Question

A river flows at a speed of 1 km/hr. A boat takes 15 hours to travel 112 km downstream and coming back the same distance upstream. Find the speed of the boat in still water. (Speed of the river flow is less than the speed of the boat in still water)

Answer 1

Let speed of boat=x
Then speed of boat in downstream=112/x+1
" " " " upstream=112/x-1
A/q,
112/x+1 + 112/x-1=15
=>112(x-1)+112(x+1)/(x+1)(x-1)=15
=>112x-112+112x+112=15(x^2-1)
=>224x=15x^2-15
=>15x^2-224x-15=0
=>15x^2-225x+x-15=0
=>15x(x-15)+1(x-15)=0
=>(15x+1)(x-15)=0
Therefore,x=15,-1/15(×)
Speed of boat=15 km/hrs.
Hope it helps

Answer 2

Speed of Boat is 15 $kmhr^{-1}$

Step-by-step explanation:

• Let speed of boat in still water be x km/hr.

∵ speed of river is given as 1 $kmhr^{-1}$

⇒ speed of boat downstream = (x + 1) $kmhr^{-1}$ .

∴ let time taken to go 112 km downstream = t

hence, t =$\frac {112}{x+1}$hours (As Time = $\frac {Distance}{Speed}$)

• Now, speed of boat upstream will be (x – 1) $kmhr^{-1}$

∴ let time taken to go 112 km upstream = T

t = $\frac {112}{x-1}\ hours$

• Given that total journey time was 15 hr

⇒ t + T = 15

$\frac {112}{x+1} + \frac {112}{x-1} = 15$

⇒$\frac {112(x-1)+ 112(x+1)}{(x-1)(x+1)} = 15$ (By LCM)

⇒ $112x -112 + 112x + 112 = 15x^2-15$

⇒ $15x^2 -224x - 15 = 0$

⇒ $15x^2 - 224x - 15 = 0$

⇒ $15x^2 - (225 - 1)x - 15 = 0$

⇒ $15x^2 - 225x + x - 15 = 0$

⇒ $15x(x - 15) + 1(x - 15) = 0$

⇒ $(x-15)(x+\frac {1}{15}) = 0$

x = 15 or x = $-\frac {1}{15}$

Since speed cannot be negative.

x = 15 $kmhr^{-1}$