 Hi and welcome to the session. Let's work out the following question. The question says a passenger train takes three hours less for a journey of 360 kilometer if its speed is increased by 10 kilometer per hour from its usual speed. What is its usual speed? Let us see the solution to this question first of all let the usual speed of the train be x kilometer per hour the increased speed of train will be equal to x plus 10 kilometer per hour distance is given to be 360 kilometer. Now from this question 360 by x minus 360 upon x plus 10 is equal to 3 hours because time is equal to distance upon speed and it's given to us that it takes three hours less for a journey of 360 kilometer if speed is increased by 10 kilometer per hour So their difference will be three hours this implies 360 into x plus 10 minus 360 x divided by x into x plus 10 is equal to 3 upon 1 this implies 360 x plus 360 minus 360 x divided by x into x plus 10 is equal to 3 upon 1 this implies 360 x s cancelled with minus 360 x we have 3600 divided by x into x plus 10 is equal to 3 on cross multiplication we get 3 x into x plus 10 is equal to 3600 this implies 3 x square plus 30 x minus 3600 is equal to 0 dividing throughout by 3 we get x square plus 10 x minus 1200 is equal to 0 Now we factorize this and we get x square plus 40 x minus 30 x minus 1200 is equal to 0 this implies x into x plus 14 minus 30 into x plus 40 is equal to 0 this implies x minus 30 into x plus 40 is equal to 0 This implies x is equal to minus 40 or x is equal to 30 But speed cannot be negative so we neglect x equals to minus 40 therefore our answer to this question is that speed is or the usual speed is 30 kilometer per hour So this is our answer to this question I hope that you understood the solution and enjoyed the session have a good day