Question

a boat goes 30 k upstream and 44 km dowmstsam in 10 hours in 13 hours it can go 40 km upstream and the 55 km down-stream and determine the speed of the stream and that of the bot in still water​

Answer 1

Answer:

Step-by-step explanation:

Let the speed of the stream be = u kmph

let the speed of the boat be = v kmph

speed upstream will be = v - u  kmph

speed downstream will be =  v + u  kmph

30 km  upstream in time duration  =   30 / (v - u)  hrs

44 km  down stream in time duration  =  44 / (v + u)  hrs

      44 / (v + u)  + 30 / (v -u)  = 10 hrs     --- (1)

Similarly,

     40 /(v - u)  + 55 / (v +u)  = 13 hrs         Multiply with 3/4:  

     30/(v-u) + 165 / 4(v+u) =    39/4        --- (2) 

Now (1) - (2)  =>    [44 - 165/4] / (v+u) = 10 - 39/4 = 1/4

                  =>     v + u = 11      --- (3)

Substitute this in (1) to get: 

            44/11 + 30/(v-u) = 10

          =>  v - u = 30/6 = 5      --- (4)

Solving  (3) and (4) ,  we get :  v = 8 kmph    and  u = 3 kmph

Answer 2

Answer:OK

Step-by-step explanation:

LET THE SPEED OF STILL WATER=x km/hr

& THE SPEED OF STREM= y km/hr

BOAT SPEED IN UPSTREAM=(x-y)km/hr

BOAT SPEED IN DOWNSTREAM=(x+y)km/hr

CASE1

30/x-y+44/x+y=10

30a+44b=10              (eq 1)        (let 1/x-y=a&1/x+y=b)

CASE2

40/x-y+55/x+y=13

40a+55b=13               (eq 2)

ATQ

eq 1*4=120a+176b=40

eq 2*3=120a+165b=39

11b=1

b=1/11

putting value of b in eq 1

30a+44b=10

30a+44*1/11=10

30a+4=10

30b=6

b=1/5

like we gave that

1/5=a

1/5=1/x-y

x-y=5              eq 1

1/11=b

1/11=1/x+y

x+y=11              eq 2

by elemination

x-y=5

x+y=11

-y=-6

y=6

putting value of y in eq 1

x-y=11

x-6=11

x=5

speed of boat in still water=5km/hr

speed of boat in stream=6km/hr