Question

A thief is running at a speed of 6 km per hour and a police constable is chasing him with a speed of 8 km per hour. If originally, the distance between thief and the constable is 500 m. After what distance, will the constable catch the thief?
(a) 0.7 km
(b) 1.5 km
(c) 0.9 km
(d) 1.6 km

Answer 1

Hint: First of all find the difference of the speeds of the thief and constable then find the time taken to cover the distance of 500 m with that speed which can be calculated by using the formula of speed, distance and time that is written as $ \text{Speed}=\dfrac{\text{Distance}}{\text{Time}} $ . Solving this formula will give us the time required to cover the gap between thief and constable. Now, substitute this time, speed of the constable in the above formula will give the distance that need to travel to catch the thief.

Complete step-by-step answer:
The speed of a thief and a constable is given as 6 km/hr and 8 km/hr respectively. The initial distance between the thief and the constable is 500 m. We are asked to find the distance travelled by constable after 500 m so that he can catch the thief.
The difference of the speeds of a constable and a thief is equal to:
 $ \begin{align}
  & \left( 8-6 \right)\text{km/hr} \\
 & =\text{2km/hr} \\
\end{align} $
The time taken to cover the distance of 500 m with the above speed of 2 km/hr is calculated using the following formula:
 $ \text{Speed}=\dfrac{\text{Distance}}{\text{Time}} $ ……….. Eq. (1)
Converting the distance of 500 m into km by dividing this distance to 1000 we get,
 $ \begin{align}
  & \dfrac{500}{1000}km \\
 & =\dfrac{1}{2}km \\
\end{align} $
Substituting the speed as 2 km/hr and distance as $ \dfrac{1}{2}km $ in eq. (1) we get,
 $ \begin{align}
  & 2=\dfrac{\dfrac{1}{2}}{\text{Time}} \\
 & \Rightarrow 2=\dfrac{1}{2\left( \text{Time} \right)} \\
\end{align} $
On cross multiplying the above equation we get,
 $ 4\left( \text{Time} \right)=1 $
Dividing 4 on both the sides of the above equation we get,
 $ \text{Time}=\dfrac{1}{4}hr $
Now, the distance traversed by the constable to catch the thief is calculated by substituting the time as $ \dfrac{1}{4}hr $ and speed of the constable as 8 km/hr in eq. (1) we get,
 $ \begin{align}
  & 8=\dfrac{\text{Distance}}{\dfrac{1}{4}} \\
 & \Rightarrow 8=\dfrac{4\left( \text{Distance} \right)}{1} \\
\end{align} $
Dividing 4 on both the sides of the above equation we get,
 $ \begin{align}
  & \dfrac{8}{4}=\text{Distance} \\
 & \Rightarrow \text{Distance}=2km \\
\end{align} $
From the above, the distance travelled by the police constable is 2 km but we have to find the distance travelled by the police after 500 m so subtracting 500 m (or 0.5 km) from the distance 2 km we get,
 $ \left( 2-0.5 \right)km $
 $ =1.5km $
Hence, the police traversed 1.5 km after 500 m to catch the thief.
Hence, the correct option is (b).

Note: The point of blunder that could happen in the above question is that you might have thought that the distance travelled by the police constable is 2 km which is wrong because we are asked to find the distance after traversing 500 m so we have to subtract 500 m from 2 km. Another point of blunder is to forget to convert the units of 500 m into km. In the rush of solving the question paper on time, students tend to make such mistakes.