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$(X*Y*Z =192$ and $Z=4)$ $\Rightarrow$ $X*Y= 48$

$\rightarrow$Given : $P = \frac{X+Y}{2}$

$\rightarrow$Average of $2$ numbers is minimum when the difference of that $2$ numbers is minimum.

$\rightarrow$We can write $48$ as product of $1*48,2*24,3*16,4*12,6*8$

$\rightarrow$Among all these $6*8$ is the expression which contains the numbers $6$ and $8$ and they have the minimum difference as compared to other expressions.

$\rightarrow$So $X=6$ and $Y=8$

$\Rightarrow$ $P = \frac{X+Y}{2} = \frac{6+8}{2} = 7 $

$\therefore$ Option B. $7$ is the correct answer.

$\rightarrow$Given : $P = \frac{X+Y}{2}$

$\rightarrow$Average of $2$ numbers is minimum when the difference of that $2$ numbers is minimum.

$\rightarrow$We can write $48$ as product of $1*48,2*24,3*16,4*12,6*8$

$\rightarrow$Among all these $6*8$ is the expression which contains the numbers $6$ and $8$ and they have the minimum difference as compared to other expressions.

$\rightarrow$So $X=6$ and $Y=8$

$\Rightarrow$ $P = \frac{X+Y}{2} = \frac{6+8}{2} = 7 $

$\therefore$ Option B. $7$ is the correct answer.