$32 \cdot 37$ A fraction has both its numerator and denominator as positive integers, and the numerator is 1 less than the denominator. If both the numerator and the denominator are decreased by 1, the resulting fraction is a positive number less than $\frac{6}{7}$. Then the number of fractions that satisfy the above conditions is
(A) 5.
(B) 6.
(C) 7.
(D) 8.
(2nd "Hope Cup" National Mathematics Invitational, 1991)
[Solution] Let $a$ be a positive integer, then the fraction can be expressed as $\frac{a}{a+1}$. According to the problem, $0<\frac{a-1}{a}<\frac{6}{7}$, which means $1<a<7$, so $a=2,3,4,5,6$. Therefore, the fractions that meet the conditions are $\frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7}$. Hence, the answer is $(A)$.