A, B, and C have a few chocolates among themselves. A gives to each of the other two half the number of chocolates they already have. Similarly B and C (in that order) give each of the other two half the number of chocolates each of them already has. Now, if each of them has the same number of chocolates, what could be the minimum number of chocolates they have among themselves?

Concept Used:-

Following expression can be used to solve the given problem,

Initial number of chocolates = Number of chocolates remain + Given chocolates to other 

Explanation:-

Suppose A, B and C have x number of chocolates in the last.

Draw a table for the number of chocolate A, B and C has and allot x chocolate to each of them in final step.

Now, in the last C has given A and B. Suppose A and B has y number of chocolates each. Sum of chocolates of A and half of B will be equal to number of chocolates of C (x).

Mathematically,

\(\begin{aligned} &\Rightarrow y+\frac{y}{2}=x \\ & \Rightarrow y=\frac{2 x}{3} \end{aligned}\)

So, A and B each has 2x/3 chocolates.

Now, the number of sums of half of chocolates of A, B and total chocolates of C will be,

\(\begin{aligned} & =\frac{2 x}{3}+\frac{2 x}{3}+x \\ & =\frac{x}{3}+\frac{x}{3}+x \\ & =\frac{5 x}{3} \end{aligned}\)

Now before C, B has double the amount of each. Since, B has given half chocolates to A. So

\(\begin{aligned} & y+\frac{y}{2}=\frac{2 x}{3} \\ & \frac{3 y}{2}=\frac{2 x}{3} \\ & y=\frac{4 x}{9} \end{aligned}\)

This is the number of chocolate A has at this stage. The number of chocolates C has in this stage,

\(\begin{aligned} & =\frac{2 \times 5 x}{9} \\ & =\frac{10 x}{9} \end{aligned}\)

And after distributing, B remains with number of chocolates,

\( \begin{aligned} & =\frac{6 x}{9}+\frac{2 x}{9}+\frac{5 x}{9} \\ & =\frac{13 x}{9} \end{aligned} \)

In the same way, A has given to all before B. So, combine value of B and C,

\(\begin{aligned} & y+\frac{y}{2}=\frac{13 x}{9} \\ & y=\frac{26 x}{27} \end{aligned}\)

The number of chocolates C has in this stage,

\( \begin{aligned} & =\frac{2 \times 10 x}{9} \\ & =\frac{20 x}{9} \end{aligned} \)

After distributing, A remain with number of chocolates,

\( \begin{aligned} & =\frac{13 x}{27}+\frac{10 x}{27}+\frac{12 x}{27} \\ & =\frac{35 x}{27} \end{aligned} \)

So, the table we get,

As we get 27 common division in each of first column. So, the chocolate each have with them is 27.

So, the value of x will be 3. Now, the total minimum number of chocolates they have among themselves,

\(=3x \\ =3\times 27 \\ =81\)

So, the correct option is 2.