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# A father runs after his son who is $1000m$ ahead. The father runs at the speed of $1km$ every $8\min$ and the son runs at the speed of $1km$ every $12\min$. How much distance has the son covered at the point when the father overtakes him?

Last updated date: 12th Aug 2024
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Hint: Some formulae that we need to know to solve this problem are, Speed $= \dfrac{\text{Distance}}{\text{Time}}$, Distance $=$ Speed $\times$ Time. Since the speed of both son and father is given in kilometers it is good to convert the distance covered by the son before his father to kilometers, $1000m = 1km$.

Complete step-by-step solution:
It is given that the father is running at the speed of $1km$every $8\min$and the son running at the speed of $1km$ every $12\min$.
Calculation for father:
We know that, Speed $= \dfrac{\text{Distance}}{\text{Time}}$.
Therefore, Speed of the father $= \dfrac{1}{8}$.
Let $t\min$ be the time taken by the father to over-take his son.
We know that, Distance $=$Speed $\times$Time.
Therefore, the distance covered by the father $= \dfrac{1}{8} \times t$.
Calculation for son:
We know that, Speed $=$Distance /Time.
Therefore, Speed of the son $= \dfrac{1}{{12}}$.
The distance covered by the son $= \dfrac{1}{{12}} \times t$ (Since, time taken by both of them are the same).
Our aim is to find the distance covered by the son when his father over-takes him. First let us find the value of $t$.
We have, $\dfrac{1}{8}t - \dfrac{1}{{12}}t = 1$
Simplifying this we get,
$\Rightarrow \dfrac{{12t - 8t}}{{96}} = 1$
On further simplification,
$\Rightarrow t = \dfrac{{96}}{4}$
$\Rightarrow t = 24$
Thus, we got the value of $t$. Now let us find the distance covered by the son when his father overtakes him.
We have the distance covered by the son $= \dfrac{1}{{12}} \times t$.
Therefore, the distance covered by the son in $24\min$ $= \dfrac{1}{{12}} \times 24$ $= 2$.
Thus, the distance covered by the son when his father overtakes him is $2km$.

Note: We can also solve this problem by equating the time taken by son and the time taken by father, since both have taken the same time to cover the distance. But here we have solved it by taking the difference between the distance covered by both of them, because it is given that the son is $1000m = 1km$ ahead from his father.