Question

A man walks a certain distance with a certain speed. If he walks $\dfrac{1}{2}$ km an hour faster, he takes 1 hour less. But, if he walks 1 km an hour slower, he takes 3 more hours. Find the distance covered by the man and his original rate of walking.

Answer 1

Hint- Here, we will be assuming the original rate of walking (or the original speed of the man) and the original time taken by the man as two variables and then apply the formula Distance = ${\text{Speed}} \times {\text{Time}}$ for all the cases given in the problem.

Complete step-by-step answer:
Let the original speed of the man = x km per hour and the original total time taken by the man = y hours.
As we know, Distance = ${\text{Speed}} \times {\text{Time }} \to {\text{(1)}}$
So, original distance covered by the man $= x \times y = xy$ km
Given, when the speed of the man is increased by $\dfrac{1}{2}$ km an hour i.e., Speed of the man = $\left( {x + \dfrac{1}{2}} \right)$ km per hour, the time taken is decreased by 1 hour i.e., Time taken = (y-1) hours
Here, the distance which the man needs to cover will be same i.e., Distance = xy km
Using equation (1), we get
$xy = \left( {x + \dfrac{1}{2}} \right) \times \left( {y - 1} \right) \\\ \Rightarrow xy = xy + \dfrac{y}{2} - x - \dfrac{1}{2} \\\ \Rightarrow 0 = \dfrac{y}{2} - x - \dfrac{1}{2} \\\ \Rightarrow x = \dfrac{y}{2} - \dfrac{1}{2}{\text{ }} \to {\text{(2)}} \\\$
Also, given that when the speed of the man is decreased by 1 km an hour i.e., Speed of the man = (x-1) km per hour, the time taken is increased by 3 hours i.e., Time taken = (y+3) hours
Here, the distance which the man needs to cover will be same i.e., Distance = xy kms
Using equation (1), we get
$xy = \left( {x - 1} \right) \times \left( {y + 3} \right) \\\ \Rightarrow xy = xy - y + 3x - 3 \\\ \Rightarrow 0 = - y + 3x - 3 \\\ \Rightarrow y = 3x - 3 \to {\text{(3)}} \\\$
By substitute the value of x in terms of y from the equation (2), equation (3) becomes

$$\Rightarrow y = 3\left( {\dfrac{y}{2} - \dfrac{1}{2}} \right) - 3 \\\ \Rightarrow y = \dfrac{{3y}}{2} - \dfrac{3}{2} - 3 \\\ \Rightarrow \dfrac{{3y}}{2} - y = \dfrac{3}{2} + 3 \\\ \Rightarrow \dfrac{{3y - 2y}}{2} = \dfrac{{3 + 6}}{2} \\\ \Rightarrow \dfrac{y}{2} = \dfrac{9}{2} \\\ \Rightarrow y = 9 \\\$$

Put y=9 in equation (2), we will get the value of x as
$\Rightarrow x = \dfrac{9}{2} - \dfrac{1}{2} = \dfrac{{9 - 1}}{2} = \dfrac{8}{2} \\\ \Rightarrow x = 4 \\\$
Original distance covered by the man $= xy = 4 \times 9 = 36$ km
Therefore, the original rate of walking or the original speed of the man is 4 km per hour and the original time taken by the man is 9 hours.
Also, the distance covered by man is 36 km.

Note- In this particular problem, it is very important to note that the distance covered by the man will always remain the same. When the man walks faster, its speed or rate of walking is increased whereas the time taken is decreased because the distance always needs to be constant. Similarly, when the man walks slowly, its speed or rate of walking is decreased whereas the time taken is increased because the distance always needs to be a constant.