Let stock price = s
Let inflation = i
Let diesel price = d
$$Given\ that\ s=\frac{k}{i}.......eqn\ 1\ \ \ \ \ \ \ where\ k=cons\tan t$$
$$and\ i\ =l\cdot d..........eqn\ 2\ \ \ \ where\ l=cons\tan t$$
$$for\ eqn\ 1\ when\ i=10,\ s=100$$
$$100=\frac{k}{10}and\ k=100\cdot10=1000$$
$$for\ eqn\ 2\ when\ i=12,\ d=60$$
$$12=l\cdot60\ and\ l=\frac{12}{60}=\frac{1}{5}$$
Substituting the value of i in eqn 2 with the value of i in eqn 1
$$s=\frac{k}{i}where\ i=l\cdot d$$
$$s=\frac{k}{l\cdot d}..........eqn\ 3$$
$$currently\ s=200;\ then\ d\ =??$$
$$u\sin g\ eqn\ 3$$
$$200=\frac{1000}{\frac{1}{5}d}=>\ \frac{200}{5}d=1000$$
$$d=1000\cdot\frac{5}{200}=25$$
With 25% increase to the value of s then s becomes
$$s=200+25\%\ of\ 200\ =>\ s=200+50=250$$
$$and\ d=??\ \ \ \ \ u\sin g\ eqn\ 3;$$
$$250=\frac{1000}{\frac{1}{5}d}=>\frac{200}{5}d=1000$$
$$d=1000\cdot\frac{5}{250}=20$$
The change or fall in diesel price from a stock price 200 to 250 => 25 - 20 = 5

Answer = A