 Hello and welcome to the session. In this session we discussed the following question which says, a train left one hour later than its scheduled time and in order to reach its destination 600 km away in time, it had to increase its speed by 20 km per hour from its usual speed So we have to find the usual speed of the train according to the conditions which are given in the question. So let's see the solution. First of all let's assume that the usual speed of the train be equal to x km per hour. Now in the question as you can see it's given that the train increases its speed by 20 km per hour in order to reach its destination in time. So the actual speed would be equal to x plus 20 km per hour. Now in the question as you can see it's given that the destination is 600 km away. So the distance traveled by the train is equal to 600 km. So let's find out the time taken by the train at usual speed which is x km per hour is equal to distance which is 600 upon the usual speed which is x. So time taken by the train at usual speed is equal to 600 upon x hours and the time taken by the train at actual speed is equal to the distance traveled which is 600 upon the actual speed that is x plus 20. So 600 upon x plus 20 hours is the time taken by the train at actual speed. In the question we have that the train left one hour later than its scheduled time. So the difference between the two times is equal to one hour or you can say 600 upon x minus 600 upon x plus 20 is equal to one. Now we solve this. So taking LCM on the left hand side we have x multiplied by x plus 20 in the denominator and in the numerator we have 600 into x plus 20 minus 600 into x. This is equal to one. Further we have 600x plus 12000 minus 600x upon x square plus 20x is equal to one. Now 600x and minus 600x cancels. So we have x square plus 20x is equal to 12000. Further x square plus 20x minus 12000 is equal to zero. Now we can solve this quadratic equation by splitting the middle term. So we get x square minus 100x plus 120x minus 12000 is equal to zero. By this we have written the middle term 20x as 120x minus 100x. So now we make the pairs and from the first pair we take out x common inside the bracket we have x minus 100 plus from the second pair we take out 120 common inside the bracket we have x minus 100 this is equal to zero. So x minus 100 whole multiplied by x plus 120 is equal to zero which means either x minus 100 equal to zero or x plus 120 equal to zero that is we have x equal to 100 or x equal to minus 120. Now since the speed cannot be negative therefore x is not equal to minus 120 and hence x is equal to 100 and we had assumed x to be the usual speed of the train. So we now say that the usual speed of the train is equal to 100 kilometers per hour. 100 kilometers per hour is our final answer. This completes the session hope you have understood the solution of this question.