Question

Shubham travels 760km to his home partly by train and partly by car. He takes 8 hours if he travels 160km by train and rest by car. He takes 12 minutes more if he travels 240km by train and rest by car. Find the speed of the train and car respectively (in kmph).
(a) 40, 80
(b) 60, 120
(c) 80, 100
(d) 100, 120

Answer 1

We solve this problem by using the formula of speed. The formula of speed is given as
$\text{Speed}=\dfrac{\text{Distance}}{\text{Time}}$
By using this formula we find the time taken by Shubham to travel in train and car based on the given conditions by assuming the speeds of train and car as variables. Then we get two equations in two variables which are easy to find the unknown values.

Complete step-by-step answer:
We are given that the total distance covered by Shubham is 760km.
Let us assume that the total distance as
$\Rightarrow D=760$
Let us take the given two conditions one by one.
First we are given that he takes 8 hours if he travels 160km by train and rest by car
Let us assume that the speed of the train is ${{V}_{t}}$ in kmph.
Let us assume that the time taken by him to travel in train as ${{t}_{1}}$
We know that the formula for speed is given as
$\text{Speed}=\dfrac{\text{Distance}}{\text{Time}}$
Now, by using this formula to train we get

& \Rightarrow {{V}_{t}}=\dfrac{160}{{{t}_{1}}} \\\ & \Rightarrow {{t}_{1}}=\dfrac{160}{{{V}_{t}}} \\\ \end{aligned}$$ Now, let us assume that the speed of car as $${{V}_{c}}$$ in kmph Here, we are given that he travels 160km by train We know that the total distance is 760km. So, the distance travelled by car is given as $$\begin{aligned} & \Rightarrow D-160 \\\ & \Rightarrow 760-160=600 \\\ \end{aligned}$$ So, the distance covered by car is 600km Let us assume that the time taken by car as $${{t}_{2}}$$ By using the speed formula we get $$\begin{aligned} & \Rightarrow {{V}_{c}}=\dfrac{600}{{{t}_{2}}} \\\ & \Rightarrow {{t}_{2}}=\dfrac{600}{{{V}_{c}}} \\\ \end{aligned}$$ Now, we are given that he takes a total of 8 hours in this case. So, by converting this statement into mathematical equation we get $$\Rightarrow {{t}_{1}}+{{t}_{2}}=8$$ By substituting the required values in above equation we get $$\Rightarrow \dfrac{160}{{{V}_{t}}}+\dfrac{600}{{{V}_{c}}}=8.....equation(i)$$ Now, let us take the second condition that is he takes 12 minutes more if he travels 240km by train and rest by car Let us assume that the speed of the train is $${{V}_{t}}$$ in kmph. Let us assume that the time taken by him to travel in train as $${{t}_{3}}$$ Now, by using the speed formula to train we get $$\begin{aligned} & \Rightarrow {{V}_{t}}=\dfrac{240}{{{t}_{3}}} \\\ & \Rightarrow {{t}_{3}}=\dfrac{240}{{{V}_{t}}} \\\ \end{aligned}$$ Now, let us assume that the speed of car as $${{V}_{c}}$$ in kmph Here, we are given that he travels 240km by train We know that the total distance is 760km. So, the distance travelled by car is given as $$\begin{aligned} & \Rightarrow D-240 \\\ & \Rightarrow 760-240=520 \\\ \end{aligned}$$ So, the distance covered by car is 520km Let us assume that the time taken by car as $${{t}_{4}}$$ By using the speed formula we get $$\begin{aligned} & \Rightarrow {{V}_{c}}=\dfrac{520}{{{t}_{4}}} \\\ & \Rightarrow {{t}_{4}}=\dfrac{520}{{{V}_{c}}} \\\ \end{aligned}$$ Now, we are given that he takes 12 minutes more in this case. Let us convert the minutes to hours We know that the conversion of hours to minutes that is $$\begin{aligned} & \Rightarrow 1hr=60\min \\\ & \Rightarrow 1\min =\dfrac{1}{60}hr \\\ \end{aligned}$$ By using this conversion we get the value of 12 minutes as $$\begin{aligned} & \Rightarrow 12\min =12\times \dfrac{1}{60}hr \\\ & \Rightarrow 12\min =\dfrac{1}{5}hr \\\ \end{aligned}$$ So, by converting the given statement of time into mathematical equation we get $$\Rightarrow {{t}_{3}}+{{t}_{4}}=8+\dfrac{1}{5}$$ By substituting the required values in above equation we get $$\Rightarrow \dfrac{240}{{{V}_{t}}}+\dfrac{520}{{{V}_{c}}}=\dfrac{41}{5}.....equation(ii)$$ Now, let us subtract equation (ii) by multiplying with 2 from equation (i) by multiplying with 3 then we get $$\begin{aligned} & \Rightarrow 3\left( \dfrac{160}{{{V}_{t}}}+\dfrac{600}{{{V}_{c}}} \right)-2\left( \dfrac{240}{{{V}_{t}}}+\dfrac{520}{{{V}_{c}}} \right)=3\left( 8 \right)-2\left( \dfrac{41}{5} \right) \\\ & \Rightarrow \dfrac{1800}{{{V}_{c}}}-\dfrac{1040}{{{V}_{c}}}=24-\dfrac{82}{5} \\\ & \Rightarrow \dfrac{760}{{{V}_{c}}}=\dfrac{38}{5} \\\ \end{aligned}$$ Now, by cross multiplying the terms from LHS to RHS in above equation we get $$\begin{aligned} & \Rightarrow {{V}_{c}}=760\times \dfrac{5}{38} \\\ & \Rightarrow {{V}_{c}}=100kmph \\\ \end{aligned}$$ Now, by substituting the value of $${{V}_{c}}$$ in equation (i) we get $$\begin{aligned} & \Rightarrow \dfrac{160}{{{V}_{t}}}+\dfrac{600}{100}=8 \\\ & \Rightarrow \dfrac{160}{{{V}_{t}}}=8-6 \\\ & \Rightarrow {{V}_{t}}=80kmph \\\ \end{aligned}$$ Therefore the speeds of trains and cars are 80 and 100 respectively. **So, the correct answer is “Option C”.** **Note:** Students may make mistakes in taking the second condition. We are given that he takes 12 minutes more if he travels 240km by train and rest by car Here, 12 minutes more is with respect to the first condition so, we get the mathematical equation as $$\Rightarrow {{t}_{3}}+{{t}_{4}}=8+\dfrac{1}{5}$$ But students may assume that the time 12 minutes more is within the condition and take the equation as $$\Rightarrow {{t}_{3}}-{{t}_{4}}=\dfrac{1}{5}$$ This gives the wrong answer.