct option is B $1000$km/hr

Let the usual time taken by the aeroplane $=$x hr.

Distance $=1500$km

In first case-

speed$=\frac{1500}{x}$ km/h ………..$\left(1\right)$

In second case-

time taken by aeroplane$=x-\frac{1}{2}$ hr

$\therefore$ Speed$=\frac{1500}{x-\frac{1}{2}}$km/h ………$\left(2\right)$

Increased speed$=250$ km/h

So, $\frac{1500}{x-\frac{1}{2}}-\frac{1500}{x}=250$

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

$\frac{3000}{2x-1}-\frac{1500}{x}=250$

$\frac{3000x-3000x+1500}{(2x-1)x}=250$

$\frac{1500}{2x^2-x}=250$

$6=2x^2-x$

$2x^2-x=6$

$2x^2-x-6=0$

$2x^2-4x+3x-6=0$

$2x(x-2)+3(x-2)=0$

$(x-2)(2x+3)=0$

$x=2,x=\frac{-3}{2}$

So time can not be negative

$x=2$

then speed$=\frac{1500}{x-\frac{1}{2}}=\frac{1500}{2-\frac{1}{2}}=1000$ km/h.