Question

Roohi travels $300{\text{ km}}$ to her home partly by train and partly by bus. She takes $4{\text{ hours}}$ if she travels $60{\text{ km}}$ by train and the remaining by bus. If she travels $100{\text{ km}}$ by train and the remaining by bus, she takes $10$ minutes longer. Find the speed of the train and the bus separately.

Answer 1

Hint: In this question, we have to find the speed of the train and bus. For that, we are going to solve the required result by using the multiple variables as the attributes in the question. And also we are going to calculate the speed of the train and bus by the formula for calculating time with distance and speed. Substituting the values we will get the final solution. Here given below the complete step-by-step solution.

Formula used: ${\text{Time = }}\dfrac{{{\text{Distance}}}}{{{\text{Speed}}}}$

Complete step-by-step solution:
Given Roohi travels $300{\text{ km}}$ to her home partly by train and partly by bus.
She takes $4{\text{ hours}}$ if she travels $60{\text{ km}}$ by train and the remaining by bus.
And If she travels $100{\text{ km}}$ by train and the remaining by bus, she takes $10$ minutes longer.
Let the speed of the train be ${\text{x km/hr}}$ and the speed of the bus is ${\text{y km/hr}}{\text{.}}$
So according to question and using ${\text{Time = }}\dfrac{{{\text{Distance}}}}{{{\text{Speed}}}}$
Roohi travels $60{\text{ km}}$ by train and $300 - 60 = 240$ by bus in $4$ minutes
$\dfrac{{60}}{{\text{x}}} + \dfrac{{240}}{{\text{y}}} = 4$
Roohi travels $100{\text{ km}}$ by train, $300 - 100 = 200$ by bus, and takes $10$ minutes,
$\dfrac{{100}}{{\text{x}}} + \dfrac{{200}}{{\text{y}}} = 4 + \dfrac{{10}}{{60}}$
Simplifying the constant term we get,
$ \Rightarrow \dfrac{{100}}{{\text{x}}} + \dfrac{{200}}{{\text{y}}} = \dfrac{{25}}{6}$
Now let us consider the terms,
$\dfrac{1}{{\text{x}}} = {\text{u}}$ and $\dfrac{1}{{\text{y}}} = {\text{v}}$
Substituting the values of u and v we get,
$ \Rightarrow {\text{60u + 240v = 4}} - - - - - - - \left( 1 \right)$
$ \Rightarrow 100{\text{u + 200v = }}\dfrac{{25}}{6} - - - - - - \left( 2 \right)$
Multiply equation (1) by $5$ and equation (2) by $6$, we get
$ \Rightarrow 300{\text{u + 1200v = 20}} - - - - - \left( 3 \right)$
$ \Rightarrow 600{\text{u + 1200v = 25}} - - - - - \left( 4 \right)$
Subtracting equation (3) and (4), we get
\[ \Rightarrow \left( {300{\text{u + 1200v = 20}}} \right) - \left( {600{\text{u + 1200v = 25}}} \right)\]
Subtracting the equation we get,
\[\begin{array}{*{20}{c}}
  { \Rightarrow \left( {\begin{array}{*{20}{c}}
  {300{\text{u + 1200v = 20}}} \\
  { - 600{\text{u}} - {\text{1200v = }} - {\text{25}}}
\end{array}} \right)} \\
  {\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_} \\
  {{\text{ }} - 300{\text{u = }} - 5} \\
  {\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}
\end{array}\]
Hence we get,
$ \Rightarrow - 300{\text{u = }} - {\text{5}}$
Solving for u,
$ \Rightarrow {\text{u = }}\dfrac{5}{{300}}$
Simplifying we get,
$ \Rightarrow {\text{u = }}\dfrac{1}{{60}}$
Putting the value of ${\text{u}}$ in equation (1), we get
$60 \times \dfrac{1}{{60}} + 240{\text{v = 4}}$
Cancelling the terms same as in numerator and denominator we get,
$ \Rightarrow 240{\text{v = 3}}$
Solving for v,
$ \Rightarrow {\text{v = }}\dfrac{3}{{240}}$
Simplifying we get,
$ \Rightarrow {\text{v = }}\dfrac{1}{{80}}$
Already we considered u and v as,
$\dfrac{1}{{\text{x}}}{\text{ = u = }}\dfrac{1}{{60}}$
$\therefore {\text{ x = 60}}$
And $\dfrac{1}{{\text{y}}}{\text{ = v = }}\dfrac{1}{{80}}$
$\therefore {\text{ y = 80}}$

Hence the speed of the train is $60{\text{ km/hr}}$ and the speed of the bus is $80{\text{ km/hr}}$.

Note: We have to mind that speed is a measure of how quickly an object moves from one place to another. It is equal to the distance travelled divided by the time. It is possible to find any of these three values using the other two. The picture is helps to understand,

The positions of the words in the triangle show where they need to go in the equations. To find the speed, distance is over time in the triangle, so speed is distance divided by time. To find distance, speed is beside time, so distance is speed multiplied by time.