Question

A motor boat whose speed is 18 km/hr in still water takes 1 hr more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.
OR
A train travels at a certain average speed for a distance of 63 km and then travels at a distance of 72 km at an average speed of 6 km/ hr more than its original speed. If it takes 3 hours to complete total journey, what is the original average speed ?

Answer 1

Given, speed of motor boat in still water = 18 km/hr.
Let speed of stream = x km/ hr.
$\therefore$ Speed of boat downstream = (18 + x) km/hr.
And speed of boat upstream = (18 - x) km/hr.
Time of the upstream journey = $\frac{24}{(18-x)}$
Time of the downstream journey = $\frac{24}{(18+x)}$
According to the question,
$\frac{24}{(18-x)}-\frac{24}{(18+x)}=1$
$\frac{24}{(18-x)}-\frac{24}{(18+x)}=1$
$\frac{24(18+x)-24(18-x)}{(18-x)(18+x)}=1$
$\frac{24\times 18+24x-24\times 18+24x}{324-x^2}=1$
$\frac{48x}{324-x^2}=1$
$48x=324-x^2$
$\Rightarrow x^2+48x-324=0$
$\Rightarrow x^2+54x-6x-324=0$
$\Rightarrow x(x+54)-6(x+54)=0$
$\Rightarrow (x+54)(x-6)=0$
Either x + 54 = 0
x = - 54
Rejected, as speed cannot be negative
or x-6 = 0
x = 6
Thus, the speed of the stream is 6 km/hr.
OR
Let original average speed of train be x km/hr.
$\therefore$ Increased speed of train = (x + 6) km/hr.
Time taken to cover 63 km with average speed $=\frac{63}{x}$ hr.
Time taken to cover 72 km with increased speed = $\frac{72}{(x+6)}$hr.
According to the question.
$\frac{63}{x}+\frac{72}{x+6}=3$
$\Rightarrow \frac{63(x+6)+72\left(x\right)}{\left(x\right)(x+6)}$
$\Rightarrow \frac{63x+378+72x}{x^2+6x}=3$
$\Rightarrow 135x+378=3(x^2+6x)$
$\Rightarrow 135x+378=3x^2+18x$
$\Rightarrow 3x^2+18x-135x-378=0$
$\Rightarrow 3x^2-117x-378=0$
$\Rightarrow 3(x^2-39x-126)=0$
$\Rightarrow x^2-39x-126=0$
$\Rightarrow x^2-42x+3x-126=0$
$\Rightarrow x(x-42)+3(x-42)=0$
$\Rightarrow (x-42)(x+3)=0$
Either x - 42 = 0
x = 42
or x + 3 = 0
x = -3
Rejected (as speed cannot be negative)
Thus, average speed of train is 42 km/hr.