List price of a toy increase …

Let the original list price of the toy be Rs. x.
$\therefore$ Number of toys can be bought for Rs 360 = $\frac{360}{x}$ toys
Now, Reduced list price of the toy = Rs(x - 2)
$\therefore$ Number of toys can be bought with new reduced list price for Rs 360 = $\frac{360}{x-2}$ toys
According to the question:
$\frac{\displaystyle360}{\displaystyle x-2}=2+\frac{\displaystyle360}{\displaystyle x}$ (2 extra toys can be bought if price reduces by 2 rupees)
$\therefore \frac{360}{x - 2 } - \frac{360}{x }$ = 2
$\Rightarrow \frac{ 360x - 360(x - 2 )}{x (x - 2)}$ = 2
$\Rightarrow \frac{360x - 360x + 720}{x^2 - 2x }$ = 2
$\Rightarrow$ 720 = 2(x² - 2x)
$\Rightarrow x^2 - 2x = \frac{720 }{2}$
$\Rightarrow$ x² - 2x = 360
$\Rightarrow$ x² - 2x - 360 = 0
$\Rightarrow$ x² - 20x + 18x - 360 = 0
$\Rightarrow$ x(x - 20) + 18(x - 20) = 0
$\Rightarrow$ (x - 20)(x + 18) = 0
$\Rightarrow$ x - 20 = 0 [ Since, Cost cannot be negative. $\therefore$,x + 18 $\neq$ 0]
$\Rightarrow$ x = 20
Hence, the original list price of the toy is x = Rs 20.