Question

A man travels to a car park at an average speed of \[40\]km/h, then he walks to his office at an average speed of \[6\]km/h. The total time for the journey is \[25\] minutes. One day his car breaks down and he walks three times as far thereby arriving at his office \[17\] minutes late. How far is the office?

Answer 1

Hint: At first, with the given information we will try to find the total time taken by the person to reach his office with both conditions. Then by solving the equation we will find the distance of the office.

Complete step by step answer:
It is given that a man travels to a car park at an average speed of \[40\]km/h, then he walks to his office at an average speed of \[6\]km/h.
Let us consider that the distance between the house and the car park is \[x\]km and the distance between the car park and office is \[y\]km.
The total time of the journey is \[25\]minutes\[ = \dfrac{{25}}{{60}}\] hours
We know that, the relationship among distance, time and speed is
Distance \[ = \] Speed \[ \times \]Time
From this we can find time using the formula:${\text{time =}}\dfrac{{{\text{distance}}}}{{{\text{speed}}}}$
So, the time taken by the car is \[ = \dfrac{x}{{40}}\] hour
And, the time taken by his walk \[ = \dfrac{x}{6}\] hour
So, total time of journey is given by,
\[\dfrac{x}{{40}} + \dfrac{y}{6} = \dfrac{{25}}{{60}}\]
Let us simplify the above equation therefore we get,
\[3x + 20y = 50\]….. (1)
One day his car breaks down so he walked three times as far thereby arriving at his office \[17\]minutes late.
So, the distance travelled by the car is \[x - 2y\] km.
So, the time taken by him to cover the journey is \[25 + 17\]minute\[ = 42\]min \[ = \dfrac{{42}}{{60}}\] hours
So, now the total time of journey is given by the equation
\[\dfrac{{x - 2y}}{{40}} + \dfrac{{3y}}{6} = \dfrac{{42}}{{60}}\]
Let us simplify the above equation then we get,
\[3x + 54y = 84\]… (2)
Let subtract the equation (1) and (2) to get the value of $y$we get,
\[3x + 54y - 3x - 20y = 84 - 50\]
On simplifying the above equation we get,
\[34y = 34\]\[3x + 20(1) = 50\]
Let us cancel the common terms in the above equation we get,
\[y = 1\]
Let us substitute \[y = 1\]in (1) then we get,
On further solving we get the value of$x$,
\[x = \dfrac{{30}}{3} = 10\]
So, the total distance from home to office is \[10 + 1 = 11\]km.

Hence, the office is \[11\] km far from the house.

Note:
We know that, the relationship among distance, time and speed is
Distance \[ = \] Speed \[ \times \] Time
Also it is given that he walked three times as far thereby arriving at his office but in the problem we subtract only \[2y\]from $x$ and added \[2y\]to $y$ because already he walks $y$ distance from car park to office, so from the given three times is reduced one time and use two times the distance.