We start by rewriting each expression so that each has the same numerator:

$$
\begin{aligned}
& \frac{7 x+1}{2}=\frac{2+(7 x-1)}{2}=1+\frac{7 x-1}{2} \\
& \frac{7 x+2}{3}=\frac{3+(7 x-1)}{3}=1+\frac{7 x-1}{3} \\
& \frac{7 x+300}{301}=\frac{301+(7 x-1)}{301}=1+\frac{7 x-1}{301}
\end{aligned}
$$

For each of these expressions, the original fraction will be in lowest terms only when the fraction in the new expression is in lowest terms, ie. $\frac{7 x+1}{2}$ is in lowest terms only when $\frac{7 x-1}{2}$ is in lowest terms.

So the original problem is equivalent to determining the number of positive integers $x$ with $x \leq 60$ such that each of

$$
\frac{7 x-1}{2}, \frac{7 x-1}{3}, \ldots, \frac{7 x-1}{301}
$$

is in lowest terms.

This is equivalent to determine the number of positive integers $x$ with $x \leq 60$ for which $7 x-1$ has no common factor with any of the integers from 2 to 301 , inclusive.

For $x$ from 1 to $43,7 x-1$ will be actually equal to one of the integers from 2 to 301 , so there will be a common factor.

So we must examine the integers from 44 to 60 .

If $x$ is odd, then $7 x-1$ is even, and so has a common factor of 2 , for example.

So we must examine the integers $44,46,48,50,52,54,56,58$, and 60.

The values of $7 x-1$ for these integers are 307, 321, 335, 349, 363, 377, 391, 405, and 419, respectively. We would like to determine how many of these have no common factors with any of the integers from 2 to 301 .

The integers 321, 363 and 405 are divisible by 3, so they can be removed. The integer 335 is divisible by 5 and so can be removed. This leaves us with 307, 349, 377, 391 and 419 .

307 is a prime number.

349 is a prime number.

377 is divisible by 13 , so can be removed.

391 is divisible by 17 , so can be removed.

419 is a prime number.

Each of these prime numbers is divisible only 1 and itself, so has no common factor with any of the integers from 2 to 301 .

Therefore, there are 3 integers $x$ with $x \leq 60$ for which the fractions are all in lowest terms.

Answer: (C)
The number of positive integers $x$ with $x \leq 60$ such that each of the rational expressions

$$
\frac{7 x+1}{2}, \frac{7 x+2}{3}, \frac{7 x+3}{4}, \cdots, \frac{7 x+300}{301}
$$

is in lowest terms (i.e. in each expression, the numerator and denominator have no common factors) is
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5