Question

Ramesh travels 760Km to his home partly by train and partly by car. He takes 8hr, if he travels 160Km by train and the rest by car. He takes 12 minutes more, if he travels 240Km by train and the rest by car. Find the speed of the train and of the car.
A. Speed of train \[ = 80Km/hr\] and speed of car \[ = 100Km/hr\]
B. Speed of train \[ = 40Km/hr\] and speed of car \[ = 50Km/hr\]
C. Speed of train \[ = 60Km/hr\] and speed of car \[ = 120Km/hr\]
D. Speed of train \[ = 70Km/hr\] and speed of car \[ = 130Km/hr\]

Answer 1

Hint: We assume the speed of train and speed of car as different variables. Using the formula of speed we write the formula of time in terms of speed and distance. Form two equations using time as a factor. Solve for the values of variables using the required method.
* If ‘d’ distance is covered in time ‘t’, then speed ‘s’ is given by \[s = \dfrac{d}{t}\]

Complete step-by-step solution:
We are given the total distance covered by Ramesh on her way home is 760 Km.
Since, he travels some distance by train and some distance by car, we form two cases which depict two situations in the question.
Let us denote the speed of the train by ‘x’ and speed of the car by ‘y’.
Since, we know speed is given by the formula \[s = \dfrac{d}{t}\]
Then by cross multiplying the values
\[ \Rightarrow t = \dfrac{d}{s}\].................................… (1)
Let time taken by train be \[{t_1}\] and time taken by car be \[{t_2}\]; Then \[{t_1} + {t_2} = T\].........................… (2)
Where, T is the total time taken to travel.
Case 1: 160 Km by train and rest by car
We are given that Ramesh travels 160 Km by train
Since, total distance \[ = 760\]Km
\[ \Rightarrow \]Distance travelled by car \[ = 760 - 160\]
\[ \Rightarrow \]Distance travelled by car \[ = 600\]Km
Also, we are given the total time taken in this case 8 hours.
\[\therefore T = 8\]
We find the value of \[{t_1}\]and \[{t_2}\]
From equation (1): \[t = \dfrac{d}{s}\]
\[ \Rightarrow \]Time taken by train, \[{t_1} = \dfrac{{160}}{x}\]
\[ \Rightarrow \]Time taken by car, \[{t_2} = \dfrac{{600}}{y}\]
Substitute values of \[{t_1}\]and \[{t_2}\]in equation (2)
\[ \Rightarrow \dfrac{{160}}{x} + \dfrac{{600}}{y} = 8\]........................… (3)
Case 2: 240 Km by train and rest by car
We are given that Ramesh travels 240 Km by train
Since, total distance \[ = 760\]Km
\[ \Rightarrow \]Distance travelled by car \[ = 760 - 240\]
\[ \Rightarrow \]Distance travelled by car \[ = 520\]Km
Also, we are given the total time taken in this case as 8 hours plus 12 minutes
\[ \Rightarrow T = (8 \times 60 + 12)\]minutes
\[ \Rightarrow T = \left( {480 + 12} \right)\]minutes
\[ \Rightarrow T = 492\]minutes
Since, 1 hour has 60 minutes
\[ \Rightarrow T = \dfrac{{492}}{{60}}\]hour
Cancel same factors from numerator and denominator
\[\therefore T = \dfrac{{41}}{5}\]hour
We find the value of \[{t_1}\] and \[{t_2}\]
From equation (1): \[t = \dfrac{d}{s}\]
\[ \Rightarrow \]Time taken by train, \[{t_1} = \dfrac{{240}}{x}\]
\[ \Rightarrow \]Time taken by car, \[{t_2} = \dfrac{{520}}{y}\]
Substitute values of \[{t_1}\]and \[{t_2}\]in equation (2)
\[ \Rightarrow \dfrac{{240}}{x} + \dfrac{{520}}{y} = \dfrac{{41}}{5}\].....................… (4)
Now we solve the two equations, equation (3) and (4).
Let us assume \[\dfrac{1}{x} = a,\dfrac{1}{y} = b\]
Then, from equation (3); \[\dfrac{{160}}{x} + \dfrac{{600}}{y} = 8\]
\[ \Rightarrow 160a + 600b = 8\]………………………..… (5)
Also, from equation (4); \[\dfrac{{240}}{x} + \dfrac{{520}}{y} = \dfrac{{41}}{5}\]
\[ \Rightarrow 240a + 520b = \dfrac{{41}}{5}\].....................… (6)
From equation (5): \[160a + 600b = 8\]
Shift all the values except one with variable ‘a’ to one side of the equation
\[ \Rightarrow 160a = 8 - 600b\]
Divide both sides by 8
\[ \Rightarrow \dfrac{{160a}}{8} = \dfrac{{8 - 600b}}{8}\]
Cancel same factors from numerator and denominator on both sides
\[ \Rightarrow 20a = 1 - 75b\]
Now multiply both sides by 12
\[ \Rightarrow 20a \times 12 = \left( {1 - 75b} \right) \times 12\]
\[ \Rightarrow 240a = 12 - 900b\].....................… (7)
Substitute the value of 240a from equation (7) in equation (6)
\[ \Rightarrow 12 - 900b + 520b = \dfrac{{41}}{5}\]
\[ \Rightarrow 12 - 380b = \dfrac{{41}}{5}\]
Shift all constants to one side of the equation
\[ \Rightarrow 12 - \dfrac{{41}}{5} = 380b\]
Take LCM on LHS of the equation
\[ \Rightarrow \dfrac{{60 - 41}}{5} = 380b\]
\[ \Rightarrow \dfrac{{19}}{5} = 380b\]
Divide both sides of the equation by 19
\[ \Rightarrow \dfrac{{19}}{{5 \times 19}} = \dfrac{{380b}}{{19}}\]
Cancel same factors on both sides of the equation
\[ \Rightarrow \dfrac{1}{5} = 20b\]
Cross multiply value of 20 from RHS to LHS
\[ \Rightarrow \dfrac{1}{{5 \times 20}} = b\]
\[ \Rightarrow \dfrac{1}{{100}} = b\]
Since, we know \[\dfrac{1}{y} = b\]
\[ \Rightarrow \dfrac{1}{{100}} = \dfrac{1}{y}\]
\[ \Rightarrow y = 100\]
\[\therefore \]The speed of a car is 100 Km per hour.
Substitute the value of b in equation (7)
\[ \Rightarrow 240a = 12 - 900 \times \dfrac{1}{{100}}\]
Cancel same factors from numerator and denominator in RHS
\[ \Rightarrow 240a = 12 - 9\]
Take LCM on RHS of the equation
\[ \Rightarrow 240a = 3\]
Divide both sides by 240
\[ \Rightarrow \dfrac{{240a}}{{240}} = \dfrac{3}{{240}}\]
Cancel same factors from numerator and denominator on both sides of the equation
\[ \Rightarrow a = \dfrac{1}{{80}}\]
Since, we know \[\dfrac{1}{x} = a\]
\[ \Rightarrow \dfrac{1}{x} = \dfrac{1}{{80}}\]
\[ \Rightarrow x = 80\]
\[\therefore \]Speed of train is 80 Km per hour.
Thus, the speed of the train is 80 Km per hour and the speed of the car is 100 Km per hour.

\[\therefore \]Correct option is A.

Note: Students are likely to make mistakes when they try to form equations with respect to speed, they tend to think we have to find speed in the question so equations formed will be of speed. But we are given the time clearly in which she covers the total distance in two cases so we form equations on the basis of time. Also, don’t solve the equations having the variables in the denominator using LCM as that will only cause more confusion in the solution.