If $a,b,c,d>0$ then the inequality $a/b<c/d$ is equivalent, on multiplying both sides by the positive number $bd$, to the inequality $ad<bc.$ So you just got things flipped around in your question, and the GRE book is right. (Hopefully; look again at exactly what inequalities were said to be equivalent.)
Of course no such simple rule will work in all cases where $a,b,c,d$ can have arbitrary signs, since you have to know the sign of $bd$ so as to know whether to reverse the inequality at the "multiply by common denominator" step. One could make a list of all four cases on the signs of $b,d$ and in each case get an equivalent inequality with no fractions, but this might be more trouble than it's worth.
ADDED NOTE: In the case of fractions (numerator and denominator are integers), it is usual to express a fraction as having a positive denominator. For example one wouldn't write $(-2)/(-3)$ but instead $2/3$, and one woudn't write $5/(-7)$ but rather $(-5)/7$. Under this convention, if $a/b$ and $c/d$ are fractions (with positive denominators) then we can definitely say that $a/b<c/d$ is equivalent to $ad<bc$ [evan if the numerators of either or both fractions are negative]. This is because in multiplying through by $bd$ to get to the integer inequality, we know that $bd>0$, so the inequality is not reversed.