2000 CEMC Pascal Problems/Problem 5

Problem

If $y = 6 + \frac{1}{6}$ , then $\frac{1}{y}$ is

$\text{ (A) }\ \frac{6}{37} \qquad\text{ (B) }\ \frac{37}{6} \qquad\text{ (C) }\ \frac67 \qquad\text{ (D) }\ \frac76 \qquad\text{ (E) }\ 1$

Converting $y$ into a improper fraction, we have:

$y = 6 + \frac{1}{6} = \frac{6 \times 6}{6} + \frac{1}{6} = \frac{36}{6} + \frac{1}{6} = \frac{37}{6}$

The reciprocal of $y$ is then $\frac{1}{y} = \boxed {\textbf {(A) } \frac{6}{37}}$

~anabel.disher

$\frac{1}{y} = \frac{1}{6 + \frac{1}{6}} = \frac{1 \times 6}{(6 + \frac{1}{6}) \times 6}$

$=\frac{6}{36 + 1} = \boxed {\textbf {(A) } \frac{6}{37}}$

~anabel.disher

We can see that $y$ is approximately $6$ . Since it is higher than $6$ , $\frac{1}{y}$ must be less than $\frac{1}{6} = \frac{6}{36}$ .

The only answer choice where this is true is $\boxed {\textbf {(A) } \frac{6}{37}}$

~anabel.disher