a) \large\frac{4}{30}

b) \large\frac{7}{30}

c)’ \large\frac{9}{30}

d) \large\frac{11}{30}

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:

**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.****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.****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.****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?**