Question

A and B start together from the same point on a walking watch round a circular course. After $\dfrac{1}{2}$ an hour A has walked three complete circuits and B four and a half. Assuming that each walks with uniform speed, find when B overtakes A.
A.15min
B.18min
C.22min
D.20min

Answer 1

Hint:
Speed is a measure of how fast an object is moving at a given instant of time. It is a scalar quantity. Its unit is m/s.
For a body covering equal distances in equal time intervals, the speed is called uniform speed.
When a body covers unequal distances in equal intervals of time, then that body is said to be in non uniform speed.
Average speed of a body is defined as the ratio of the total distance travelled by the body to the total time taken by it.
${\text{Average speed = }}\dfrac{{{\text{Total distance travelled}}}}{{{\text{Total time taken}}}}$
To overtake B has to walk one circle more than A.

Complete step by step solution:
$\because $ A completed 3 rounds in$\dfrac{1}{2}hours$ i.e. A complete 3 rounds in 30 minutes
∴ Distance travelled by him in $1{\text{ minute}} = \dfrac{3}{{30}}$
$ = \dfrac{1}{{10}}{\text{ of the circuits}}$
Similarly in$\dfrac{1}{2}hours$ i.e. in 30 minutes B completed 4 and half i.e. $4\dfrac{1}{2}$ round i.e. $\dfrac{9}{2}$ circuits.
∴Distance travelled by B in 1 minute $ = \dfrac{9}{2} \times \dfrac{1}{{30}} = \dfrac{3}{{20}}$ of the circuit
Let after‘t’ minutes B overtakes A
$\begin{gathered}
  \dfrac{{3t}}{{20}} - \dfrac{t}{{10}} = 1 \\
  \dfrac{{3t - 2t}}{{20}} = 1 \\
  \dfrac{t}{{20}} = 1 \\
  t = 20{\text{minute}} \\
\end{gathered} $
∴Option (D) is correct.

Note:
When two or three persons are running around a circular track, it is difficult to imagine the condition when they meet each other. Let’s see the various cases for solving circular track problems:
Case 1: When two persons A and B are running around a circular track of length L meter with speeds of a, b m/s in the same direction.
They meet each other at any point on the track is $\dfrac{{\text{L}}}{{\left( {{\text{a - b}}} \right)}}{\text{seconds}}$
They meet each other at exactly at the starting point = $\left( {\dfrac{{\text{L}}}{{\text{a}}}{\text{,}}\dfrac{{\text{L}}}{{\text{b}}}} \right){\text{seconds}}$
Case 2: When two persons A and B are running around a circular track of length L meter with speeds of a, b m/s in the opposite direction.
They meet each other at any point on the track is $\dfrac{{\text{L}}}{{\left( {{\text{a + b}}} \right)}}{\text{seconds}}$
They meet each other at exactly at the starting point = $\left( {\dfrac{{\text{L}}}{{\text{a}}}{\text{,}}\dfrac{{\text{L}}}{{\text{b}}}} \right){\text{seconds}}$