Question

$x$, $y$ and $z$ start at the same time in the same direction to run around a circular stadium. $x$ completes a round in $126$ seconds, $y$ in $154$ seconds and $z$ in $231$ seconds, all starting at the same point. After what time will they meet again at the starting point? How many rounds would have $x$, $y$ and $z$ completed by this time?

Answer 1

It is given that $x$ completes a round in $126$ seconds, $y$ in $154$ seconds and $z$ in $231$ seconds.

$126$, $154$ and $231$ can be factorised as follows:

$126=2\times 3\times 3\times 7154=2\times 7\times 11231=3\times 7\times 11$

The time after which $x,y$ and $z$ meet is the $LCM$ of the time taken by $x,y$ and $z$ to cover one round.

We know that $LCM$ is the least common multiple, therefore, the $LCM$ of $126$, $154$ and $231$ is:

$LCM=2\times 3\times 3\times 7\times 11=1386$

Thus, time after which $x,y$ and $z$ will meet is $1386$ seconds.

Now, the number of rounds covered can be obtained by dividing total time by time for each round, therefore,

Number of rounds of $x$ is $\frac{1386}{126}=11$

Number of rounds of $y$ is $\frac{1386}{154}=9$

Number of rounds of $z$ is $\frac{1386}{231}=6$

Hence, $x$ completed $11$ rounds, $y$ completed $9$ rounds and $z$ completed $6$ rounds.