Question

A streamer goes downstream from one part to other in 9 hours.It covers the same question upstreams in 10 hours. If the speed of the stream be 1km/hr find the speed of the streamer in still water and the distance between ports.

Answer 1

Answer:

brainlest pls

Step-by-step explanation:

A steamer goes downstream. The distance between the two ports is 180 km.

Solution:

Let the speed of the streamer be ‘x’ and Distance be ‘D’

Given the speed of the stream is 1 km/hr.

Thus the downstream speed is given by ‘(x+1)’km/hr and the upstream speed is given by (x-1) km/hr.

Given that time taken to travel downstream is 9 hr

Therefore downstream distance = 9(x+1) …………..(Equation 1)

Again it is given that time taken to travel upstream is 10 hour.

Therefore upstream distance = 10(x-1)  ……………(Equation 2)

Since distance is same in both cases, we equate the upstream distance and downstream distance.

9(x+1) = 10(x-1)

9x + 9 = 10x - 10

9 + 10=10x - 9x

x = 19

Now substituting the value of "x" in any of the equation 1 or 2, we get

D = 10(19-1)     [Substituting the value of x in Equation 2]

D = 190-10  = 180

Thus the distance between two ports is 180 cm

Answer 2

Correct Question:

A streamer goes downstream from one part to other in 9 hours. It covers the same distance in upstream in 10 hours. If the speed of the stream is 1 km/hr. Find the speed of the streamer in still water and the distance between parts.

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Answer:

$\sf{The \ speed \ of \ streamer \ in \ still \ water}$

$\sf{is \ 19 \ km \ hr^{-1} \ and \ distance \ between}$

$\sf{points \ is \ 180 \ km}$

Given:

$\sf{\leadsto{A \ streamer \ goes \ downstream \ from}}$

$\sf{one \ part \ to \ other \ in \ 9 \ hour}$

$\sf{\leadsto{It \ covers \ the \ same \ distance \ in}}$

$\sf{upstream \ in \ 10 \ hour}$

$\sf{\leadsto{The \ speed \ of \ stream \ is \ 1 \ km \ hr^{-1}}}$

To find:

$\sf{The \ speed \ of \ streamer \ in \ still \ water}$

$\sf{and \ the \ distance \ between \ two \ points.}$

Solution:

$\sf{Let \ the \ speed \ of \ streamer \ be \ x \ km \ hr^{-1}}$

$\sf{Distance=Speed\times \ Time}$

$\sf{For \ downstream \ speed=(x+1) \ km \ hr^{-1}}$

$\sf{\therefore{Distance_{downstream}=9\times(x+1)}}$

$\sf{For \ upstream \ speed=(x-1) \ km \ hr^{-1}}$

$\sf{\therefore{Distance_{upstream}=10\times(x-1)}}$

$\sf{According \ to \ the \ given \ condition}$

$\sf{Distance_{downstream}=Distance_{upstream}}$

$\sf{\therefore{9(x+1)=10(x-1)}}$

$\sf{\therefore{9x+9=10x-10}}$

$\sf{\therefore{x=19}}$

$\sf{Hence, \ the \ speed \ of \ streamer \ in}$

$\sf{still \ water \ is \ 19 \ km \ hr^{-1}}$

$\sf{Distance_{downstream}=9(x+1)}$

$\sf{But, \ x=19}$

$\sf{\therefore{Distance=9(19+1)}}$

$\sf{\therefore{Distance=9(20)}}$

$\sf{\therefore{Distance=180 \ km}}$

$\sf\purple{\tt{\therefore{The \ speed \ of \ streamer \ in \ still \ water}}}$

$\sf\purple{\tt{is \ 19 \ km \ hr^{-1} \ and \ distance \ between}}$

$\sf\purple{\tt{points \ is \ 180 \ km}}$