Unchanged Rectangle

Let the original length and width of the rectangle be $L_1$ and $W_1$ respectively and its area $A$ . Then $A = L_1W_1$ . Let the increased length be $L_2$ and the decreased width be $W_2$ . Then $A=L_2W_2$ and

$\begin{aligned} L_2 W_2 & = L_1 W_1 \\ \left(1+ \frac {150}{100} \right)L_1 \left(1-\frac w{100} \right) W_1 & = L_1 W_1 \\ \frac 52 \left(1-\frac w{100} \right) & = 1 \\ 1 - \frac w{100} & = \frac 25 \\ \frac w{100} & = 1 - \frac 25 = \frac 35 \\ \implies w & = \frac 35 \times 100 = \boxed{60} \end{aligned}$

Let the initial values of length and breadth of the rectangle be $l$ and $b$ , then

$lb=l\left (1+\dfrac {150}{100}\right )b \left (1-\dfrac {w}{100}\right )$

$\implies w=100\left (1-\dfrac 25\right )=\boxed {60}$ .

If $z-w\%=2$ , then $x=1$ .

$100+150=250$

$250 \times 2=500$

$100 \times 5= 500$

$100\% - \frac {2} {5}=60\%$

$w=\boxed{60}$