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Let the average height of the original $22$ toddlers be $`x\text{’}$ inches.

$\dfrac{x_1 + x_2 + \ldots + x_{20} + x_{21} + x_{22}}{22} = x \quad \longrightarrow (1)$

Let $x_{21},x_{22}$ leaves the group.

$\dfrac{x_1 + x_2 + \ldots + x_{22}}{20} = x + 2$

$\Rightarrow x_1 + x_2 + \ldots + x_{20} = 20x + 40 \quad \longrightarrow (2)$

Also, $\dfrac{x_{21} + x_{22}}{2} = \dfrac{x}{3}$

$\Rightarrow x_{21} + x_{22} = \dfrac{2x}{3}\longrightarrow(3)$

From equation $(2)$ and$(3)$ , put the value in equation $(1)$, we get.

$\dfrac{20x + 40 + \frac{2x}{3}}{22} = x$

$\Rightarrow \dfrac{60x + 120 + 2x}{66} = x$

$\Rightarrow 62x + 120 = 66x$

$\Rightarrow 4x = 120$

$\Rightarrow \boxed{x = 30}$

$\therefore$ The average height of the remaining $20$ toddlers $ = x + 2= 30 + 2= 32$ inches.

Correct Answer$:\text{C}$

$\dfrac{x_1 + x_2 + \ldots + x_{20} + x_{21} + x_{22}}{22} = x \quad \longrightarrow (1)$

Let $x_{21},x_{22}$ leaves the group.

$\dfrac{x_1 + x_2 + \ldots + x_{22}}{20} = x + 2$

$\Rightarrow x_1 + x_2 + \ldots + x_{20} = 20x + 40 \quad \longrightarrow (2)$

Also, $\dfrac{x_{21} + x_{22}}{2} = \dfrac{x}{3}$

$\Rightarrow x_{21} + x_{22} = \dfrac{2x}{3}\longrightarrow(3)$

From equation $(2)$ and$(3)$ , put the value in equation $(1)$, we get.

$\dfrac{20x + 40 + \frac{2x}{3}}{22} = x$

$\Rightarrow \dfrac{60x + 120 + 2x}{66} = x$

$\Rightarrow 62x + 120 = 66x$

$\Rightarrow 4x = 120$

$\Rightarrow \boxed{x = 30}$

$\therefore$ The average height of the remaining $20$ toddlers $ = x + 2= 30 + 2= 32$ inches.

Correct Answer$:\text{C}$