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Difficulty: Very hardHCF and LCM

Three automated robotic arms in an assembly line complete their respective sorting cycles in 1415\frac{14}{15} minutes, 2120\frac{21}{20} minutes, and 4225\frac{42}{25} minutes. If all three robotic arms start a sorting cycle simultaneously at 9:00:00 AM, after what minimum time interval will all three arms simultaneously begin a new sorting cycle again?

  1. 88 minutes 2424 secondsAnswer
  2. B
    1.41.4 seconds
  3. C
    4242 minutes 5151 seconds
  4. D
    42.8642.86 seconds

Answer

88 minutes 2424 seconds
To find the time when events with fractional period lengths coincide again, we compute the LCM of the fraction time intervals. The formula for the LCM of fractions is LCM of numeratorsHCF of denominators\frac{\text{LCM of numerators}}{\text{HCF of denominators}}. Finding LCM(14,21,42)=42\text{LCM}(14, 21, 42) = 42 and HCF(15,20,25)=5\text{HCF}(15, 20, 25) = 5 yields 425=8.4\frac{42}{5} = 8.4 minutes, which equals 88 minutes 2424 seconds.

Step-by-Step Solution

1
Identify the mathematical operation required for simultaneous recurrence
The required minimum time interval is the Least Common Multiple (LCM) of the three cycle times: LCM(1415,2120,4225)\text{LCM}\left(\frac{14}{15}, \frac{21}{20}, \frac{42}{25}\right) minutes.
Simultaneous future events occurring at periodic intervals require finding the LCM of the given time periods.
2
Apply the formula for the LCM of fractions
LCM(ab,cd,ef)=LCM(a,c,e)HCF(b,d,f)\text{LCM}\left(\frac{a}{b}, \frac{c}{d}, \frac{e}{f}\right) = \frac{\text{LCM}(a, c, e)}{\text{HCF}(b, d, f)}
For simplified fractions, the LCM is determined by dividing the LCM of all numerators by the HCF of all denominators.
3
Calculate the LCM of the numerators (14,21,42)(14, 21, 42)
14=2×714 = 2 \times 7, 21=3×721 = 3 \times 7, 42=2×3×742 = 2 \times 3 \times 7. Thus, LCM(14,21,42)=2×3×7=42\text{LCM}(14, 21, 42) = 2 \times 3 \times 7 = 42.
The smallest positive integer divisible by 14, 21, and 42 is 42.
4
Calculate the HCF of the denominators (15,20,25)(15, 20, 25)
15=3×515 = 3 \times 5, 20=22×520 = 2^2 \times 5, 25=5225 = 5^2. Thus, HCF(15,20,25)=5\text{HCF}(15, 20, 25) = 5.
The greatest integer that divides 15, 20, and 25 without remainder is 5.
5
Combine the results and convert to minutes and seconds
LCM=425=8.4\text{LCM} = \frac{42}{5} = 8.4 minutes. Converting 0.40.4 minutes to seconds gives 0.4×60=240.4 \times 60 = 24 seconds. Total time = 88 minutes 2424 seconds.
Multiplying the fractional minute by 60 converts it into exact seconds.

Key Concept

LCM of Fractions
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Topics

HCF and LCM

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