Question

A boat goes 24km upstream and 28km downstream in 6 hours. It goes 30km upstream and 21km downstream in 6 hours and 30 minutes. The speed of the boat on still water is
(a) 8kmph
(b) 9kmph
(c) 12kmph
(d) 10kmph

Answer 1

Hint: Assume that the speed of the boat in still water is x kmph and the speed of the stream is y kmph. Use the fact that the upstream speed of the boat will be \[x-y\] kmph and the downstream speed of the boat will be \[x+y\] kmph. Write equations using the fact that speed is the ratio of distance travelled to the time taken. Solve the equations using elimination method to find the speed of the boat in still water.


Complete step-by-step answer:
We have data regarding distance covered by boat upstream and downstream.
We have to calculate the speed of the boat in still water.
We will assume that the speed of the boat in still water is x kmph and the speed of the stream is y kmph.
Thus, the upstream speed of the boat is \[x-y\] kmph and the downstream speed of the boat is \[x+y\] kmph.
We know that speed is the ratio of distance covered to the time taken.
So, time taken to cover the distance will be the ratio of distance covered to the speed.
Thus, the time taken by the boat to cover 24km upstream with speed \[x-y\] kmph is \[\dfrac{24}{x-y}\] hours.
Similarly, the time taken by the boat to cover 28km upstream with speed \[x+y\] kmph is \[\dfrac{28}{x+y}\] hours.
 The total time taken by the boat to travel 24km upstream and 28km downstream is 6 hours.
Thus, we have \[6=\dfrac{24}{x-y}+\dfrac{28}{x+y}.....\left( 1 \right)\].
Similarly, the time taken by the boat to cover 30km upstream with speed \[x-y\] kmph is \[\dfrac{30}{x-y}\] hours.
The time taken by the boat to cover 21km upstream with speed \[x+y\] kmph is \[\dfrac{21}{x+y}\] hours.
The total time taken by the boat to travel 30km upstream and 21km downstream is 6 hours 30 minutes.
We know that \[1hr=60\min \].
Dividing the above equation by 2, we have \[\dfrac{1}{2}hr=30\min \].
Thus, 6 hours 30 minutes is equal to \[6+0.5=6.5\] hours.
So, we have \[6.5=\dfrac{30}{x-y}+\dfrac{21}{x+y}.....\left( 2 \right)\].
Let’s assume \[u=\dfrac{1}{x-y},v=\dfrac{1}{x+y}\].
We can rewrite equation (1) as \[6=24u+28v.....\left( 3 \right)\].
We can rewrite equation (2) as \[6.5=30u+21v.....\left( 4 \right)\].
Multiplying equation (3) by 3 and equation (4) by 4 and subtracting the two equations, we have \[72u+84v-\left( 120u+84v \right)=18-26\].
Thus, we have \[48u=8\Rightarrow u=\dfrac{1}{6}.....\left( 5 \right)\].
Substituting the value of equation (5) in equation (3), we have \[6=24\left( \dfrac{1}{6} \right)+28v\].
Simplifying the above equation, we have \[28v=2\Rightarrow v=\dfrac{1}{14}\].
Thus, we have \[u=\dfrac{1}{x-y}=\dfrac{1}{6}.....\left( 6 \right)\] and \[v=\dfrac{1}{14}=\dfrac{1}{x+y}.....\left( 7 \right)\].
Rearranging the terms of equation (6) and (7), we have \[x-y=6,x+y=14\].
Adding the two equations, we have \[2x=20\Rightarrow x=10\].
Hence, the speed of the boat in still water is \[x=10\] kmph, which is option (d).

Note: We can check if the calculated values are correct or not by substituting the values in the equations and checking if the calculated values satisfy the equations or not. It’s important to consider that the speed of the boat in still water decreases while going upstream and increases while going downstream.