Two bodies of equal mass thrown from a height-Practical physics question I have a doubt about this question:

Two bodies of masses $M$ and $m$ are allowed to fall from the same height. If the air resistance be the same for each of the bodies, will the two bodies reach the Earth simultaneously? Justify.

The answer given is not that both reach at same time. I fail to understand why?
According to the equation by Galileo:
$a=\frac{F}{m}=\frac{GM_\oplus}{R^2}$
where $M_\oplus$ is the mass of the Earth, $R$ is the radius of the Earth, and $G$ is Newton's gravitational constant.
So the acceleration is independent of the mass. So in this question both the bodies should fall at the same time. 
Is it that due to air resistance the force on $M$ will be more or something?
 A: I suspect it's all in what they meant by "air resistance is the same ." For example, if both bodies are the same size and shape, the equation for air resistance as a func of velocity is the same, but the greater mass will be less affected by this equal force magnitude.  Air resistance being "same" is not the same thing as being "neglectable" or "zero." 
A: 
So the acceleration is independent of the mass. So in this question both the bodies should fall at the same time.

Your answer is correct if you dropped those objects on the Moon. However, the Moon has no atmosphere. You are ignoring drag, or "air resistance". Suppose you drop a ping pong ball and a super ball of the same size as the ping pong ball. Both are bodies the same size and shape, making "air resistance be the same for each of the bodies." The much smaller mass of the ping pong ball means that drag has a much greater effect on the ping pong ball than it does on the super ball. The super ball hits first.
Suppose you drop the two balls from a good height (say the top of a six story building). The super ball will hit the ground hard in a couple of seconds. At this time, the ping pong ball will be less than halfway down and will be falling at its terminal velocity of about 8 meters/second.
A: Consider the force applied by air resistance $F_a$, equal for both objects. Then the equations of motion are:
$$F_a - \frac{GM_\oplus M}{r^2} = Ma$$
and
$$F_a - \frac{GM_\oplus m}{r^2} = ma$$
Solving for $a$ in each case, you'll find that the acceleration due to gravity is the same in both cases (as expected), but the acceleration due to $F_a$ is different for the two masses, so they do not fall at the same rate.
For what it's worth, I think the question is phrased vaguely/misleadingly...
