## Elementary Technical Mathematics

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