Question

A boat travels 30 km upstream and 44 km downstream in 10 hours in 13 hours it can go 40 km upstream and 55 km downstreem determine the speed of the stream and that of the biat in sitll water show solution

Answer 1

Answer:

Speed of stream = 3 km / hr.

Speed of boat in still water = 8 km / hr.

Step-by-step explanation:

Let the speed of the boat in still water be a km / hr and stream be b km / hr

For upstream = a - b

For downstream = a + b

We know :

Speed = Distance / Time

Case 1 .

10 = 30 / a - b + 44 / a + b

Let 1 / a - b = x and 1 / a + b = y

30 x + 44 y = 10 ... ( i )

Case 2 .

13 = 40 / a - b + 55 / a + b

40 x + 55 y = 13 ... ( i )

Multiply by 4 in ( i ) and by 3 in ( ii )

120 x + 176 y = 40

120 x = 40 - 176 y ... ( iii )

120 x + 165 y = 39

120 = 39 - 165 y ... ( iv )

From ( iii )  and  ( iv )

40 - 176 y = 39 - 165 y

11 y = 1

y = 1 / 11

120 x = 40 - 176 y

120 x = 40 - 176 / 11

x = 1 / 5

Now :

1 / a - b = 1 / 5

a - b = 5

a = 5 + b ... ( v )

1 / a + b = 1 / 11

a + b = 11

a = 11 - b ... ( vi )  

From ( v  ) and ( vi )

11 - b = 5  + b

2 b = 6

b = 3

a = 5 + b

a = 5 + 3

a = 8

Hence we get answer.

Answer 2

Answer:

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

Explanation:

$\text{Let the speed of the boat in still water be x kmh ^{-1} }\\\\\text{and of the stream be y kmh ^{-1} .}\\\\\\\text{ \therefore Speed of the boat upstream = (x-y) kmh ^{-1} }\\\\\text{and speed of the boat downstram = (x+y) kmh ^{-1} .}\\\\\\\text{Distance upstream = 30 km and distance downstream = 44 km}\\\\\\\text{We know that time, t = \dfrac{Distance, \: \: d}{Speed, \: \: s} .}\\\\\\\text{So, we get}\\\\\text{Time taken to go upstream = \dfrac{30}{x-y} }$

$\text{and Time taken to go downstream = \dfrac{44}{x+ y} .}\\\\\\\text{According to the question,}\\\\\text{Time taken to go upstream + Time taken to go downstream = 10 hrs}\\\\\\\implies \dfrac{30}{x-y} + \dfrac{44}{x+y} = 10 \; \; \; -----------\text{eqn 1}\\\\\\\text{But, the boat can also go}\\\\\text{40 km upstream and 55 km downstream in 13 hrs.}$

$\text{So, time taken to go upstream = \dfrac{40}{x-y} }\\\\\\\\\text{and time taken to go downstream = \dfrac{55}{x + y} .}\\\\\\\text{According to the question,}\\\\\text{Time taken to go upstream + Time taken to go downstream = 13 hrs}\\\\\\\implies \dfrac{40}{x-y} + \dfrac{55}{x+y} = 13 \; \; \; -----------\text{eqn 2}\\\\\\\text{Let \dfrac{1}{x-y} be a and \dfrac{1}{x+y} be b.}\\\\\text{Then from eqn 1 and eqn 2, we get}$

$30a + 44b = 10 \; \; \; --------\text{eqn 1}\\\\\\\text{And}\\\\\\40a + 55b = 13 \; \; \; --------\text{eqn 2}\\\\\\\text{Now, multiplying eqn 1 by 4 and eqn 2 by 3, we get}\\\\\\120a + 176b = 40 \; \; \; --------\text{eqn 1}\\\\\\\text{And}\\\\\\120a + 165b = 39 \; \; \; --------\text{eqn 2}\\\\\\\text{Now, subtracting eqn 2 from eqn 1, we get}\\\\\\120a + 176b - (120a + 165b) = 40 - (39)\\\\\implies 120a - 120 a + 176b - 165b = 40 - 39\\\\\implies 11b = 1$

$\implies \boxed{b = \dfrac{1}{11}}\\\\\\\text{Put the value of b in eqn 1 or eqn 2. I'll put in eqn 1.}\\\\\\\text{We get}\\\\\\120a + 176 \left( \dfrac{1}{11} \right) = 40\\\\\\\implies 120a + 16 = 40\\\\\implies 120a = 40 - 16\\\\\implies 120a = 24\\\\\implies a = \dfrac{24}{120}\\\\\\\implies \boxed{a = \dfrac{1}{5}}$

$\text{Now put the values of a and b in}\\\\\\\dfrac{1}{x-y} = \dfrac{1}{5}\\\\\\\text{And}\\\\\\\dfrac{1}{x+y} = \dfrac{1}{11}\\\\\\\text{OR}\\\\\\x - y = 5\\\\x + y = 11\\\\\text{Adding these two equations, we get}\\\\2x = 16\\\\\\\implies \boxed{\bold{x = 8 \: kmh^{-1}}}\\\\\\\implies 8 + y = 11\\\\\implies \boxed{\bold{y = 11-8 = 3 \: kmh^{-1}}}$

∴ The speed of the boat in still water is 8 km/h and that of the stream is 3 km/h.