Question

14. A boat covers a certain distance downstream in 2 hours. It covers the same distance upstream in 2 half hours. The speed of the boat in still water is 18 km/h. Find the speed of the water. Also find the distance covered by the boat.
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Answer 1

Answer:-

$\sf{The \ speed \ of \ water \ is \ 2 \ km \ hr^{-1} \ and}$

$\sf{distance \ is \ 40 \ km}$

Given:

• A boat covers a certain distance downstream in 2 hours.
• It covers the same distance upstream in 2 half hours.
• The speed of the boat in still water is 18 km/h

To find:

1. Speed of boat.
2. Distance covered by the boat.

Solution:

$\sf{Let \ the \ speed \ of \ water \ be \ x.}$

$\sf{2 \ and \ half \ hour=2.5 \ hr}$

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

$\sf{According \ to \ first \ condition. }$

$\sf{Distance_{downstream}=2\times(18+x)...(1)}$

$\sf{According \ to \ the \ second \ condition. }$

$\sf{Distance_{upstream}=2.5\times(18-x)...(2)}$

$\sf{But,}$

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

$\sf{...from \ (1) \ and \ (2)}$

$\sf{2\times(18+x)=2.5\times(18-x)}$

$\sf{Multiply \ both \ sides \ by \ 2, \ we \ get}$

$\sf{4\times(18+x)=5\times(18-x)}$

$\sf{\therefore{72+4x=90-5x}}$

$\sf{\therefore{4x+5x=90-72}}$

$\sf{\therefore{9x=18}}$

$\sf{\therefore{x=\frac{18}{2}}}$

$\boxed{\sf{\therefore{x=2}}}$

$\sf{\therefore{Speed \ of \ water=2 \ km \ hr^{-1}}}$

$\sf{Distance_{downstream}=2\times(18+2)}$

$\sf{\therefore{Distance_{downstream}=2\times20}}$

$\sf{\therefore{Distance_{downstream}=40 \ km \ hr^{-1}}}$

$\sf\purple{\tt{\therefore{The \ speed \ of \ water \ is \ 2 \ km \ hr^{-1} \ and}}}$

$\sf\purple{\tt{distance \ is \ 40 \ km}}$

Answer 2

Answer:

Let the speed of water be x.

\sf{2 \ and \ half \ hour=2.5 \ hr}2 and half hour=2.5 hr

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

Distance=Time× Speed

\sf{According \ to \ first \ condition. }According to first condition.

\sf{Distance_{downstream}=2\times(18+x)...(1)}Distance

downstream

=2×(18+x)...(1)

\sf{According \ to \ the \ second \ condition. }According to the second condition.

\sf{Distance_{upstream}=2.5\times(18-x)...(2)}Distance

upstream

=2.5×(18−x)...(2)

\sf{But,}But,

\sf{Distance_{downstream}=Distance_{upstream}...Given}Distance

downstream

=Distance

upstream

...Given

\sf{...from \ (1) \ and \ (2)}...from (1) and (2)

\sf{2\times(18+x)=2.5\times(18-x)}2×(18+x)=2.5×(18−x)

\sf{Multiply \ both \ sides \ by \ 2, \ we \ get}Multiply both sides by 2, we get

\sf{4\times(18+x)=5\times(18-x)}4×(18+x)=5×(18−x)

\sf{\therefore{72+4x=90-5x}}∴72+4x=90−5x

\sf{\therefore{4x+5x=90-72}}∴4x+5x=90−72

\sf{\therefore{9x=18}}∴9x=18

\sf{\therefore{x=\frac{18}{2}}}∴x=

2

18

\boxed{\sf{\therefore{x=2}}}

∴x=2

\sf{\therefore{Speed \ of \ water=2 \ km \ hr^{-1}}}∴Speed of water=2 km hr

−1

\sf{Distance_{downstream}=2\times(18+2)}Distance

downstream

=2×(18+2)

\sf{\therefore{Distance_{downstream}=2\times20}}∴Distance

downstream

=2×20

\sf{\therefore{Distance_{downstream}=40 \ km \ hr^{-1}}}∴Distance

downstream

=40 km hr

−1

\sf\purple{\tt{\therefore{The \ speed \ of \ water \ is \ 2 \ km \ hr^{-1} \ and}}}∴The speed of water is 2 km hr

−1

and

\sf\purple{\tt{distance \ is \ 40 \ km}}distance is 40 km