#### Answer

remaining fries = $\frac{1}{8}$ bags

#### Work Step by Step

Step 1.
Let us find the total amount of fries before the shift starts. We just need to add them.
total fries = 1$\frac{1}{2}$bags + 3 bags
We can write "A$\frac{x}{y}$" as A + $\frac{x}{y}$, so we can do this with the mixed numbers in our problem.
total fries = 1 bags + $\frac{1}{2}$bags + 3 bags (Use the m.c.d. to add the fractions. The m.c.d. in this case is 2)
total fries = $\frac{2(1)+1+2(3)}{2}$bags = $\frac{9}{2}$bags = 4$\frac{1}{2}$bags
Step 2.
Now let's find the total bags of fries used and wasted.
used fries = 1$\frac{3}{4}$bags + 2$\frac{1}{2}$bags + $\frac{1}{8}$bags
Repeating the procedure as in Step 1, we get:
used fries = $\frac{35}{8}$bags = 4$\frac{3}{8}$bags
Step 3.
So the total amount of fries remaining at the end of the shift is the total fries minus the used fries:
remaining fries = 4$\frac{1}{2}$bags - 4$\frac{3}{8}$bags
remaining fries = 4 bags + $\frac{1}{2}$bags - 4 bags - $\frac{3}{8}$bags
remaining fries = $\frac{1}{8}$ bags