 Hello and welcome to this session. In this session we will discuss a question which says that an airplane left 15 minutes later than its schedule time and in order to use the destination 1,250 km away in time it had to increase its speed by 250 km per hour from its usual speed. Now before starting the solution of this question we should know our result and that is speed is equal to distance over time which implies time is equal to distance over speed. Now this result will work out as a key idea by solving out this question and now we will start with the solution. Now in the question we have to find the usual speed of the airplane. Submit the usual speed of the airplane is equal to x km per hour. Now the distance which is to be covered is equal to 1,250 km and given distance is equal to 1,250 km. Now using this formula which is given as a key idea time taken to cover a distance of 1,250 km with the speed x km per hour is equal to distance over speed which is equal to 1,250 km over the speed which is x km per hour. Which is equal to 1,250 by x hours. Now in the second case the speed is increased by 250 km per hour and we have taken the usual speed of the airplane as x km per hour. So now the increased speed of the airplane is equal to x plus 250 km per hour. Now the distance is equal to 1,250 km therefore the time taken to cover the distance of 1,250 km with the speed of x plus 250 km per hour is distance by speed which is further equal to 1,250 km over x plus 250 km per hour. This is further equal to 1,250 km over x plus 250 hours. Now in the equation it is given that the airplane left 50 minutes later than its scheduled time. Now this is the time taken with the speed of x km per hour and this is the time taken by the airplane with the speed of x plus 250 km per hour. Now according to the equation 1,250 by x minus 1,250 by x plus 250 is equal to 50 by 60. That means the difference of the time taken is equal to 50 minutes and 50 minutes is equal to 50 by 60 hours. This implies on taking 1,250 common within brackets that will be 1 by x minus 1 by x plus 250 is equal to 5 by 6. This implies 1,250 within brackets taking the same which is x into x plus 250 the whole and in the numerator it will be x plus 250 minus x is equal to 5 by 6. Further this implies 1,250 and within brackets these times will be cancelled with each other and it will be 250 over x square plus 250x is equal to 5 by 6. This implies on trans-multiplying 1,250 into 250 into 6 is equal to 5 into x square plus 250x the whole. This implies 1,250 into 250 into 6 equal to 5 is equal to x square plus 250x. Further this implies now here 5 into 50 is 250 and on solving this it will be 3x 75000 which is equal to x square plus 250x. This further implies x square plus 250x minus 3x 75000 is equal to 0. Now this is a quadratic equation in x so for solving it we will split the middle term. So this implies x square plus 750x minus 500x minus 3x 75000 is equal to 0. Which further implies x into x plus 750 the whole minus 500 into x plus 750 the whole is equal to 0. Which further implies x plus 750 the whole into x minus 500 the whole is equal to 0. Now this implies either 750 is equal to 0 or 500 is equal to 0. Which further implies x is equal to minus 750 or x is equal to 500. Now we have taken x through speed of power plane. So as speed cannot be negative therefore x is equal to 500 that is x is equal to 500 kilometer per hour. Therefore, usual speed of the airplane is equal to 500 kilometer per hour. So this is the solution of the given question and that's all for this session. Hope you all have enjoyed the session.