Question

A boat goes 30 km upstream and 44 km downstream in 10 ours. In 13 hours it goes 40km upstream and 55 km downstream. Determine the speed of the stream and that boat in still water.​

Answer 1

Answer:

Let the speed of boat in still water=x=x km\hr and The speed of stream=y=y km\hr

Speed of boat at downstream

\Rightarrow \left (x+y  \right )km/hr⇒(x+y )km/hr

Speed of boat at upstream

\Rightarrow \left (x-y  \right )km/ hr⇒(x−y )km/hr

\because time =\cfrac{distance}{speed}∵time=speeddistance

Time taken to cover 30 km upstream \Rightarrow \cfrac{30}{x-y}⇒x−y30

Time taken to cover 44 km downstream\Rightarrow \cfrac{44}{x+y}⇒x+y44

According to the first condition,

\Rightarrow \dfrac{30}{x-y}=\dfrac{44}{x+y}=10⇒x−y30=x+y44=10

Time taken to cover 40 km upstream \Rightarrow \cfrac{40}{x-y}⇒x−y40

Time taken to cover 55 km downstream \Rightarrow \cfrac{55}{x+y}⇒x+y55

According to the second condition,

\Rightarrow \dfrac{40}{x-y}=\dfrac{55}{x+y}=13⇒x−y40=x+y55=13

Let \dfrac{1}{x-y}=u\quad and \quad\dfrac{1}{x+y}=vx−y1=uandx+y1=v

\Rightarrow 30u+44v=10.....eq1⇒30u+44v=10.....eq1

\Rightarrow 40u+55v=13.....eq2⇒ please mark as brain list answer

Answer 2

Hello ,

Let the speed of the boat in still water be x km/h and speed of the stream be y km/h. Then the speed of the boat downstream → ( x + y ) km/h ..

And the speed of the boat upstream = ( x - y ) km/h ..

$Time = \dfrac{Distance}{Speed}$ ..

In the First Case : -

When the boat goes 30 km upstream ,let the time taken,in hour,be ${t}_{1}$ . Then,

$\longmapsto\tt{t}_{1}=\dfrac{30}{x-y}$

Let ${t}_{2}$ be the time,taken by the boat to go 44 km downstream .. then,

$\longmapsto\tt{t}_{2}=\dfrac{44}{x+y}$

The total time taken ${t}_{1}+{t}_{2}$ is 10 hours , ∴ we get the equation :

$\longmapsto{\dfrac{30}{x-y}+\dfrac{44}{x+y}=10}$......... (1)

In the Second Case: -

In 13 hours it can go 40 km upstream and 55 km downstream . we get the equation :

$\longmapsto{\dfrac{40}{x-y}}+\dfrac{55}{x+y}=13$......(2)

Put : -

$\longmapsto\tt{\dfrac{1}{x-y}}=u\:and\:\dfrac{1}{x+y}=v$...(3)

On substituting these values in equations (1) and (2), we get the pair of Linear equations :

30u + 44v = 10 or 30u + 44v - 10 = 0 ... (4)

40u + 55v = 13 or 40u + 55v-13 = 0 .. (5)

Using cross multiplication,we get :

$\longmapsto\tt{\dfrac{u}{44(-13)-55(-10)}=\dfrac{v}{40(-10)-30(-13)}=\dfrac{1}{30(55)-44(40)}}$

i.e. ,

$\longmapsto\tt{\dfrac{u}{-22}=\dfrac{v}{-10}=\dfrac{1}{-110}$

i.e.,

$\longmapsto\tt{u=\dfrac{1}{5},v=\dfrac{1}{11}}$

Now put these values of u and v in equation (3),we get :

$\longmapsto\tt{\dfrac{1}{x-y}=\dfrac{1}{5}\:and\:\dfrac{1}{x+y}=\dfrac{1}{11}$

→x-y = 5 and x+y =  11 .... (6)

Adding these equations , we get :

2x = 16

x = 8

Subtracting the equations in(6),we get :

2y = 6

y = 3

Hence,the speed of the boat in still water is 8 km/h and the speed of the stream is 3 km/h ..