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A motor boat whose speed is 18 km/hr in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. The speed of the stream isA.6 km/hrB.5 km/hrC.3.5 km/hrD.4.5 km/hr

Last updated date: 26th Mar 2023
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Hint: To solve the question, we have to apply the upstream and downstream speed formula and the given information to obtain equations.

Complete step-by-step Solution:
Let the speed of the stream be x km/hr.
Let the time taken to travel 24 km downstream by motor boat be t hours.
$\Rightarrow$The time taken to travel 24 km upstream by motor boat = (t + 1) hours.
The given speed of the motor boat in the still water is equal to 18 km/hr.
The given distance travelled by motor boat is equal to 24 km.
We know that the formulae
The upstream speed of the motor boat = Speed of the motor boat in still water – Speed of the stream
= 18 - x
The downstream speed of the motor boat = Speed of the motor boat in the still water + Speed of the stream
= 18 + x
We know that the formula for the distance travelled by a boat = Net speed of the boat $\times$time taken to travel
By substituting given values in the above formula for the boat travelled 25 km upstream in (t + 1) hours, we get
$24=(18-x)(t+1)$
$24=18t+18-xt-x$
$xt+x=18t-6$
$x=\dfrac{18t-6}{t+1}$ …….(1)
By substituting given values in the above formula for the boat travelled 25 km downstream in t hours, we get
$24=(18+x)t$
By substituting the equation (1) in the above equation we get
$24=\left( 18+\dfrac{18t-6}{t+1} \right)t$
$24=\left( \dfrac{18t+18+18t-6}{t+1} \right)t$
$24=\left( \dfrac{36t+12}{t+1} \right)t$
$24t+24=36{{t}^{2}}+12t$
$36{{t}^{2}}-12t-24=0$
$3{{t}^{2}}-t-2=0$
$3{{t}^{2}}-3t+2t-2=0$
$\left( 3t+2 \right)\left( t-1 \right)=0$
$\Rightarrow t=1,\dfrac{-3}{2}$
The time taken to travel 24 km downstream by motor boat = 1 hour.
By substituting the t value in equation (1) we get
$x=\dfrac{18(1)-6}{1+1}$
$x=\dfrac{12}{2}=6$km/hr.
$\therefore$ The speed of the stream = 6 km/hr.
Hence, option (b) is the right choice.

Note: The alternative procedure can be forming a quadratic equation of x instead of forming a quadratic equation of t and the options can be eliminated by substituting the values in the obtained quadratic equation of x to check whether the values satisfy the equation or not. The possibility of mistake can be the calculations since the procedure of solving has multiple calculations.