Question

Four extra-large sandwiches of exactly the same size were ordered for m students, where m > 4. Three of the sandwiches were evenly divided among the students. Since 4 students did not want any of the fourth sandwich, it was evenly divided among the remaining students. If Carol ate one piece from each of the four sandwiches, the amount of sandwich that she ate would be what fraction of a whole extra-large sandwich?

• A. $\frac{m+4}{m(m+4)}$
• B. $\frac{2m-4}{m(m-4)}$
• C. $\frac{4m-4}{m(m-4)}$
• D. $\frac{4m-8}{m(m-4)}$
• E. $\frac{4m-12}{m(m-4)}$

Answer 1

Answer: (E)

$\frac{4m-12}{m(m-4)}$

We are given that four sandwiches were ordered and three of those sandwiches were divided among $m$ students. Thus, each student had $\frac{3}{m}$ sandwiches.

However, since four students did not want any of the 4th sandwich, the 4th sandwich was actually divided among the remaining students, or $(m–4)$ students. Thus, each of the remaining students received $\frac{1}{(m-4)}$ sandwiches.

We are given that Carol actually ate one piece from each of the four sandwiches. Thus, Carol ate:

$\frac{3}{m}+\frac{1}{m-4}$

Getting a common denominator of $m(m-4)$ we have:

$\frac{3(m-4)}{m(m-4)}+\frac{1\times m}{m(m-4)}=\frac{3(m-4)+m}{m(m-4)}=\frac{3m-12+m}{m(m-4)}=\frac{4m-12}{m(m-4)}$

Since a whole sandwich is equal to one, Carol had $\frac{4m-12}{m(m-4)}$ of a whole extra-large sandwich.