CLASS-9
SIMULTANEOUS LINEAR EQUATIONS - BOAT & STREAM

PROBLEM ON BOAT & STREAM -

Example.1) A sailor goes 10 km downstream in 40 minutes and returns back to the starting point in 90 minutes. Find the speed of the sailor in still water and the speed of the current.

Ans.) Let, the speed of the sailor in still water be ‘x’ km/h and the speed of the current be ‘y’ km/h

Then, speed downstream = (x + y) km/h

Speed upstream = (x – y) km/h

                                                                                                                                    10

Time taken to cover 10 km downstream = --------- hours

                                                 x + y

         10              40

so, ---------- = ----------

        x + y            60

 

         10             2

or, ---------- = --------

       x +  y           3


or,    x + y  =  15  …………………….. (i)

 

                                                                                                                                 10

Time taken to cover 10 km upstream = --------- hours

                                              x – y

 

         10             90

so, ---------- = --------

        x - y           60

 

         10             3

or, ---------- = -------

       x - y            2


or,   3x – 3y = 20 ……………………….(ii)

 

we have to multiply (i) by 3, and we get –

           3x + 3y = 45 …………………….(iii)

Now, we will add (iii) & (ii), and we get –

                 3x + 3y = 45 

                 3x – 3y =  20

             ------------------

                       6x = 65

               =>      x = 65/6

Now, we will put the value of x in (i), and we get –

             x + y = 15 

    =>   65/6 + y = 15

    =>            y =  15 - 65/6

    =>            y = (90 – 65)/6 = 25/6

Hence, the speed of the sailor in still water is 65/6 km/h and the speed of the current is 25/6 km/h.        (Ans.)