Question

A man can row 30 km upstream and 44 km downstream in 10 hours. He can also row 40 km upstream and 55 km downstream in 13 hours. Find the rate of the current.
A.3 km/hr.
B.2 km/hr.
C.4 km/hr.
D.5 km/hr.

Answer 1

Hint: In this question, we need to determine the rate of the current or the speed of the stream such that the man can row 30 km upstream and 44 km downstream in 10 hours and, also row 40 km upstream and 55 km downstream in 13 hours. For this, we will establish the relation between the speed of the boat and the stream (or current) and solve them simultaneously to get the result.

Complete step-by-step answer:
Let the speed of the boat be $ {v_b} $ and the speed of the stream (or current) be $ {v_s} $ .
Speed of the man while travelling in the downstream is $ \left( {{v_b} + {v_s}} \right) $ and speed of the man while travelling in the downstream is $ \left( {{v_b} - {v_s}} \right) $ .
Let $ 1/\left( {{v_b} - {v_s}} \right) $ be ‘u’ and $ 1/\left( {{v_b} + {v_s}} \right) $ be ‘v’.
According to the question, man can row 30 km upstream and 44 km downstream in 10 hours. So, we can write
 $
  {\text{Distance}} = velocity \times time \\
   \Rightarrow time = \dfrac{{{\text{Distance}}}}{{velocity}} \\
   \Rightarrow 10 = \dfrac{{30}}{{{v_b} - {v_s}}} + \dfrac{{44}}{{{v_b} + {v_s}}} \\
   \Rightarrow 10 = 30u + 44v - - - - (i) \;
  $
Also, it has been given that the man can row 40 km upstream and 55 km downstream in 13 hours.
 $
  {\text{Distance}} = velocity \times time \\
   \Rightarrow time = \dfrac{{{\text{Distance}}}}{{velocity}} \\
   \Rightarrow 13 = \dfrac{{40}}{{{v_b} - {v_s}}} + \dfrac{{55}}{{{v_b} + {v_s}}} \\
   \Rightarrow 13 = 40u + 55v - - - - (ii) \;
  $
Solving equations (i) and (ii), we get
 $
  \left( {30u + 44v = 10} \right) \times 4 \to 120u + 176v = 40 - - - - (iii) \\
  \left( {40u + 55v = 13} \right) \times 3 \to 120u + 165v = 39 - - - - (iv) \;
  $
Subtracting equation (iii) and (iv), we get
 $
  120u + 176v - 120u - 165v = 40 - 39 \\
   \Rightarrow 11v = 1 \\
   \Rightarrow v = \dfrac{1}{{11}}\;km/hr \;
  $
Substituting the value of ‘v’ in equation (i), we get
 $
  10 = 30u + 44 \times \dfrac{1}{{11}} \\
   \Rightarrow 30u = 10 - 4 \\
   \Rightarrow u = \dfrac{6}{{30}} \\
   \Rightarrow u = \dfrac{1}{5}\;km/hr \;
  $
Now, bask substituting the values of ‘u’ and ‘v’, we get
 $
  \dfrac{1}{{\left( {{v_b} - {v_s}} \right)}} = \dfrac{1}{5} \\
   \Rightarrow \left( {{v_b} - {v_s}} \right) = 5 \;
  $
And,
  $
  \dfrac{1}{{\left( {{v_b} + {v_s}} \right)}} = \dfrac{1}{{11}} \\
   \Rightarrow \left( {{v_b} + {v_s}} \right) = 11 \;
  $
Adding the above two equations, we get
 $
  {v_b} - {v_s} + {v_b} + {v_s} = 5 + 11 \\
   \Rightarrow 2{v_b} = 16 \\
   \Rightarrow {v_b} = 8\;km/hr \;
  $
Also,
Subtracting the above two equations, we get
 $
  {v_b} - {v_s} - {v_b} - {v_s} = 5 - 11 \\
   \Rightarrow - 2{v_s} = - 6 \\
   \Rightarrow {v_s} = 3\;km/hr \;
  $
Hence, the speed of the stream (or current) is 3km/hr.
So, the correct answer is “Option A”.

Note: It is interesting to note here that the relative velocity of the man in the downstream is more than the upstream as in the downstream the velocity of the stream supports the movement of the boat while in the upstream the velocity of the stream is against the movement of the boat. Also, all the units in the problem are the same i.e., in kilometers and hours so we don’t need to transform any unit.