Question

A man travels 370km partly by train and partly by car. If he covers 250 km by train and the rest by car, the total time taken is 4 hours. If he covers 130km by train and the rest by car, it takes him 18 minutes more. Find the speed of the train and the car, respectively.
[a] 80km/hr, 100 km/hr
[b] 98km/hr,100km/hr
[c] 100km/hr,98km/hr
[d] 100km/hr,80 km/hr

Answer 1

Hint: Assume that the speed of the train is x and the speed of the car is y. Using $\text{speed=}\dfrac{\text{Distance}}{\text{time}}$, determine the time take to cover 250 km by train, 120 km by car, 130 km by train and 140 km by car. Use the statement of the question that it takes a total 4 hours to cover 250km by train and 120 km car and it takes a total of 4 hours and 18 minutes to cover 130 km by train and 240 km by car to form two equations in x and y. Put $\dfrac{1}{x}=u$ and $\dfrac{1}{y}=v$ and hence convert the two equations into two linear equations in u and v. Solve for u and v and hence find the value of x and y and hence the speed of the train and the speed of the car.

Complete step-by-step answer:
Let the speed of the train be x km/hr, and the speed of the car be y.
Hence the time taken to cover 250km by train is $\dfrac{250}{x}hrs$ and the remaining 120 km by car is $\dfrac{120}{y}hrs$
Hence the total time taken is equal to $\left( \dfrac{250}{x}+\dfrac{120}{y} \right)hrs$
We know that 1 hr = 60 min
Hence, the total time taken is equal to $\left( \dfrac{250}{x}+\dfrac{120}{y} \right)60\min $
But according to the question, total time taken $=4hrs=4\times 60\min $
Hence, we have
$\left( \dfrac{250}{x}+\dfrac{120}{y} \right)60=4\times 60$
Dividing both sides by 60, we get
$\dfrac{250}{x}+\dfrac{120}{y}=4\text{ }\left( i \right)$
Also, we have
Time taken to cover 130 km by train is $\dfrac{130}{x}hrs$ and the time taken to cover 240km by car is $\dfrac{240}{y}hrs$
Hence the total time taken is equal to $\left( \dfrac{130}{x}+\dfrac{240}{y} \right)hrs$
We know that 1 hr = 60 min
Hence, the total time taken is equal to $\left( \dfrac{130}{x}+\dfrac{240}{y} \right)60\min $
But according to the question, total time taken $=4hrs18\min =\left( 4\times 60+18 \right)\min =258\min $
Hence, we have
$\left( \dfrac{130}{x}+\dfrac{240}{y} \right)60=258$
Divding both sides by 6, we get
$\dfrac{1300}{x}+\dfrac{2400}{y}=43\text{ }\left( ii \right)$
Put $\dfrac{1}{x}=u$ and $\dfrac{1}{y}=v$, we have
Equation (i) becomes
$250u+120v=4\text{ }\left( iii \right)$
Equation (ii) becomes
$1300u+2400v=43\text{ }\left( iv \right)$
Multiplying equation (iii) by 20, and subtracting equation (iv) from the resulting equation, we get
$\begin{align}
  & 5000u+2400v-1300u-2400v=80-43 \\
 & \Rightarrow 3700u=37 \\
\end{align}$
Dividing both sides by 3700, we get
$u=\dfrac{37}{3700}=\dfrac{1}{100}$
Substituting the value of u in equation (iii), we get
$\begin{align}
  & 250\left( \dfrac{1}{100} \right)+120v=4 \\
 & \Rightarrow 120v+2.5=4 \\
\end{align}$
Subtracting 2.5 from both sides, we get
$120v=4-2.5=1.5$
Dividing both sides by 120, we get
$v=\dfrac{1.5}{120}=\dfrac{15}{1200}=\dfrac{1}{80}$
Hence, we have
$u=\dfrac{1}{100},v=\dfrac{1}{80}$
Reverting to original variables, we get
$\dfrac{1}{x}=\dfrac{1}{100},\dfrac{1}{y}=\dfrac{1}{80}$
Taking reciprocals, we get
$x=100,y=80$
Hence, the speed of the train is 100 km/hr, and the speed of the car is 80km/hr
Hence option [d] is correct.

Note: Verification:
Time taken to cover 250km by train and 120km by car $=\dfrac{250}{100}+\dfrac{120}{80}=2.5+1.5=4$
Time taken to cover 130km by train and 240 km by car $=\dfrac{130}{100}+\dfrac{240}{80}=1.3+3=4.3hrs=4hrs18\min $
Hence our answer is verified to be correct.