A plane left 30 minutes late …

Let the usual speed of the plane b x km/hr.
Distance to the destination = 1500 km
Case (i):
 {tex}\text{we know that,} \ Speed = {Distance\over Time} \\ \Rightarrow Time = {Distance\over speed}{/tex}
So, in case(i) Time = {tex}1500 \over x{/tex}Hrs
Case (iI)
Distance to the destination = 1500 km
Increased speed = 100 km/hr
So, speed = x+100
So, in case(ii) Time = {tex}1500 \over {x+100}{/tex}Hrs
So, according to the question
{tex}\therefore{/tex} {tex}\frac{1500}{x}{/tex} - {tex}\frac{1500}{x + 100}{/tex} = {tex}\frac{30}{60}{/tex}
{tex}\Rightarrow{/tex} x² + 100x - 300000 = 0
{tex}\Rightarrow{/tex} x² + 600x - 500x - 300000 = 0
{tex}\Rightarrow{/tex} (x + 600)(x - 500) = 0
{tex}\Rightarrow{/tex}x = 500 or x = -600
Since, speed can not be negative, x = 500
Therefore, Speed of plane = 500 km/hr.