Question

A boat goes \[{\text{30}}\] km upstream and \[{\text{44}}\] km downstream in \[{\text{10}}\] hours. In \[{\text{13}}\] hours it can go \[{\text{40}}\] km upstream and \[{\text{55}}\] km downstream. Then determine the speed of the stream and the speed of the boat in still water.

Answer 1

Hint: Here we use the concept of Distance and speed involving Boats and streams.
Upstream means the boat moves against the stream of water. Because of which
\[{\text{Upstream speed = speed of the boat - speed of the stream}}{\text{.}}\]
Downstream means a boat moves along the stream of water. Because of which
\[{\text{Downstream speed = speed of the boat + speed of the stream}}{\text{.}}\]
Required Formula:
 $ {Distance = Speed \times Time} $

Complete step-by-step answer:
Given :
Time taken for \[{\text{30}}\] km upstream and \[{\text{44}}\] km downstream is \[{\text{10}}\] hours.
Time taken for \[{\text{40}}\] km upstream and \[{\text{55}}\] km downstream is \[{\text{13}}\] hours.
We need to determine the speed of the stream and the speed of the boat in still water.
Let “s” be the speed of the stream.
Let “b” be the speed of the boat in still water.
Therefore,
\[{\text{Upstream speed = b - s}}\]
\[{\text{Downstream speed = b + s}}\]
According to the question,
Time taken for \[{\text{30}}\] km upstream and \[{\text{44}}\] km downstream is \[{\text{10}}\] hours.
As we know that $ {\text{time = }}\dfrac{{{\text{distance}}}}{{{\text{speed}}}} $
Time taken is the sum of time taken by upstream plus the time taken by downstream
 $ \Rightarrow \dfrac{{{\text{30}}}}{{{\text{b - s}}}}{\text{ + }}\dfrac{{{\text{44}}}}{{{\text{b + s}}}}{\text{ = 10}} $ ………………..(1)
Time taken for \[{\text{40}}\] km upstream and \[{\text{55}}\] km downstream is \[{\text{13}}\] hours.
 $ \Rightarrow \dfrac{{{\text{40}}}}{{{\text{b - s}}}}{\text{ + }}\dfrac{{{\text{55}}}}{{{\text{b + s}}}}{\text{ = 13}} $ …………….(2)
Solving 1 and 2 we get,
Multiply (1) with 4 and Multiply (2) with 3 we get,
 $ \dfrac{{{\text{120}}}}{{{\text{b - s}}}}{\text{ + }}\dfrac{{{\text{176}}}}{{{\text{b + s}}}}{\text{ = 40}} $ ……………….(3)
 $ \dfrac{{{\text{120}}}}{{{\text{b - s}}}}{\text{ + }}\dfrac{{{\text{165}}}}{{{\text{b + s}}}}{\text{ = 39}} $ ………………(4)
(3) – (4) ,
 $ \dfrac{{{\text{11}}}}{{{\text{b + s}}}}{\text{ = 1}} $
 $ \Rightarrow {\text{b + s = 11}} $ ……………………..(5)
Substituting (5) in (4) we get,
 $ \dfrac{{{\text{120}}}}{{{\text{b - s}}}}{\text{ + }}\dfrac{{{\text{165}}}}{{{\text{11}}}}{\text{ = 39}} $
 $
   \Rightarrow \dfrac{{{\text{120}}}}{{{\text{b - s}}}}{\text{ + 15 = 39}} \\
   \Rightarrow \dfrac{{{\text{120}}}}{{{\text{b - s}}}}{\text{ = 39 - 15}} \\
   \Rightarrow \dfrac{{{\text{120}}}}{{{\text{b - s}}}}{\text{ = 24}} \\
   \Rightarrow {\text{b - s = }}\dfrac{{{\text{120}}}}{{{\text{24}}}} \\
  $
   $ \Rightarrow {\text{b - s = 5}} $ …………………….(6)
Adding (5) and (6) we get,
 $
  {\text{b + s + b - s = 11 + 5}} \\
   \Rightarrow {\text{2b = 16}} \\
   \Rightarrow {\text{b = }}\dfrac{{{\text{16}}}}{{\text{2}}} \\
  $
 $ \Rightarrow {\text{b = 8}} $ …………..(7)
Substituting (7) in (6) we get,
 $
   \Rightarrow {\text{b - s = 5}} \\
   \Rightarrow {\text{8 - s = 5}} \\
   \Rightarrow {\text{8 - 5 = s}} \\
   \Rightarrow {\text{s = 3}} \;
  $
Therefore, speed of the boat “b” in still water is $ {\text{8 kmph}} $
Speed of the stream “s” is $ {\text{3 kmph}} $
So, the correct answer is “$ {\text{3 kmph}} $”.

Note: Questions that involve the concept of Boats and streams are involved with distance and speed concepts. We need to have knowledge about the terms involved in the questions and the appropriate formulae needed to solve the questions. Assigning variables to the unknown terms and framing the equations as per the given conditions helps us to solve for the required quantity.