Question

There are three kinds of liquids X, Y, Z. Three jars $J_1,J_2,J_3$ contain $100$ ml of liquids X, Y, Z, respectively. By an operation we mean three steps in the following order: -stir the liquid in $J_1$ and transfer $10$ml from $J_1$ into $J_2$;
- Stir the liquid in $J_2$ and transfer $10$ml from $J_2$ into $J_3$;
- Stir the liquid in $J_3$ and transfer $10$ml from $J_3$ into $J_1$.
After performing the operation four times, let x, y, z be the amounts of X, Y, Z, respectively, in $J_1$. Then.

• A. x $>$ y $>$ z
• B. x $>$ z $>$ y
• C. y $>$ x $>$ z
• D. z $>$ x $>$ y

Answer 1

The correct option is B x $>$ z $>$ y

When $10ml$ of $X$ is transferred from$J_1\to J_2$

Then concentarion of $X$ in $J_2=\frac{10}{110}$

When $10ml$ of this is transferred from$J_2\to J_3$

Then concentarion of $X$ in $J_3=\frac{\frac{10}{110}\times 10}{110}=\frac{1}{11}\times \frac{1}{11}$

And when 10 ml of this is transferred from $J_3\to J_1$

Then total amount of $X$ in $J_1=90+(\frac{1}{11}\times \frac{1}{11})\times 10=90+\frac{10}{{11}^2}$

Concentration of $Y$ in $J_2=\frac{10}{110}$ after $10ml$ is transferred from $J_1\to J_2$

Concentraion of $Y$ in $J_3=\frac{\frac{100}{110}\times 10}{110}=\frac{10}{11\times 11}$

Amount of $Y$ in $J_1=\frac{10}{11\times 11}\times 10=\frac{100}{121}$

Amount of $Z$ in $J_1=\frac{100}{110}\times 10=\frac{100}{11}$

After one operation $x>z>y$

The order will remain same after four such operations

So option $B$ is correct.