Question

Q. A boat goes 30 km upstream and 44 km downstream in 10hrs. In 12hrs it can go 40km upstream and 55km downstream. Determine the speed of the stream and that of the boat.​

Answer 1

Given:-

In case I,

Distance covered in upstream =30km

Distance covered in downstream = 44km

Total time taken = 10hours

In case II,

Distance covered in upstream = 40km

Distance covered in downstream = 55km

Total time taken = 13hours

As we know:-

Let speed of boat in still water be v km hr & speed of current(stream) is u km/hr.

Speed of upstream = (v - u) km/hr

Speed of downstream = (v + u) km/hr

As:-

$\sf{Speed = \frac{Distance}{Time}} \\ \\ \sf{Time = \frac{Distance}{Speed}}$

Therefore:-

[Time taken in upstream + Time taken in downstream =Total time]

$\sf{\frac{30}{(v - u)} + \frac{44}{(v + u)} = 10 \: \: ...... {eq} \: 1}$

$\sf{\frac{40}{(v - u)} + \frac{55}{(v + u)} = 13 \: \: ......eq \: 2}$

$\sf{Now \: let \: us \: consider \: \frac{1}{(v - u)} \: be \: x \: and \: \frac{1}{(v + u)} \: be \: y.}$

Then the required equation becomes,

30x + 44y = 10.........eq 3

40x + 55y = 13....... eq 4

Now by using the cross multiplication method that is,

$\sf{ \frac{1}{(a_{1}b_{2} - a_{2}b_{1})} = \frac{x}{(b_{1}c_{2} - b_{2}c_{1})} = \frac{y}{(c_{1}a_{2} - c{2}a_{1})}}$

$\sf{\frac{1}{30 \times (55) - 40 \times 44} = \frac{x}{44 \times ( - 13) - 55( - 10)} = \frac{y}{40(- 10) - 30( - 13)}}$

$\sf{\frac{x}{( - 22)} = \frac{y}{( - 10)} = \frac{1}{( - 110)}}$

Solving for x,

$\sf{\frac{x}{( - 22)} = \frac{1}{( - 110)}}$

$\sf{x \: = \frac{1}{5}}$

Solving for y,

$\sf{\frac{y}{( - 10)} = \frac{1}{( - 110)}}$

$\sf{y \: = \frac{1}{11}}$

Now:-

$\sf{ \frac{1}{(v - u)} = \frac{1}{5} } \\ \\ \sf{\frac{1}{(v + u)} = \frac{1}{11}}$

Taking the reciprocal we have,

$\sf{(v - u) = 5} \\ \\ \sf{(v + u) = 11}$

Now on further solving we have,

$\sf{v = 5 + u} \\ \\ \sf{ \bold{Puting \: the \: value \: of \: v \: we \: get}} \\ \\ = > \sf{v + u = 11} \\ \\ \sf{= > 5 + u + u = 11} \\ \\ \sf{= > 5 + 2u = 11} \\ \\ \sf{= > 2u = 6} \\ \\ \sf{= > u = 3}$

Also:-

$\sf{= > v - u = 5} \\ \\ \sf{= > v - 3 = 5} \\ \\ \sf{= > v = 5 + 3} \\ \\ = > \sf{v = 8}$

Therefore:-

The speed of boat in still water v is 8km/hour and speed of stream u is 3 km/hour.

Answer 2

Answer:

Solution is in the given attachment.

Hope it will help you mate.