Question

A boat goes 35 km. upstream and 55 km down-stream in 12 hours. In 10
hours it can go 30 km. upstream and 44 km. down-stream. Determine the
speed of the stream and that of the boat in still water.​

Answer 1

$\sf\red{\underline{\underline{Answer:}}}$

$\sf{The \ speed \ of \ the \ stream \ and \ boat \ in}$

$\sf{still \ water \ are \ 3 \ km \ per \ hour \ and}$

$\sf{8 \ km \ per \ hour \ respectively.}$

$\sf\orange{Given:}$

$\sf{\implies{A \ boat \ goes \ 35 \ km \ upstream \ and}}$

$\sf{55 \ km \ downstream \ in \ 12 \ hours.}$

$\sf{\implies{In \ 10 \ hours \ it \ can \ go \ 30 \ km}}$

$\sf{upstream \ and \ 44 \ km \ downstream.}$

$\sf\pink{To \ find:}$

$\sf{The \ speed \ of \ the \ stream \ and \ the \ speed}$

$\sf{of \ the \ boat \ in \ still \ water.}$

$\sf\green{\underline{\underline{Solution:}}}$

$\sf{Let \ the \ speed \ of \ boat \ in \ still \ water \ be}$

$\sf{x \ km \ hr^{-1} \ and \ speed \ of \ stream \ be}$

$\sf{y \ km \ hr^{-1}}$

$\sf{\therefore{Speed \ of \ boat_{upstream}=(x-y) \ km \ hr^{-1}}}$

$\sf{Speed \ of \ boat_{downstream}=(x+y) \ km \ hr^{-1}}$

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

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

$\sf{\frac{35}{x-y}+\frac{55}{x+y}=12...(1)}$

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

$\sf{\frac{30}{x-y}+\frac{44}{x+y}=10....(2)}$

$\sf{Substitute \ \frac{1}{x-y} \ by \ a \ and \ \frac{1}{x+y} \ by \ b}$

$\sf{The \ equations \ (1) \ and \ (2) \ become}$

$\sf{35a+55b=12...(3)}$

$\sf{30a+44b=10}$

$\sf{\therefore{2(15a+22b)=10}}$

$\sf{\therefore{15a+22b=\frac{10}{2}}}$

$\sf{\therefore{15a+22b=5...(4)}}$

$\sf{Multiply \ equation (3) \ by \ 2, \ we \ get}$

$\sf{70a+110b=24...(5)}$

$\sf{Multiply \ equation (4) \ by \ 5, \ we \ get}$

$\sf{75a+110b=25...(6)}$

$\sf{Subtract \ equation (5) \ from \ equation (6)}$

$\sf{75a+110b=25}$

$\sf{-}$

$\sf{70a+110b=24}$

____________________

$\sf{5a=1}$

$\boxed{\sf{\therefore{a=\frac{1}{5}}}}$

$\sf{Substitute \ a=\frac{1}{5} \ in \ equation (3)}$

$\sf{35\times\frac{1}{5}+55b=12}$

$\sf{\therefore{7+55b=12}}$

$\sf{\therefore{55b=12-7}}$

$\sf{\therefore{55b=5}}$

$\sf{\therefore{b=\frac{5}{55}}}$

$\boxed{\sf{\therefore{b=\frac{1}{11}}}}$

$\sf{Resubstituting \ a=\frac{1}{x-y} \ and \ b=\frac{1}{x+y}}$

$\sf{\therefore{\frac{1}{x-y}=a}}$

$\sf{\therefore{\frac{1}{x-y}=\frac{1}{5}}}$

$\sf{\therefore{x-y=5...(7)}}$

$\sf{\frac{1}{x+y}=b}$

$\sf{\therefore{\frac{1}{x+y}=\frac{1}{11}}}$

$\sf{\therefore{x+y=11...(8)}}$

$\sf{Add \ equations (7) \ and \ (8)}$

$\sf{x-y=5}$

$\sf{+}$

$\sf{x+y=11}$

_________________

$\sf{\therefore{2x=16}}$

$\sf{\therefore{x=\frac{16}{2}}}$

$\boxed{\sf{\therefore{x=8}}}$

$\sf{Substitute \ x=8 \ in \ equation (8)}$

$\sf{8+y=11}$

$\sf{\therefore{y=11-8}}$

$\boxed{\sf{\therefore{y=3}}}$

$\sf\purple{\tt{\therefore{The \ speed \ of \ the \ stream \ and \ boat \ in}}}$

$\sf\purple{\tt{still \ water \ are \ 3 \ km \ per \ hour \ and}}$

$\sf\purple{\tt{8 \ km \ per \ hour \ respectively.}}$