Because, \( \Large \frac{x}{2}+\frac{y}{3}=4 \)

= \( \Large \frac{3x+2y}{6}=4 \)

= \( \Large 3x+2y=24 \) ...(i)

and \( \Large \frac{2}{x}+\frac{3}{y}=1 = \frac{2y+3x}{xy}=1 \)

= \( \Large 2y+3x=xy \) ...(ii)

From Eqs. (i) and (ii),

xy = 24

There are 6 possibilities for x and y, respectively. 

2 *12, 3*8,4*6, 6*4,8*3,12*2

2 and 12 cannot be the values of x and y as their sum is 14 and it is not given in options.

Now, we check both 3 and 8 as well as 4 and 6 as values of x and y or value of y and x.

Only 4 as a value of x and 6 as a value of

y satisfied the given condition \( \Large \frac{x}{2}+\frac{y}{3}=4. \)

So, \( \Large x = 4 \  and \  y = 6 \)

Therefore, \( \Large x+y = 4+6 = 10 \)