It takes 8 , text{hours} for a 600 , text{km} | vedclass

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(D) Let the speed of the train be $x \, 	ext{km/h}$ and the speed of the car be $y \, 	ext{km/h}$.According to the problem:Case $1$: $\frac{120}{x}

Vedclass · Accepted Answer

$3:4$

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(D) Let the speed of the train be $x \, 	ext{km/h}$ and the speed of the car be $y \, 	ext{km/h}$.According to the problem:Case $1$: $\frac{120}{x} + \frac{480}{y} = 8$ --- $(i)$Case $2$: $\frac{200}{x} + \frac{400}{y} = 8 + \frac{20}{60} = 8 + \frac{1}{3} = \frac{25}{3}$ --- $(ii)$Multiply equation $(i)$ by $5$ and equation $(ii)$ by $3$:$5 	imes (\frac{120}{x} + \frac{480}{y}) = 5 	imes 8 \Rightarrow \frac{600}{x} + \frac{2400}{y} = 40$ --- $(iii)$$3 	imes (\frac{200}{x} + \frac{400}{y}) = 3 	imes \frac{25}{3} \Rightarrow \frac{600}{x} + \frac{1200}{y} = 25$ --- $(iv)$Subtracting equation $(iv)$ from $(iii)$:$(\frac{600}{x} - \frac{600}{x}) + (\frac{2400}{y} - \frac{1200}{y}) = 40 - 25$$\frac{1200}{y} = 15 \Rightarrow y = \frac{1200}{15} = 80 \, 	ext{km/h}$.Substitute $y = 80$ into equation $(i)$:$\frac{120}{x} + \frac{480}{80} = 8 \Rightarrow \frac{120}{x} + 6 = 8$$\frac{120}{x} = 2 \Rightarrow x = 60 \, 	ext{km/h}$.The ratio of the speed of the train to that of the car is $x:y = 60:80 = 3:4$.

It takes $8 \, \text{hours}$ for a $600 \, \text{km}$ journey if $120 \, \text{km}$ is covered by train and the rest by car. It takes $20 \, \text{minutes}$ more if $200 \, \text{km}$ is covered by train and the rest by car. The ratio of the speed of the train to that of the car is:

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