Question

A motor boat whose speed is 20 km/h is still water, takes 1 hour more to go 48 km upstream than to return downstream to the same spot. Find the speed of the steam.

OR

Find the roots of the equation $\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30},x\ne -4,7$

Answer 1

Let the speed of the stream be x km/h

Speed of the boat while going upstream = (20 -x) km/h

Speed of the boat while going downstream = (20 +x) km/h
Time taken for the upstream journey $=\frac{48km}{(20-x)km/h}=\frac{48}{20-x}h$
Time taken for the downstream journey $=\frac{48km}{(20+x)km/h}=\frac{48}{20+x}h$
It is given that,
Time taken for the upstream Journey = Time taken for the downstream journey + 1 hour
$\Rightarrow \frac{48}{20-x}-\frac{48}{20+x}=1$
$\Rightarrow \frac{960+48x-960+48x}{(20-x)(20+x)}=1$
$\Rightarrow \frac{96x}{400-x^2}=1$
$\Rightarrow 400-x^2=96x$
$\Rightarrow x^2+96x-400=0$
$\Rightarrow x^2+100x-4x-400=0$
$\Rightarrow x(x+100)-4(x+100)=0$
$\Rightarrow (x+100)(x-4)=0$
$\Rightarrow x+100=0orx-4=0$
$\Rightarrow x=-100prx=1$
$\therefore$ x=4 [Since speed cannot be negative]
Thus, the speed of the stream is 4 km/h
OR
$\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}$
$\Rightarrow \frac{(x-7)-(x+4)}{(x+4)(x-7)}=\frac{11}{30}$
$\Rightarrow \frac{-11}{x^2-3x-28}=\frac{11}{30}$
$\Rightarrow x^2-3x-28=-30$
$\Rightarrow x^2-3x+2=0$
$\Rightarrow x^2-2x-x+2=0$
$\Rightarrow x(x-2)-1(x-2)=0$
$\Rightarrow (x-1)(x-2)=0$
$\Rightarrow x-1=0orx-2=0$
$\Rightarrow x=1orx=2$
Hence, the roots of the given equation are 1 and 2