correct answer is: b) \large\frac{7}{30}

Explanation

To find the required greatest quantity we need to HCF the given quotients.
Here, the given quantities are =2\large\frac{1}{3} kg, 4\large\frac{1}{5} kg and 8\large\frac{1}{6} kg.
Or, \large\frac{7}{3},\frac{21}{5} and \large\frac{49}{6} kg. _{[covert mixed to improper fraction]}
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 7,21 and 49.
\therefore The HCF of 4,16 and 8 is =7
And given denominators are 3,5 and 6.
\therefore The LCM of 3,5 and 6 is =3\times5\times2=30
So that, the HCF of 2\large\frac{1}{3} kg, 4\large\frac{1}{5} kg and 8\large\frac{1}{6} kg is =\large\frac{7}{30}
\therefore\large\frac{7}{30} is the greatest quantity that must be divided into 2\large\frac{1}{3} kg, 4\large\frac{1}{5} kg and 8\large\frac{1}{6} kg so that the quotients are integers.
Ans: \large\frac{7}{30} is the greatest quantity.

Other variants of this particular math question are:

1. Determine the largest quantity by which 2 1/3 kg, 4 1/5 kg, and 8 1/6 kg must be divided so that the resulting quotients are integers.
2. Find the maximum amount by which 2 1/3 kg, 4 1/5 kg, and 8 1/6 kg must be divided to ensure that the quotient is an integer.
3. Calculate the greatest divisor by which 2 1/3 kg, 4 1/5 kg, and 8 1/6 kg must be divided to yield integer quotients.
4. What is the largest number by which 2 1/3 kg, 4 1/5 kg, and 8 1/6 kg must be divided to obtain whole-number quotients?