Examkin

Question

Difficulty: HardHCF and LCM

In a chemical manufacturing plant, three automated valves release specific additives into a continuous mixing tank. Valve A opens every 92\frac{9}{2} minutes, Valve B opens every 154\frac{15}{4} minutes, and Valve C opens every 458\frac{45}{8} minutes. The system is programmed to record a 'synchronization event' whenever all three valves open at the exact same instant.

If the mixing process runs continuously for exactly 1212 hours, how many synchronization events will be recorded during this period? (Assume a synchronization event is recorded at the very beginning of the process, which counts as the first event).

  1. A
    129129
  2. B
    3232
  3. C
    271271
  4. 3333Answer

Answer

The correct answer is 3333 synchronization events.
The synchronization interval is the LCM of the fractions 92\frac{9}{2}, 154\frac{15}{4}, and 458\frac{45}{8}. Using the formula LCM of numeratorsHCF of denominators\frac{\text{LCM of numerators}}{\text{HCF of denominators}}, we get LCM(9,15,45)HCF(2,4,8)=452=22.5\frac{\text{LCM}(9, 15, 45)}{\text{HCF}(2, 4, 8)} = \frac{45}{2} = 22.5 minutes. Over a 1212-hour period (720720 minutes), there are precisely 72022.5=32\frac{720}{22.5} = 32 complete intervals. Since a synchronization event is recorded at the very beginning of the process, the total number of recorded events is 32+1=3332 + 1 = 33.

Step-by-Step Solution

1
Identify the mathematical concept required to find the synchronization interval.
The shortest time interval for all valves to open together is the Least Common Multiple (LCM) of their individual time cycles.
Synchronization occurs at common multiples of the cycle times, and the next event always occurs at the least common multiple.
2
Calculate the LCM of the fractions 92\frac{9}{2}, 154\frac{15}{4}, and 458\frac{45}{8}.
LCM of fractions=LCM(9,15,45)HCF(2,4,8)\text{LCM of fractions} = \frac{\text{LCM}(9, 15, 45)}{\text{HCF}(2, 4, 8)}.
The standard formula for the LCM of fractions requires finding the LCM of the numerators divided by the HCF of the denominators.
3
Compute the LCM of numerators and HCF of denominators.
LCM(9,15,45)=45\text{LCM}(9, 15, 45) = 45 and HCF(2,4,8)=2\text{HCF}(2, 4, 8) = 2. Thus, the synchronization interval is 452=22.5\frac{45}{2} = 22.5 minutes.
This establishes that every 22.522.5 minutes, all three valves will open simultaneously.
4
Calculate the total time in minutes for the 1212-hour shift.
12 hours×60 minutes/hour=720 minutes12 \text{ hours} \times 60 \text{ minutes/hour} = 720 \text{ minutes}.
Units must be consistent to accurately determine the number of intervals.
5
Determine the number of full intervals and total synchronization events.
Number of intervals=72022.5=32\text{Number of intervals} = \frac{720}{22.5} = 32. Total events =32+1=33= 32 + 1 = 33.
The division yields the number of periods between events. We add 11 to account for the initial synchronization event explicitly mentioned in the problem (the fencepost principle).

Key Concept

Calculating the Least Common Multiple (LCM) of fractions and applying it to a temporal boundary condition (fencepost counting).
Estimated Time:2m 0s
Rate this question

Topics

HCF and LCM

More questions from this topic