If a journey of 120 km is done by train and the rest by car, then it takes 8 hours for total 600 km journey. And if 200 km is done by train and the rest by car, then it takes 20 minutes more. Then what would be the ratio of the speed of the train to that of the car?

Asked in GPSC Prelims CSP 1 2023 · Topic: Mechanics (Newton's Laws of Motion and their applications)

1. 4 : 3
2. 4 : 5
3. 3 : 4
4. 5 : 4

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• $\frac{120}{v_T} + \frac{480}{v_C} = 8$. (Equation 1)Case 2: 200 km by train, remaining by car. Remaining distance $= 600 - 200 = 400$ km by car. Total time = 8 hours 20 minutes. 8 hours 20 minutes $= 8 + \frac{20}{60}$ hours $= 8 + \frac{1}{3}$ hours $= \frac{25}{3}$ hours. Time by train + Time by car = $\frac{25}{3}$. $ \frac{200}{v_T} + \frac{400}{v_C} = \frac{25} {3}$. (Equation 2)Let $x = \frac{1}{v_T}$ and $y = \frac{1}{v_C}$. The equations become: $120x + 480y = 8$ (Equation 1') $200x + 400y = \frac{25} {3}$ (Equation 2')Divide Equation 1' by 8: $15x + 60y = 1$. (Equation 1'')From Equation 1'', $15x = 1 - 60y$.Substitute this into Equation 2': $200 \left(\frac{1 - 60y}{15}\right) + 400y = \frac{25} {3}$. $\frac{40}{3} (1 - 60y) + 400y = \frac{25} {3}$. Multiply the entire equation by 3: $40(1 - 60y) + 1200y = 25$. $40 - 2400y + 1200y = 25$. $40 - 1200y = 25$. $1200y = 40 - 25 = 15$. $y = \frac{15}{1200} = \frac{1}{80}$. Since $y = \frac{1}{v_C}$, $v_C = 80$ km/h.Substitute the value of $y$ back into Equation 1'': $15x + 60\left(\frac{1}{80}\right) = 1$. $15x + \frac{60} {80} = 1$. $15x + \frac{3}{4} = 1$. $15x = 1 - \frac{3}{4} = \frac{1}{4}$. $x = \frac{1}{4 \times 15} = \frac{1}{60}$. Since $x = \frac{1}{v_T}$, $v_T = 60$ km/h.The ratio of the speed of the train to that of the car is $v_T : v_C = 60 : 80$. Divide both parts by 20: $60 : 80 = \frac{60}{20} : \frac{80}{20} = 3 : 4$. 278. A speedboat runs 6 km upstream in a river and comes back to the starting point in 33 minutes. The stream of the river is running at 2 km/hr. What is the speed of speedboat in still water?
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