Question

A man walks a certain distance with a certain speed. If he walks $\dfrac{1}{2}km$ an hour faster, he takes $1$ an hour less. If he walks $1km$ an hour slower, he takes $3$ more hours. Find the distance covered by the man and his original rate of walking.
(A) \[\text{Distance = 45 km,original speed = 5 kmph}\]
(B) \[\text{Distance = 36 km,original speed = 4 kmph}\]
(C) \[\text{Distance = 63 km,original speed = 7 kmph}\]
(D) \[\text{Distance = 72 km,original speed = 8 kmph}\]

Answer 1

Hint: For answering this question we will use the concept that the rate of walking is given as the distance covered by total time taken which can be mathematically given as $x=\dfrac{d}{y}$ and write expressions for all the 3 cases and equate them and derive the equations which we will solve to obtain the values of the original rate of walking and time taken and use them obtain the distance.

Complete step-by-step answer:
Now let us assume that the original rate of walking of a man is $x\text{ kmph}$ and time taken for covering the distance $d$ is $y\text{ hours}$ .
As we know that the rate of walking is given as the distance covered by total time taken which can be mathematically given as $x=\dfrac{d}{y}$ .
As here $d$ is constant for all the given cases we can say $d=xy$.
As given in the question if he walks $\dfrac{1}{2}km$ an hour faster, he takes $1$ an hour less.
For this case the rate of walking of the man is $\left( x+\dfrac{1}{2} \right)kmph$ and the time taken for covering the same distance $d$ is $\left( y-1 \right)hours$ .
For this case we can say $d=\left( x+\dfrac{1}{2} \right)\left( y-1 \right)$ .
Similarly for the other given case that is if he walks $1km$ an hour slower, he takes $3$ more hours.
This can be mathematically expressed as the rate of walking of the man is $\left( x-1 \right)kmph$ and the time taken for covering the same distance $d$ is $\left( y+3 \right)hours$.
Similarly for the above case we can say $d=\left( x-1 \right)\left( y+3 \right)$ .
By equating the expressions we will have $xy=\left( x-1 \right)\left( y+3 \right)$.
So we will have
$\begin{align}
  & xy=xy+3x-y-3 \\
 & \Rightarrow 3x-y-3=0 \\
\end{align}$ .
And for the expression $xy=\left( x+\dfrac{1}{2} \right)\left( y-1 \right)$ after simplifying we will have
 $\begin{align}
  & xy=xy-x+\dfrac{1}{2}y-\dfrac{1}{2} \\
 & \Rightarrow x-\dfrac{1}{2}y+\dfrac{1}{2}=0 \\
 & \Rightarrow 2x-y+1=0 \\
\end{align}$ .
After solving the two obtained expressions $3x-y-3=0$ and $2x-y+1=0$ we will have $x=4$ and $y=9$ .
As the rate at which man is walking is $\text{4 kmph}$ and the time taken to cover the distance is $9\text{ hours}$.
The distance covered will be equal to the product of the rate of walking and the time taken which will have $36\text{ km}$.
Hence, option B is correct.

So, the correct answer is “Option B”.

Note: While answering questions of this type we should be careful with expressions we choose to derive the values of $x$ and $y$. For example in case if we had chosen the other two expressions we will have $\left( x+\dfrac{1}{2} \right)\left( y-1 \right)=\left( x-1 \right)\left( y+3 \right)$ we will obtain $\dfrac{1}{2}y-x-\dfrac{1}{2}=3x-y-3\Rightarrow 8x-3y-5=0$ which would leave us with one equation and two unknowns. This type of mistake will lead to wastage of time in exams.