Four navigation buoys in a harbor flash at regular intervals of 3 \frac{1}{3} minutes, 4 \… — State PSC Exam

Four navigation buoys in a harbor flash at regular intervals of $3 \frac{1}{3}$ minutes, $4 \frac{1}{6}$ minutes, $5 \frac{5}{9}$ minutes, and $7 \frac{1}{2}$ minutes, respectively. If all four buoys flash simultaneously at exactly 12:00 noon, how many additional times will they all flash together again within the next $12$ hours?

Cevap: 4 times

Cevap

The correct answer is derived by finding the Least Common Multiple of the fractional intervals to determine the duration between simultaneous flashes (150 minutes). Then, by converting the 12-hour period into 720 minutes, dividing 720 by 150 yields 4.8, indicating the buoys will flash together 4 additional times within that exact timeframe.

Adım Adım Çözüm

Convert the mixed numbers into improper fractions.

The intervals are

\frac{10}{3}

\frac{25}{6}

\frac{50}{9}

, and

\frac{15}{2}

minutes.

Mathematical operations involving LCM and HCF of fractions require them to be in proper or improper fractional forms, rather than mixed numbers.

Determine the mathematical formula required for simultaneous events.

We must calculate the Least Common Multiple (LCM) of the intervals using the formula:

\text{LCM of fractions} = \frac{\text{LCM of numerators}}{\text{HCF of denominators}}

Simultaneous cyclic events synchronize at the lowest common multiple of their individual intervals.

Calculate the LCM of the numerators.

The numerators are

10, 25, 50

, and

15

. Their LCM is

150

150

is the smallest integer divisible by all four numerators (

150 = 10 \times 15 = 25 \times 6 = 50 \times 3 = 15 \times 10

Calculate the HCF of the denominators.

The denominators are

3, 6, 9

, and

2

. Their HCF is

1

The numbers

2

and

3

are coprime (they share no common factors greater than

1

), forcing the overall HCF of the set to be

1

Calculate the exact time interval between simultaneous flashes.

The LCM of the intervals is

\frac{150}{1} = 150

minutes.

Applying the results from steps 3 and 4 directly into the fractional LCM formula.

Determine how many times this 150-minute interval fits into the 12-hour window.

12 \text{ hours} = 720 \text{ minutes}

\lfloor \frac{720}{150} \rfloor = \lfloor 4.8 \rfloor = 4

Dividing the total available time by the simultaneous interval length gives the exact count of additional flashes.

Anahtar Kavram

LCM of fractions in simultaneous cyclic events

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