2003 CEMC Pascal Problems/Problem 3

Problem

The value of $\frac{6 + 6 \times 3 - 3}{3}$ is

$\text{ (A) }\ 11 \qquad\text{ (B) }\ 7 \qquad\text{ (C) }\ 3 \qquad\text{ (D) }\ 9 \qquad\text{ (E) }\ 17$

We can just simply use order of operations, and then divide by $3$ at the end:

$\frac{6 + 6 \times 3 - 3}{3} = \frac{6 + 18 - 3}{3} = \frac{24 - 3}{3} = \frac{21}{3} = \boxed {\textbf {(B) } 7}$

~anabel.disher

We can also combine $6 + 6 \times 3$ in the original problem:

$\frac{6 + 6 \times 3 - 3}{3} = \frac{6 \times 4 - 3}{3} = \frac{24 - 3}{3} = \frac{21}{3} = \boxed {\textbf {(B) } 7}$

~anabel.disher

We can factor out $3$ from the top equation to cancel out the $3$ in the denominator of the fraction:

$\frac{6 + 6 \times 3 - 3}{3} = \frac{3(2 + 6 - 1)}{3} = 2 + 6 - 1 = 8 - 1 = \boxed {\textbf {(B) } 7}$

~anabel.disher