Question

Three persons begin to walk around a circular track. The first completes revolution in \[15\dfrac{1}{6}\] seconds, and the second in \[16\dfrac{1}{4}\] seconds and the third in \[18\dfrac{2}{3}\] seconds, respectively. After what time will they be together at the starting point again?
(A) 1 hr 40 min
(B) 1 hr 40 sec
(C) 1.4 hrs
(D) 1 hr 3 min 40 sec

Answer 1

Hint: The time of revolution for the first person, second person, and the third person are \[\dfrac{91}{6}\] seconds, \[\dfrac{65}{4}\] seconds, and \[\dfrac{56}{3}\] seconds respectively. Use the logic that the lowest multiple of the time of the revolution for first person, second person, and third person will be the time at which all three persons will again meet at the starting point. For LCM of fractional numbers, use the formula, \[\text{LCM=}\dfrac{\text{LCM}\,\text{of}\,\text{numerator}}{\text{HCF}\,\text{of}\,\text{denominator}}\] . Now, solve it further and get the LCM of are \[\dfrac{91}{6}\] seconds, \[\dfrac{65}{4}\] seconds, and \[\dfrac{56}{3}\] seconds.

Complete step-by-step answer:
According to the question, we are given that
The time of the revolution for first person = \[15\dfrac{1}{6}\] seconds = \[\dfrac{91}{6}\] seconds ………………………………………(1)
The time of the revolution for second person = \[16\dfrac{1}{4}\] seconds = \[\dfrac{65}{4}\] seconds ………………………………………(2)
The time of the revolution for third person = \[18\dfrac{2}{3}\] seconds = \[\dfrac{56}{3}\] seconds ………………………………………(3)
We have to find the time when they will be again meeting at the starting point.
Here, we can think of the logic that the lowest multiple of the time of the revolution for first person, second person, and third person will be the time at which all three persons will again meet at the starting point.
Using this logic, we need the lowest common factor of the time of the revolution for the first person, second person, and the third person that are \[\dfrac{91}{6}\] seconds, \[\dfrac{65}{4}\] seconds, and \[\dfrac{56}{3}\] seconds ………………………………………………..(4)
For LCM of fractional numbers, we have the formula, \[\text{LCM=}\dfrac{\text{LCM}\,\text{of}\,\text{numerator}}{\text{HCF}\,\text{of}\,\text{denominator}}\] ………………………………………..(5)
Now, from equation (4) and equation (5), we get
 \[\text{LCM=}\dfrac{\text{LCM}\,\text{of}\,\text{91,65,56}}{\text{HCF}\,\text{of}\,\text{6,4,3}}\] …………………………………………….(6)
LCM of 91, 65, and 56 = 3640 …………………………………………..(7)
HCF of 6, 4, and 3 = 1 ………………………………………………….(8)
Now, from equation (6), equation (7), and equation (8), we get
\[\text{LCM=}\dfrac{3640}{1}\]
So, the LCM of \[\dfrac{91}{6}\] seconds, \[\dfrac{65}{4}\] seconds, and \[\dfrac{56}{3}\] seconds is 3640 seconds …………………………………………..(9)
We know the relation between seconds and hour, i.e., \[\text{1}\,\text{hr=3600}\,\text{seconds}\] ………………………………..(10)
Now, from equation (9) and equation (10), we get
\[\begin{align}
  & \text{=3640}\,\text{seconds} \\
 & \text{=3600}\,\text{seconds+}\,\text{40}\,\text{seconds} \\
 & \text{=1}\,\text{hr}\,\text{40}\,\text{seconds} \\
\end{align}\]
Therefore, after 1 hr 40 seconds they all will be again meeting at the starting point.

So, the correct answer is “Option (B)”.

Note: In this question, one might a silly mistake and miss to divide the LCM of the numerators of \[\dfrac{91}{6}\] seconds, \[\dfrac{65}{4}\] seconds, and \[\dfrac{56}{3}\] seconds by the HCF of the denominators. To rectify this mistake always keep the formula, \[\text{LCM=}\dfrac{\text{LCM}\,\text{of}\,\text{numerator}}{\text{HCF}\,\text{of}\,\text{denominator}}\] in mind.