Question

a motorboar whose speed is 18km/hour in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.

appropriate answer wid correct procedure plz​

Answer 1

QUESTION :-

★ A motorboat whose speed in still water is 18km/h,takes 1hour more to go 24km upstream than to return downstream to the same spot. Find the speed of the stream.

ANSWER:-

☞ Let speed of the stream to be 'x' km/hr.

☞ Net upstream speed = (18-x) km/hr

☞ Net downstream speed = (18+x) km/hr

☞ Distance = 24 km

☞Time taken to go upstream = 24 / (18-x)

☞Time taken to go downstream = 24 /(18+x)

Given,

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

So,

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

That is,

24 (18+x) = (42+x) (18-x)

Solving brackets of both sides,

☞ ${x}^{2} + 48x - 34$

☞ By Splitting The Middle Term Method, we get

X = 6 or -54

☞ Speed of the stream cannot be negative, so neglect -54.

Therefore, 6 km/hr is the speed of the stream.

Answer 2

Given:-

Speed of boat =18km/hr

Distance =24km

Let x be the speed of stream.

Let $t_1\:\:and\:\: t_2$

be the time for upstream and downstream.

As we know that,

speed= distance/time

⇒time= distance/speed

For upstream,

Speed =(18−x)km/hr

Distance =24km

Time =$t_1$

Therefore,

$t_1$ = $\large\frac{24}{18 - x}$

For downstream,

Speed =(18+x)km/hr

Distance =24km

Time =$t_2$

Therefore,

$t_2$ = $\large\frac{24}{18 + x}$

Now according to the question-

$t_1$ = $t_2$ + 1

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

⇒ $\large\frac{1}{18 - x}\:-\:\large\frac{1}{18 + x} = \huge\frac{1}{24}$

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

⇒48x = (18 − x)(18 + x)

⇒48x = 324 + 18x − 18x − $x^2$

⇒$x^2$ + 48x − 324 = 0

⇒$x^2$ +54x−6x−324=0

⇒x(x + 54) −6 (x + 54) = 0

⇒(x + 54)(x − 6) = 0

⇒x = −54 or x = 6

Since speed cannot be negative.

⇒x ≠ −54

∴ x = 6

Thus the speed of stream is 6km/hr

Hence the correct answer is 6km/hr.