Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.
Given:
Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer.
To do:
We have to find the speed of the train and the bus separately.
Solution:
Total distance to the home $=300\ km$.
Let the speed of the train be $x$ km/hr and the speed of the bus be $y$ km/hr.
We know that,
Time $=$ Distance $\div$ Speed
In the first case, she takes 4 hours if she travels 60 km. by train and the remaining by bus.
Time taken $=\frac{60}{x}+\frac{300-60}{y}$
$\Rightarrow \frac{60}{x}+\frac{240}{y}=4$.....(i)
In the second case, she takes 10 minutes more if she travels 100 km by train and the remaining by bus.
Time taken $=\frac{100}{x}+\frac{300-100}{y}$
$\Rightarrow \frac{100}{x}+\frac{200}{y}=4+\frac{10}{60}$
$\Rightarrow \frac{100}{x}+\frac{200}{y}=4+\frac{1}{6}$
$\Rightarrow \frac{100}{x}+\frac{200}{y}=\frac{4\times6+1}{6}$
$\Rightarrow \frac{100}{x}+\frac{200}{y}=\frac{25}{6}$......(ii)
Multiplying equation (i) by 5 and equation (ii) by 3, we get,
$5(\frac{60}{x}+\frac{240}{y})=5(4)$
$\frac{300}{x}+\frac{1200}{y}=20$....(iii)
$3(\frac{100}{x}+\frac{200}{y})=3(\frac{25}{6})$
$\frac{300}{x}+\frac{600}{y}=\frac{25}{2}$.....(iv)
Subtracting equation (iv) from (iii), we get,
$\frac{300}{x}+\frac{1200}{y}-\frac{300}{x}-\frac{600}{y}=20-\frac{25}{2}$
$\frac{1200-600}{y}=\frac{20(2)-25}{2}$
$\frac{600}{y}=\frac{15}{2}$
$y=\frac{600\times2}{15}$
$y=40\times2=80$
Substituting $y=80$ in equation (i), we get,
$\frac{60}{x}+\frac{240}{80}=4$
$\frac{60}{x}+3=4$
$\frac{60}{x}=4-3=1$
$x=60(1)$
$x=60$
Therefore, the speed of the train is $60\ km/hr$ and the speed of the bus is $80\ km/hr$ respectively.
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