Question

A plane left 30 minutes late than its schedule time and in order to reach the destination 1500 km away in time it had to increase its speed by 100km/hr from the usual speed find its usual speed (class 10th)

Answer 1

Let x= the usual speed
1500/x would equal how long it usually took the plane to read the destination
1500/(x+100) would equal how long it would take the plane now
So$\frac{1500}{x} - \frac{1500}{x+100} = \frac{1}{2}$ because it would lose 30 minutes already
Simplify:
$\frac{1500(x+100)-1500x}{x(x+100)} = \frac{1}{2}$
$\frac{1500x+150000-1500x}{x^2+100x} = \frac{1}{2}$
$\frac{150000}{x^2+100x} = \frac{1}{2}$
$\frac{300000}{x^2+100x} =1$
$300000= x^{2} +100x x^2+100x-300000=0 x^2+600x-500x-300000=0 x(x+600)-500(x+600) (x-500)(x+600) x=500, x \neq -600$

The usual speed was 500 hm/hr