A circular path is 30 km long. Four racers start at the same time in the same direction to run around this circular path with the speed 4, 5, 6 and 10 km/ hours respectively. After what time they meet again at the starting point?

Options

A. A)30 hours
B. B)42 hours
C. C)15 hours
D. D)20 hours

Answer

To solve the problem of when the four racers will meet again at the starting point on a circular path of 30 km, we can follow these steps:

### Step 1: Calculate the time taken by each racer to complete one lap.
To find the time taken by each racer to complete the circular path, we can use the formula:

\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} \]

- For the first racer (speed = 4 km/h):
\[ \text{Time}_1 = \frac{30 \text{ km}}{4 \text{ km/h}} = 7.5 \text{ hours} \]

- For the second racer (speed = 5 km/h):
\[ \text{Time}_2 = \frac{30 \text{ km}}{5 \text{ km/h}} = 6 \text{ hours} \]

- For the third racer (speed = 6 km/h):
\[ \text{Time}_3 = \frac{30 \text{ km}}{6 \text{ km/h}} = 5 \text{ hours} \]

- For the fourth racer (speed = 10 km/h):
\[ \text{Time}_4 = \frac{30 \text{ km}}{10 \text{ km/h}} = 3 \text{ hours} \]

### Step 2: Find the least common multiple (LCM) of the times taken.
Now we need to find the LCM of the times taken by the racers to determine when they will all meet again at the starting point.

The times are:
- 7.5 hours (which can be expressed as \( \frac{15}{2} \))
- 6 hours (which can be expressed as \( \frac{6}{1} \))
- 5 hours (which can be expressed as \( \frac{5}{1} \))
- 3 hours (which can be expressed as \( \frac{3}{1} \))

To find the LCM of the numerators (15, 6, 5, 3) and the HCF of the denominators (2, 1, 1, 1):

**Finding LCM of 15, 6, 5, and 3:**
- Prime factorization:
- 15 = \( 3 \times 5 \)
- 6 = \( 2 \times 3 \)
- 5 = \( 5 \)
- 3 = \( 3 \)

The LCM takes the highest power of each prime:
- LCM = \( 2^1 \times 3^1 \times 5^1 = 30 \)

**Finding HCF of 2, 1, 1, 1:**
- HCF = 1

### Step 3: Calculate the time when they will meet again.
Using the LCM and HCF, we can calculate the time:
\[ \text{Time} = \frac{\text{LCM of numerators}}{\text{HCF of denominators}} = \frac{30}{1} = 30 \text{ hours} \]

### Conclusion
The four racers will meet again at the starting point after **30 hours**.

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