The circumferences of the fore and hind wheels of a carriage are `(6)3/(14)`metres and `(8)1/(18)` meters respectively .At any given moment a chalk mark is put on the point of contact of each wheel with the ground. Find the distance travelled by the carriage so that both the chalk marks are again on the ground at the same time-

Options

A. 218 m
B. `217.5` m
C. `218.5` m
D. 217 m

Answer

To solve the problem of finding the distance traveled by the carriage so that both chalk marks are again on the ground at the same time, we will follow these steps:

### Step 1: Convert the Circumferences to Improper Fractions
The circumferences of the fore and hind wheels are given as:
- Fore wheel: \(6 \frac{3}{14} = \frac{6 \times 14 + 3}{14} = \frac{84 + 3}{14} = \frac{87}{14}\) meters
- Hind wheel: \(8 \frac{1}{18} = \frac{8 \times 18 + 1}{18} = \frac{144 + 1}{18} = \frac{145}{18}\) meters

### Step 2: Find the LCM of the Circumferences
To find the distance traveled by the carriage when both wheels have completed a whole number of rotations, we need to calculate the LCM of the two circumferences:
- LCM of \(\frac{87}{14}\) and \(\frac{145}{18}\)

Using the formula for LCM of two fractions:
\[ \text{LCM}\left(\frac{a}{b}, \frac{c}{d}\right) = \frac{\text{LCM}(a, c)}{\text{HCF}(b, d)} \]
where \(a = 87\), \(b = 14\), \(c = 145\), and \(d = 18\).

### Step 3: Calculate LCM of Numerators
We need to find the LCM of 87 and 145.
- Prime factorization:
- \(87 = 3 \times 29\)
- \(145 = 5 \times 29\)

The LCM will be:
\[ \text{LCM}(87, 145) = 3 \times 5 \times 29 = 435 \]

### Step 4: Calculate HCF of Denominators
Next, we find the HCF of 14 and 18:
- Prime factorization:
- \(14 = 2 \times 7\)
- \(18 = 2 \times 3^2\)

The HCF will be:
\[ \text{HCF}(14, 18) = 2 \]

### Step 5: Calculate the LCM of the Circumferences
Now we can calculate the LCM of the circumferences:
\[ \text{LCM}\left(\frac{87}{14}, \frac{145}{18}\right) = \frac{435}{2} \]

### Step 6: Convert to Decimal
To find the distance in meters:
\[ \frac{435}{2} = 217.5 \text{ meters} \]

### Final Answer
The distance traveled by the carriage so that both chalk marks are again on the ground at the same time is **217.5 meters**.

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