Question

a motor boat can travel 30 km upstream and 28 km down stream in 7 hours . it can travel 21 km upstream and return in 5 hours . find the speed of the boat in still water and the speed of the stream . use cross multiplication method.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{speed}\mathrm{of}\mathrm{boat}\mathrm{be}\mathrm{x} \\ \mathrm{Let}\mathrm{the}\mathrm{speed}\mathrm{of}\mathrm{stream}\mathrm{be}\mathrm{y} \\ \mathrm{time}=\frac{\mathrm{distance}}{\mathrm{speed}} \\ \mathrm{A}.\mathrm{T}.\mathrm{Q} \\ \frac{30}{\mathrm{x}-\mathrm{y}}+\frac{28}{\mathrm{x}+\mathrm{y}}=7 \\ \frac{21}{\mathrm{x}-\mathrm{y}}+\frac{21}{\mathrm{x}+\mathrm{y}}=5 \\ \mathrm{Let}\left(\mathrm{x}-\mathrm{y}\right)=\mathrm{u}\mathrm{and}\left(\mathrm{x}+\mathrm{y}\right)=\mathrm{v} \\ 30\mathrm{u}+28\mathrm{v}-7=0...\left(1\right) \\ 21\mathrm{u}+21\mathrm{v}-5=0...\left(2\right) \\ \mathrm{Using}\mathrm{cross}\mathrm{multiplication}\mathrm{method} \\ \frac{\mathrm{u}}{28\left(-5\right)-21\left(-7\right)}=\frac{\mathrm{v}}{21\left(-7\right)-30\left(-5\right)}=\frac{1}{30\left(21\right)-21\left(28\right)} \\ \frac{\mathrm{u}}{-140+147}=\frac{\mathrm{v}}{-147+150}=\frac{1}{630-588} \\ \frac{\mathrm{u}}{7}=\frac{\mathrm{v}}{3}=\frac{1}{42} \\ \mathrm{u}=\frac{1}{6}\mathrm{and}\mathrm{v}=\frac{1}{14} \\ \mathrm{So}, \\ \mathrm{x}-\mathrm{y}=6...\left(3\right) \\ \mathrm{and} \\ \mathrm{x}+\mathrm{y}=14...\left(4\right) \\ \mathrm{Adding}\left(3\right)\mathrm{and}\left(4\right),\mathrm{we}\mathrm{get} \\ 2\mathrm{x}=20 \\ \mathrm{x}=10\mathrm{km}/\mathrm{hour} \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{x}\mathrm{in}\left(3\right) \\ 10-\mathrm{y}=6 \\ \mathrm{y}=4\mathrm{km}/\mathrm{hour} \\ \mathrm{Spped}\mathrm{of}\mathrm{boat}\mathrm{in}\mathrm{still}\mathrm{water}\mathrm{is}10\mathrm{km}/\mathrm{hour}\mathrm{and}\mathrm{speed}\mathrm{of}\mathrm{stream}\mathrm{is}4\mathrm{km}/\mathrm{hour}. \end{gathered}$$