1999 CEMC Gauss (Grade 8) Problems/Problem 2

Problem

$\frac{1}{2} + \frac{1}{3}$ is equal to

$\text{ (A) }\ \frac{2}{5} \qquad\text{ (B) }\ \frac{1}{6} \qquad\text{ (C) }\ \frac{1}{5} \qquad\text{ (D) }\ \frac{3}{2} \qquad\text{ (E) }\ \frac{5}{6}$

Solution 1

We can first find the least common denominator, convert each fraction, and then add them.

We can see that both $2$ and $3$ are coprime (i.e. $2$ and $3$ 's greatest common factor is $1$ ), because both are prime numbers. This means the least common denominator will just be $2 \times 3 = 6$ .

Converting these, we get: $\frac{1}{2} = \frac{1 \times 3}{2 \times 3} = \frac{3}{6}$

$\frac{1}{3} = \frac{1 \times 2}{3 \times 2} = \frac{2}{6}$

Adding these together, we have:

$\frac{3}{6} + \frac{2}{6} = \boxed {\textbf {(E) } \frac{5}{6}}$

~anabel.disher

Solution 2 (answer choices)

The answer must be greater than $0.5$ because $\frac{1}{2}$ is equal to $0.5$ , and $\frac{1}{3} > 0$ . However, we also know that it is less than $1$ because $\frac{1}{2} + \frac{1}{3} < \frac{1}{2} + \frac{1}{2} = \frac{2}{2} = 1$ due to $\frac{1}{3}$ having a larger denominator than $\frac{1}{2}$ .

$\frac{2}{5} = 0.4$ , which is less than or equal to $0.5$ , so answer choice A is incorrect. This also eliminates answer choices B and C because $\frac{1}{6} < \frac{1}{5} < \frac{2}{5}$ for similar logic and the numerator being larger in $\frac{2}{5}$ than in $\frac{1}{5}$ .

$\frac{3}{2} > \frac{2}{2} = 1$ . However, we know that the result must be less than $1$ , so answer choice D is eliminated (we could have also seen that $\frac{3}{2} = 1.5$ )

Since all of the other answer choices have been eliminated, the answer is $\boxed {\textbf {(E) } \frac{5}{6}}$ .

~anabel.disher

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1999 CEMC Gauss (Grade 8) Problems/Problem 2

Problem

Solution 1

Solution 2 (answer choices)