correct answer is: a) \large\frac{4}{147}

Explanation

To find the required greatest length we need to HCF the given metres.
Here, given metres are =\large\frac{4}{7},\frac{16}{21} and \large\frac{8}{49}
The formula we use to get the HCF of two or more fractions is:
\large\frac{The\ HCF\ of\ Numerator}{The\ LCM\ of\ Denominator}

Here, given numerators are 4,16 and 8.
\therefore The HCF of 4,16 and 8 is =4
And given denominators are 7,21 and 49.
\therefore The LCM of 7,21 and 49 is =7\times3\times7=147
So that, the HCF of the metres \large\frac{4}{7},\frac{16}{21} and \large\frac{8}{49} is =\large\frac{4}{147}
\therefore\large\frac{4}{147} is the greatest length which is contained a whole number of times exactly in \large\frac{4}{7},\frac{16}{21} and \large\frac{8}{49} metres.
Ans: \large\frac{4}{147} is the greatest length.

These particular math questions can also be asked in the following ways:

1. Finding the maximum length that evenly divides into 4/7, 16/21, and 8/49 meters to ensure integer divisions.
2. What is the greatest common factor of length for 4/7, 16/21, and 8/49 meters to ensure integer results?
3. Finding the largest whole number divisor for evenly fitting into lengths of 4/7, 16/21, and 8/49 meters.