total weight experienced in Five ball Cascade 
An expert juggler, carrying five juggling balls, has to cross a swing bridge which has a maximum load rating of 50 kg. The juggler weighs 47kg and each of his 5 balls weigh 2kg. He believes that he can make it across safely in one trip by juggling balls, so that he is never holding more than one ball. his skill enables him to juggle smoothly without any jerking.
He is
A) Correct because total weight will never exceed 49kg
b) Correct, because total weight can be made exceed 49kg by arbitrary small amount.
C) Incorrect because total weight is 47 (5*2) =. 57kg

I am confused as answer is given C. but I believe most appropriate option is B.
Thanks in advance
 A: Intuitive approach: Imagine there is a black box around the juggler and the balls. Gravity exerts a force on all the objects inside the box; there must be an equal and opposite force on the outside of the box to keep equilibrium. It doesn't matter what is in the box - you can't see, it can't matter.
Mathematically, the change in momentum for each ball is $F\Delta t$ for a given force $F$ during an interval $\Delta t$. Now over a complete juggling cycle, the change of momentum must be zero. If we put the time that the ball is in the hands of the juggler as $\Delta t$ and the time it is in the air as $T-\Delta t$, then we know the net change in momentum is
$$\Delta P = F_j \Delta t + F_g T=0$$
Where $F_j$ is the instantaneous force of the juggler while holding the ball, and $F_g$ is the force of gravity on the ball. The force of gravity acts all the time. It follows that
$$F_j = -\frac{F_g T}{\Delta t}$$
The time-averaged force of the juggler is of course the integral of force and time, divided by the total time:
$$F_{j,av} = \frac{\int F~dt}{T} = \frac{F_j \Delta t}{T} =  -F_g$$
So it doesn't matter how fast he juggles. That bridge can't hold him.
A: In order to propel a juggling ball into the air 4/5 of the time so that he only has 1 ball at a time in his hands he must accelerate it at much higher acceleration than the acceleration due to gravity. It is assumed that the juggler is skilled enough to be able to exert a constant force on a ball while in contact with it.
Useful equations: 
eqn_1) $F = Ma$ 
eqn_2) $velocity = v_0 + at$
Each ball is stationary at the bottom (and at the top) of it's travel so $v_0$ is 0. The ball is accelerated upwards by the juggler's arm at $a_j$ and accelerated downwards by gravity at $a_g$. While traveling upwards, the time the ball is being accelerated downwards due to gravity ($t_g$) must be 4x the time it is being accelerated upwards by the juggler ($t_j$) so $t_g = 4t_j$ (we know the 2 accelerations operate in opposite directions so we are only concerned with the magnitudes). The velocity as it leaves the juggler's hand = the velocity when it starts being decelerated by gravity so using eqn_2):
$a_jt_j$ = $a_gt_g$ but $t_g = 4t_j$ so  $a_jt_j$ = $a_g4t_j$ (we know the 2 accelerations operate in opposite directions so we are only concerned with the magnitudes).
rearranging this: $a_j=\frac{a_g4t_j}{t_j}$ = $4a_g$ 
Just to prevent the ball from falling the juggler must push it upwards with a force equal to the ball's weight ($F_{stationary} = Ma_g$) and to accelerate it upwards fast enough to be in the air $\frac45$ of the time requires an additional force of $F_j = Ma_j = M(4a_g)$ so $F_{total} = Ma_g + M(4a_g) = 5Ma_g$. The same is true in reverse for the ball traveling back downwards.
The force to keep a ball in the air for $\frac45$ of the time is therefore 5 times the force required to carry the ball, and to keep  4 of the 5 balls in the air at any moment requires this to be done continuously, this is the same as carrying the 5 balls!
