Question

There are 2 baskets with n identical balls. Half the balls from basket 1 are put into basket 2. The ratio of balls in Basket 1 to Basket 2 gets reversed. Next ${\frac{1}{3}}^{rd}$ and ${\frac{2}{3}}^{rd}$ of the balls from the $1^{st}$ basket and $2^{nd}$ basket respectively are exchanged. The ratio of balls n basket 2 to basket 1 is?

• A. 3 : 1
• B. 2 : 1
• C. 1 : 1
• D. 1 : 2

Answer 1

The correct option is B 2 : 1

$\begin{array}{cc}Basket1& Basket2\\ xballs& yballs\\ \frac{x}{2}balls& (y+\frac{x}{2})balls\end{array}$
$\{\frac{\frac{x}{2}}{(y+\frac{x}{2})}=\left(\frac{y}{x}\right)......(1\left)\right[Theratiogetsreversed\left]\right\}$
x to need of this statement. You can get answer directly
${\frac{1}{3}}^{rd}$ from $1^{st}$ basket & ${\frac{2}{3}}^{rd}$ from $2^{nd}$ basket are ex changed.
$\frac{x}{2}-\frac{x}{6}+\frac{2}{3}(y+\frac{x}{2})\frac{1}{3}(y+\frac{x}{2})+\frac{x}{6}=\frac{\frac{x}{3}+\frac{2}{3}(y+\frac{x}{2})}{\frac{x}{6}+\frac{1}{3}(y+\frac{x}{2})}=\frac{\frac{2}{3}(y+x)}{\frac{1}{3}(y+x)}=2:1$