91Ó°ÊÓ

The small box contains $1$dozen oatmeal and $1$dozen chocolate chip.

The medium box contains $2$dozen oatmeal raisin, $1$dozen chocolate chip and $1$dozen shortbread.

The large contains $2$dozen oatmeal raisin, $2$ dozen chocolate chip, and$3$ dozen shortbread.

Let $x,y,z$denote the small, medium and large boxes respectively.

So the total number of raisins required is given by,

$x+2y+2z=15$

The number of chocolate chip required is given by,

$x+y+2z=10$

The number of short bread required is given by,

$y+3z=11$

The system of equation will be,

$\left\{\begin{array}{l}x+2y+2z=15\\ x+y+2z=10\\ y+3z=11\end{array}\right.$

Subtract second equation from first equation, we get,

$y=5$

Substitute $y=5$in the third equation, we get,

$$\begin{gathered} y+3z=11 \\ \left(5\right)+3z=11 \\ 3z=11-5 \\ 3z=6 \\ z=2 \end{gathered}$$

Substitute $y=5;z=2$in the first equation, we get,

$$\begin{gathered} x+2y+2z=15 \\ x+2\left(5\right)+2\left(2\right)=15 \\ x+14=15 \\ x=1 \end{gathered}$$

The number of small box required is $1$, the medium box is $5$ and the large box is $2$