Question

a boat takes 7 hours to travel 30 km upstream and 28 km downstream .it takes 5 hours to travel 21 km upstream and to return back. find the speed of the boat in still water.

Answer 1

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✺Answer:

♦️GiveN:

• Covers 30 km upstream and 28 km downstream in 7 hours.
• Then covers 21 km upstream and 21 downstream in 5 hours.

♦️To FinD:

• Speed of boat and speed of still water.

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✺Explanation of Q.

The above question is a type of time and distance problem which includes the data of a boat speed and speed of stream. When the boat is travelling against the stream ,we can say it is travelling upstream. Ans when it is travelling with the stream ,we can say it is travelling downstream.

⏺ Refer to the attachment.

So, we can say,

Speed Upstream

$\large{ \dag{ \red{ \boxed{ \rm{ = speed \: of \: boat \: in \: still \: water - speed \: of \: steam}}}}}$

Speed Downstream

$\large{ \dag{ \red{ \boxed{ \rm{ = speed \: of \: boat \: in \: still \: water + speed \: of \: stream}}}}}$

Basically it can be solved using linear equations in two variables. So, here is how we can solve it...

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✺Solution:

Let the speed of boat in still water = x

And, speed of stream = y

Then, Speed upstream =( x - y)km/hr

Speed downstream = (x + y)km/hr

Now, Using the Distance,speed and time formula:

$\large{ \rightarrow{ \boxed{ \green{ \rm{speed = \frac{distance}{time}}}}}}$

This can be written as,

$\large{ \rightarrow{ \boxed{ \rm{ \green{time = \frac{distance}{speed}}}}}}$

In the first journey,

Now, Distance travelled upstream = 30 km

Distance travelled downstream = 28 km

$\large{ \rm{ \rightarrow \: time \: taken(upstream) = \frac{32}{x - y}hrs}} \\ \\ \large{ \rm{ \rightarrow \: time \: taken(downstream) = \frac{28}{x + y} \: hrs}}$

Total time given 7 hours

$\Large{ \rm{ \therefore \: \frac{30}{x - y} + \frac{28}{x + y} = 7 \: hrs............(1) }}$

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In the 2nd journey.

Distance travelled upstream = 21 km

Distance travelled downstream = 21 km

$\large{ \rm{ \rightarrow \: time \: taken(upstream) = \frac{21}{x - y} hrs}} \\ \\ \large{ \rm{ \rightarrow \: time \: taken(downstream) = \frac{21}{x + y}hrs}}$

Total time given 5 hours

$\Large{ \rm{ \therefore \: \frac{21}{x - y} + \frac{21}{x + y} = 5 \: hrs.............(2)}}$

Now putting 1/x - y = a and 1/x + y = b in equations (1 ) and (2).

$\large{ \rm{ \rightarrow \: 30a + 28b = 7}} \\ \\ \large{ \rm{ \rightarrow \: 21a + 21b = 5}}$

Using Elimination method:

Multiplying eq. (1) with 3 and eq.(2) with 4,

$\large{ \rm{ \rightarrow \: 90a + 84b = 21}} \\ \\ \large{ \rm{ \rightarrow \: 84a + 84b = 20}}$

Subtracting these equations ,we get

$\large{ \rm{ \rightarrow \: 6a = 1}} \\ \\ \large{ \rm{ \rightarrow \: a = \frac{1}{6} }}$

Putting this value of a in Eq. (1)

$\large{ \rm{ \rightarrow \: \: 5\: \cancel{30} \times \frac{1}{ \cancel{6} } + 28b = 7}} \\ \\ \large{ \rm{ \rightarrow \: 28b = 2}} \\ \\ \large{ \rm{ \rightarrow \: b = \frac{1}{14}}}$

So, from here, we got, x-y = 6

And other one, x+y=14

By simply adding this equation,

$\large{ \rm{ \rightarrow \: 2x = 20}} \\ \\ \large{ \rightarrow \boxed{ \rm{\: x = 10 \: km {h}^{ - 1} }}} \\ \\\large{ \rm{ then\: we\: have \: 10 + y = 14}} \\ \\ \large{ \rm{ \rightarrow \: \boxed{ \rm{y = 4 \: km {h}^{ - 1} }}}}$

Hence, we got,

$\large{ \boxed{ \rm{ \therefore{ \pink{speed \: of \: boat = 10km {h}^{ - 1}}}}}} \\ \\ \large{ \boxed{ \rm{ \therefore{ \pink{speed \: of \: stream = 4km {h}^{ - 1} }}}}}$

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