Question

a motorboat was period 24 km per hour in still water takes 1 hour more to go 32 km upstream then to return downstream to the same spot find the speed of the stream

Answer 1

let speed of the stream be x km/h
distance = 32 km
time going upstream = t + 1, hrs going downstream = t hrs
therefore,
$24 + x = \frac{32}{t} \: \: \: \: \: \: \: (i)\\ 24 - x = \frac{32}{t + 1} \: \: \: \: \: (ii) \\ (i) + (ii) - - - > \\ 24 + x + 24 - x = \frac{32}{t} + \frac{32}{t + 1} \\ 48 = \frac{32t + 32 + 32t}{t {}^{2} + t} \\ 48 {t}^{2} + 48t = 64t + 32$
48t²-16t-32 = 0
3t²-t-2 = 0
3t³-3t+2t-2 = 0
3t(t-1) + 2(t-1) = 0
(3t+2)(t-1) = 0
t = -⅔, 1
but time can't be -ve
therefore t = 1 hr
therefore
24 + x = 32
x = 32 - 24 = 8
speed of stream is 8 km/h

Answer 2

Let, the speed of the stream be x km/hr

Speed of boat in still water =20 km/hr

∴Speed of boat with downstream 20+x km/hr

∴ Speed of boat with upstream 20−x km/hr

As per given condition

20−x48−20+x48=1

⟹48[20−x1−20+x1]=1

⟹[(20−x)(20+x)20+x−20+x]=481

⟹400−x22x=481

⟹96x=400−x2

⟹x2+96x−400=0

⟹x2+100x−4x−400

⟹x(x+100)−4(x+100)=0

⟹(x−4)(x+100)=0

Either, x=4 or x=−100

∵ Speed cannot be negative ∴x=4 km/hr is considered.

∴ the speed of the stream =4 km/hr