Question

A motor boat whose speed in still water is 18 km/hr takes 1 hour more to go 24 km up stream that to return down stream to the same spot. Find the speed of the stream.

Answer 1

SOLUTION :  

Given : Speed of the motor boat in still water= 18 km/h and Distance = 24km.  

Let the speed of the stream be 'x' km/h

speed of the boat in upstream =(18 - x)km/h

speed of the boat in downstream =(18 + x)km/h

Time taken by the motorboat to cover 24 km upstream  = 24/(18- x)

[Time = distance/speed]

Time taken by the motorboat to cover 24 km downstream = 24/(18 + x)

A. T.Q

24/(18 - x) =1 + 24/(18 + x)

24/(18 - x) - 24/(18 + x) =1  

24(18 + x) - 24(18 - x) / [(18 + x)(18 - x)] = 1

24[ 18 + x - 18 + x) / [(18 + x)(18 - x)] = 1

24[2x] = 1  [(18 + x)(18 - x)]  

48x = 18² - x²

48x = 324 - x²

x²  + 48x - 324 = 0

x²  + 54x - 6x - 324 = 0

[By middle term splitting]

x(x + 54) - 6(x + 54) = 0

(x - 6) (x + 54) = 0

(x - 6)  = 0 (x + 54) = 0

x = 6  or x = - 54

Since, the speed can’t be negative, so x ≠ - 54 Therefore , x = 6

Hence, the speed of the stream is 6 km/h.

HOPE THIS ANSWER WILL HELP YOU…

Answer 2

$\huge{\boxed{\mathbb{\mathfrak{ANSWER}}}}$

given==>>

speed of boat in still water...=> 18$\frac{km}{h}$

then speed of stream=>

speed , at upstream= 18 - x

at downstream= 18 +x..

then time taken for moving upstream....

t(u) =$\frac{24}{18-x}$==(1)

time taken , while streaming down..

t(d) = $\frac{24}{18+x}$==(2)

according to the condition....

t(d) + 1 = t(u)

substituting....

$\frac{24}{18-x}$ = $\frac{24}{18+x}$ + 1

1 = $\frac{24}{18-x}$-$\frac{24}{18+x}$

1 = $\frac{24(18+x) - 24(18-x)}{(18+x)(18-x)}$

1 = $\frac{24(18 + x - 18 + x)}{ 324 - [tex] x^{2}$

324 - $x^{2}$ = 48x

$x^{2}$ + 48x - 324 = 0

we have =>

d =$b^{2}$ - 4 a c

where,, a = 1, b = 48, c = -324...

d = $48^{2}$ - 4 (1)(-324)

d = 2304 + 1296

d = 3600..

,,

since,,,,

x = $\frac{-b+-{d}^1/2}{2a}$

x = $\frac{-48+-{3600}^1/2}{2(1)}$

x = $\frac{-48 +-60}{2}$

1..

when,,,

x = $\frac{-48 +60}{2}$

x = $\frac{12}{2}$

x =6 $\frac{km}{h}$

2..

when,,

x = $\frac{-48-60}{2}$

x = $\frac{-108}{2}$

x = -54$\frac{km}{h}$

since speed can't be negative,, hence, x =6 $\frac{km}{h}$..is the required answer...

then speed while downstream =18 +x = 18 +6 = 24$\frac{km}{h}$

upstream speed =18 - x = 18 - 6 = 12$\frac{km}{h}$...